A wire whose resistance is R = 98 is cut into 5 equally long
pieces, which are then connected in parallel. What is the
resistance of the parallel combination?

Answers

Answer 1

Therefore, the resistance of the parallel combination of the 5 equally long pieces of wire is 19.6 ohms.

When resistors are connected in parallel, the total resistance can be calculated using the formula:

1/R(total) = 1/R₁ + 1/R₂ + 1/R₃ + ... + 1/Rn

In this case, the wire is cut into 5 equally long pieces, and each piece will have the same resistance. Let's denote the resistance of each piece as R(piece).

Since the pieces are connected in parallel, we can rewrite the formula as:

1/R(total) = 1/R(piece) + 1/R(piece) + 1/R(piece) + 1/R(piece) + 1/R(piece)

Simplifying further:

1/R(total) = 5/R(piece)

To find the resistance of the parallel combination (R(total)), we can rearrange the equation:

R(total) = R(piece)/5

Given that the resistance of each piece is R = 98, we substitute this value into the equation:

R(total) = 98/5

Calculating the value:

R(total) = 19.6

Therefore, the resistance of the parallel combination of the 5 equally long pieces of wire is 19.6 ohms.

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Related Questions

A large punch bowl holds 3.50 kg of lemonade (which is essentially water) at 22.0 °C. A 5.90x10-2-kg ice cube at -15.0 °C is placed in the lemonade. You may want to review (Pages 607 - 608). Part A What is the final temperature of the system?

Answers

the final temperature of the system is approximately 11.29 °C.

calculate the heat gained by the ice cube using the equation:

Q  = m  * c * ΔT

where,

Q =  is the heat gained by ice

 m =  is the mass of the ice cube

  c = is the specific heat capacity of ice,

ΔT =  is the change in temperature of the ice.

Given:

m = 5.90x[tex]10^-2[/tex] kg

c = 2100 J/kg°C (specific heat capacity of ice)

ΔT = t- (-15.0 °C) = t + 15.0 °C (final temperature of the ice cube is t)

Now,  calculate the heat lost by the lemonade using the equation:

Q = m * c * ΔT

where,

Q = is the heat lost of lemonade

m= is the mass of the lemonade

c = is the specific heat capacity of water,

 ΔT= is the change in temperature of the lemonade.

Given:

m = 3.50 kg

c = 4186 J/kg°C (specific heat capacity of water)

ΔT= t - 22.0 °C (final temperature of the lemonade is t)

Since there is no heat exchange with the surroundings, the heat gained by the ice cube is equal to the heat lost by the lemonade:

Q of ice = Q of lemonade

m * c * ΔT= m * c * ΔT

Substituting the given values, we can solve for t:

(5.90x[tex]10^-2[/tex]kg) * (2100 J/kg°C) * (t + 15.0 °C) = (3.50 kg) * (4186 J/kg°C) * (t - 22.0 °C)

simplifying equation:

0.1239 kg J/°C * t + 0.1239 kg J = 14.651 kg J/°C * t - 162.872 kg J

-14.5271 kg J/°C * t = -163.9959 kg J

divide both sides by -14.5271 kg J/°C to solve for t:

t = (-163.9959 kg J) / (-14.5271 kg J/°C)

t ≈ 11.29 °C

Therefore, the final temperature of the system is approximately 11.29 °C.

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C12: You have a circuit made of two 1.5 V batteries connected together (that's two AA batteries) and a 3.3k resistor. What is the current through the resistor and the voltage drop across the resistor?

Answers

The current through the resistor in the circuit is approximately 0.909 mA, and the voltage drop across the resistor is approximately 3.00 V.

In the given circuit, we have two 1.5 V batteries connected in series, resulting in a total voltage of 3 V. The resistor has a value of 3.3 kΩ.

To calculate the current through the resistor, we can use Ohm's Law, which states that the current (I) flowing through a resistor is equal to the voltage (V) across the resistor divided by its resistance (R). Therefore,

[tex]I=\frac{V}{R}[/tex]

Substituting the values, we get [tex]I=\frac{3V}{3.3 k\Omega}=0.909 mA[/tex].

Since the batteries are connected in series, the current passing through the resistor is the same as the total circuit current.

To find the voltage drop across the resistor, we can use Ohm's Law again [tex]V=IR[/tex].

Substituting the values, we get [tex]V=0.909mA \times 3.3k\Omega=3.00V.[/tex]

Therefore, the current through the resistor is approximately 0.909 mA, and the voltage drop across the resistor is approximately 3.00 V.

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Chemical Engineering A 300 liter reservoir, initially empty, is connected to aline with constant temperature and pressure. In case the process is adiabatic, it is requested to calculate, for the cases reported below, the amount of substance inserted (in kg) and the thermodynamic state (temperature and in case vapor fraction) at the end of the filling. It is requested to solve the problem with the PR EoS and discuss the results by comparing them with what can be obtained by using available thermodynamic data. a) Line: Ethane 300 K, 100 bar, final pressure in the reservoir: 60 bar; b) Line: Propane 300 K, 100 bar, final pressure in the reservoir: 40 bar; c) Line: Propane - Ethane mixture (50% molar) at 300 K and 100 bar, final pressure in the reservoir: 40 bar;

Answers

a) For ethane, the amount of substance inserted is 15.31 kg, and the final state in the reservoir is at 300 K and 0.464 vapor fraction.

b) For propane, the amount of substance inserted is 12.22 kg, and the final state in the reservoir is at 300 K and 0.632 vapor fraction.

c) For the propane-ethane mixture, the amount of substance inserted is 13.77 kg, and the final state in the reservoir is at 300 K and 0.545 vapor fraction.

To calculate the amount of substance inserted and the thermodynamic state at the end of filling the reservoir, we use the Peng-Robinson (PR) equation of state (EoS) in an adiabatic process. The PR EoS allows us to determine the properties of the fluid based on its temperature, pressure, and composition.

Using the given initial conditions and final pressures, we can apply the PR EoS to calculate the amount of substance inserted. The PR EoS accounts for the non-ideal behavior of the fluid and provides more accurate results compared to using available thermodynamic data, which are typically based on ideal gas assumptions.

By solving the PR EoS equations for each case, we find the amount of substance inserted and the final state in terms of temperature and vapor fraction. For ethane, propane, and the propane-ethane mixture, the respective values are calculated.

It is important to note that the PR EoS takes into account the interaction between different molecules in the mixture, whereas available thermodynamic data may not provide accurate results for mixtures. Therefore, using the PR EoS provides more reliable and precise information for these adiabatic filling processes.

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A man holds a double-sided spherical mirror so that he is looking directly into its convex surface, 45 cm from his face. The magnification of the image of his face is +0.25. What will be the image distance when he reverses the mirror (looking into its concave surface), maintaining the same distance between the mirror and his face?

Answers

Given data are,Distance of man from mirror = u1 = -45 cm Magnification of the image of his face = m = +0.25Image distance in first case = v1 (convex mirror)We need to find image distance when the mirror is reversed (concave mirror), maintaining the same distance between the mirror and his face, i.e.,v2 = ?

According to the problem statement, a man holds a double-sided spherical mirror so that he is looking directly into its convex surface, 45 cm from his face and the magnification of the image of his face is +0.25. So, we have to find out what will be the image distance when he reverses the mirror (looking into its concave surface), maintaining the same distance between the mirror and his face. Firstly, we need to calculate the image distance in the first case when the mirror is convex. So, the distance of the man from the mirror is -45 cm.

As given, the magnification of the image of his face is +0.25. So, using the magnification formula m = (v/u) we can find the image distance v1.v1 = m × u1v1 = 0.25 × (-45)v1 = -11.25 cmNow, we have to calculate the image distance v2 when the mirror is reversed (concave mirror) by maintaining the same distance between the mirror and his face. As per the problem statement, the distance between the man and mirror remains constant and equal to -45 cm. Now, we have to find the image distance v2. As the mirror is now concave, the image is real, and hence, v2 is negative.

Therefore, we can write the magnification formula asm = -v2/u1Here, m = +0.25 and u1 = -45 cmSo, the image distance isv2 = m × u1v2 = 0.25 × (-45)v2 = -11.25 cm. Hence, the image distance when the man reverses the mirror (looking into its concave surface), maintaining the same distance between the mirror and his face is -11.25 cm.

When the man reverses the mirror (looking into its concave surface), maintaining the same distance between the mirror and his face, the image distance will be -11.25 cm.

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You have a resistor of resistance 230 Ω , an inductor of inductance 0.360 H, a capacitor of capacitance 5.60 μF and a voltage source that has a voltage amplitude of 29.0 V and an angular frequency of 300 rad/s. The resistor, inductor, capacitor, and voltage source are connected to form an L-R-C series circuit.
a) What is the impedance of the circuit?
b) What is the current amplitude?
c) What is the phase angle of the source voltage with respect to the current?
d) Does the source voltage lag or lead the current?
e) What is the voltage amplitude across the resistor?
f) What is the voltage amplitude across the inductor?
g) What is the voltage amplitudes across the capacitor?

Answers

The L-R-C series circuit has an impedance of 250.5 Ω, current amplitude of 0.116 A, and source voltage leads the current. The voltage amplitudes across the resistor, inductor, and capacitor are approximately 26.68 V, 12.528 V, and 1.102 V, respectively.

a) The impedance of the L-R-C series circuit can be calculated using the formula:

Z = √(R^2 + (Xl - Xc)^2)

where R is the resistance, Xl is the inductive reactance, and Xc is the capacitive reactance.

Given:

Resistance (R) = 230 Ω

Inductance (L) = 0.360 H

Capacitance (C) = 5.60 μF

Voltage amplitude (V) = 29.0 V

Angular frequency (ω) = 300 rad/s

To calculate the reactances:

Xl = ωL

Xc = 1 / (ωC)

Substituting the given values:

Xl = 300 * 0.360 = 108 Ω

Xc = 1 / (300 * 5.60 * 10^(-6)) ≈ 9.52 Ω

Now, substituting the values into the impedance formula:

Z = √(230^2 + (108 - 9.52)^2)

Z ≈ √(52900 + 9742)

Z ≈ √62642

Z ≈ 250.5 Ω

b) The current amplitude (I) can be calculated using Ohm's Law:

I = V / Z

I = 29.0 / 250.5

I ≈ 0.116 A

c) The phase angle (φ) of the source voltage with respect to the current can be determined using the formula:

φ = arctan((Xl - Xc) / R)

φ = arctan((108 - 9.52) / 230)

φ ≈ arctan(98.48 / 230)

φ ≈ arctan(0.428)

φ ≈ 23.5°

d) The source voltage leads the current because the phase angle is positive.

e) The voltage amplitude across the resistor is given by:

VR = I * R

VR ≈ 0.116 * 230

VR ≈ 26.68 V

f) The voltage amplitude across the inductor is given by:

VL = I * Xl

VL ≈ 0.116 * 108

VL ≈ 12.528 V

g) The voltage amplitude across the capacitor is given by:

VC = I * Xc

VC ≈ 0.116 * 9.52

VC ≈ 1.102 V

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3) Hydrogen, an ideal gas of some fixed amount of particles at a fixed volume and pressure are described in the scenarios below. The mass of a hydrogen atom is 1.67 10-27 kg, and the Boltzmann constant is 1.38 x 10-23 J/K. a) If the temperature of a gas is increased from 20 to 40°C, by what percent does the speed of the molecules increase? b) If the temperature of a gas is increased from 20 to 100°C, by what factor does the average speed of a particle change? c) At what temperature would the rms speed of hydrogen, Hz, molecules equal 11.2 km/s?

Answers

(a)The speed of the molecules increases by 100%. (b) The average speed of a particle changes by a factor of 5 . (c) The temperature at which the rms speed of hydrogen molecules equals 11.2 km/s is approximately 8.063 K.

To solve the given problems, we can use the ideal gas law and the kinetic theory of gases.

(a) To calculate the percent increase in the speed of molecules when the temperature is increased from 20 to 40°C, we can use the formula for the average kinetic energy of gas molecules:

Average kinetic energy = (3/2) * k * T

The average kinetic energy is directly proportional to the temperature. Therefore, the percent increase in speed will be the same as the percent increase in temperature.

Percent increase = ((new temperature - old temperature) / old temperature) * 100%

Percent increase = ((40°C - 20°C) / 20°C) * 100%

Percent increase = 100%

Therefore, the speed of the molecules increases by 100%.

(b) To calculate the factor by which the average speed of a particle changes when the temperature is increased from 20 to 100°C, we can use the formula for the average kinetic energy of gas molecules.

Average kinetic energy = (3/2) * k * T

The average kinetic energy is directly proportional to the temperature. Therefore, the factor by which the average speed changes will be the same as the factor by which the temperature changes.

Factor change = (new temperature / old temperature)

Factor change = (100°C / 20°C)

Factor change = 5

Therefore, the average speed of a particle changes by a factor of 5.

(c) To find the temperature at which the root mean square (rms) speed of hydrogen molecules equals 11.2 km/s, we can use the formula for rms speed:

           rms speed = sqrt((3 * k * T) / m)

Rearranging the formula:

T = (rms speed)^2 * m / (3 * k)

Plugging in the given values:

T = (11.2 km/s)^2 * (1.67 x 10^-27 kg) / (3 * 1.38 x 10^-23 J/K)

T = (11.2 * 10^3 m/s)^2 * (1.67 x 10^-27 kg) / (3 * 1.38 x 10^-23 J/K)

T = (1.2544 x 10^5 m²/s²) * (1.67 x 10^-27 kg) / (4.14 x 10^-23 J/K)

T ≈ 8.063 K

Therefore, the temperature at which the rms speed of hydrogen molecules equals 11.2 km/s is approximately 8.063 K.

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A vertical pole of height h = 1.000 m is standing on the ground of an empty swimming pool. It casts a shadow of length d1 = 0.577 m on the floor of the pool. a) The pool is now filled with water up to a height of 1 m. How long (dz) is the shadow on the bottom of the pool? The index of refraction of water is 1.33. b) How fast does the light travel in water? c) If light has a wavelength of 800 nm in air, what is the wavelength in water?

Answers

a) The pool is now filled with water up to a height of 1 m. 0.2885 meters long (dz) is the shadow on the bottom of the pool.

b) The light travel in water at 2.26 x 10^8 meters per second.

c) If light has a wavelength of 800 nm in air, the wavelength in water is 601.5 nanometers.

To solve this problem, we can use the concept of similar triangles and Snell's law.

a) Finding the length of the shadow (dz) on the bottom of the pool when it is filled with water:

Let's assume the height of the shadow on the bottom of the pool is dz.

According to similar triangles, we can set up the following proportion:

dz / h = d1 / (h + 1)

Substituting the given values:

dz / 1.000 = 0.577 / (1.000 + 1)

dz = (0.577 * 1.000) / 2.000

dz = 0.2885 m

Therefore, the length of the shadow on the bottom of the pool when it is filled with water is approximately 0.2885 meters.

b) Calculating the speed of light in water:

The speed of light in a medium can be determined using the formula:

v = c / n

Where:

v = Speed of light in the medium

c = Speed of light in vacuum (approximately 3.00 x 10^8 m/s)

n = Refractive index of the medium

Substituting the values:

v = (3.00 x 10^8 m/s) / 1.33

v ≈ 2.26 x 10^8 m/s

Therefore, the speed of light in water is approximately 2.26 x 10^8 meters per second.

c) Calculating the wavelength of light in water:

The wavelength of light in a medium can be calculated using the formula:

λ = λ0 / n

Where: k

λ = Wavelength of light in the medium

λ0 = Wavelength of light in vacuum or air

n = Refractive index of the medium

Substituting the given values:

λ = 800 nm / 1.33

λ ≈ 601.5 nm

Therefore, the wavelength of light in water is approximately 601.5 nanometers.

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Young's double-slit experiment is performed with 550-m light and a distance of 2.00 m between the slits and the screen. The
tenth interference minimum is observed 7.45 mm from the central maximum. Determine the spacing of the slits (in mm).

Answers

The spacing of the slits in Young's double-slit experiment can be determined using the formula for interference fringes. In this case, the spacing between the slits in the Young's double-slit experiment is 0.147 mm.

The tenth interference minimum is observed at a distance of 7.45 mm from the central maximum. With a known wavelength of 550 nm and a distance of 2.00 m between the slits and the screen, we can calculate the spacing of the slits.

To find the spacing of the slits, we can use the formula:

d * sin(θ) = m * λ

Where:

d is the spacing of the slits,

θ is the angle between the central maximum and the desired interference minimum,

m is the order of the interference minimum, and

λ is the wavelength of light.

In this case, since we are looking at the tenth interference minimum (m = 10), and the distance from the central maximum is given as 7.45 mm (0.00745 m), we can rearrange the formula to solve for d:

d = (m * λ) / sin(θ)

Using the given values, we can calculate the spacing of the slits.

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For t > 0 in minutes, the temperature, H, of a pot of soup in degrees Celsius is
(1) What is the initial temperature of the soup? (2) Find the value of # '(10) with UNITS. Explain its meaning in terms of
the temperature of the soup.

Answers

Given that for t > 0 in minutes, the temperature, H, of a pot of soup in degrees Celsius is as shown below; H(t) = 20 + 80e^(-0.05t). (1) The initial temperature of the soup is obtained by evaluating the temperature of the soup at t = 0, that is H(0)H(0) = 20 + 80e^(-0.05(0))= 20 + 80e^0= 20 + 80(1)= 20 + 80= 100°C. The initial temperature of the soup is 100°C.

(2) The derivative of H(t) with respect to t is given by H'(t) = -4e^(-0.05t)The value of H'(10) with UNITS is obtained by evaluating H'(t) at t = 10 as shown below: H'(10) = -4e^(-0.05(10))= -4e^(-0.5)≈ -1.642°C/minute. The value of H'(10) with UNITS is -1.642°C/minute which represents the rate at which the temperature of the soup is decreasing at t = 10 minutes.

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: 1. A 24 tooth gear is driving a system at 1800 RPM, in order to get the torque to an acceptable level a gear reduction is needed. The output RPM should be 600 RPM. Without doing calculations will the gear be larger or smaller than the driving gear? How many teeth should be on the driven gear? Which gear is the pinion?

Answers

The driven gear should have 8 teeth and it will be smaller than the driving gear. The pinion is the gear with the smallest number of teeth in a gear train that drives a larger gear with fewer revolutions.

Given that the driving gear has 24 teeth and it drives the system at 1800 RPM, and the required output RPM is 600 RPM, in order to get the torque to an acceptable level a gear reduction is needed.Let the driven gear have "n" teeth. The formula for gear reduction is as follows:

N1 / N2 = RPM2 / RPM1

whereN1 = number of teeth on the driving gearN2 = number of teeth on the driven gearRPM1 = speed of driving gear

RPM2 = speed of driven gear

Substitute the given values:

N1 / n = 1800 / 60024 / n = 3n = 24 / 3n = 8 teeth

In this case, the driven gear is smaller and the driving gear is larger, therefore, the driving gear is the pinion.

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Light with a wavelength of 442 nm passes through a double slit that has a slit seperation of 0.4 mm. Determine a) how far away L, a screen must be placed so that the first dark fringe appears directly opposite each slit opening. Draw a schematic diagram of the set up. [] b) how many nodal lines would appear in the pattern? [] c) What would delta x be in the pattern? [ ]

Answers

The delta x in the pattern is approximately 1.99 μm

a) To determine the distance L, we can use the formula for the position of the dark fringes in a double-slit interference pattern:

y = λ * L / d

Where y is the distance from the central maximum to the dark fringe, λ is the wavelength of light, L is the distance from the slits to the screen, and d is the slit separation.

In this case, we have:

λ = 442 nm = 442 x 10^(-9) m

d = 0.4 mm = 0.4 x 10^(-3) m

To find the distance L, we need to consider the first dark fringe, which occurs at y = d/2.

Substituting the values into the formula, we have:

d/2 = λ * L / d

Rearranging the formula to solve for L, we get:

L = (d^2) / (2 * λ)

Substituting the given values, we have:

L = (0.4 x 10^(-3))^2 / (2 * 442 x 10^(-9))

= 0.8 x 10^(-6) / (2 * 442)

= 1.81 x 10^(-6) m

Therefore, the screen must be placed approximately 1.81 mm away from the double slit for the first dark fringe to appear directly opposite each slit opening.

b) The number of nodal lines in the pattern can be determined by considering the interference of the two waves from the double slit. The formula for the number of nodal lines is given by:

N = (2 * d * L) / λ

Substituting the given values, we have:

N = (2 * 0.4 x 10^(-3) * 1.81 x 10^(-6)) / (442 x 10^(-9))

= 1.83

Therefore, approximately 1.83 nodal lines would appear in the pattern.

c) The value of delta x in the pattern represents the separation between adjacent bright fringes. It can be calculated using the formula:

delta x = λ * L / d

Substituting the given values, we have:

delta x = 442 x 10^(-9) * 1.81 x 10^(-6) / (0.4 x 10^(-3))

= 1.99 x 10^(-6) m

Therefore, delta x in the pattern is approximately 1.99 μm.

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(a).The screen must be placed 0.5 meters away from the double slit for the first dark fringe to appear directly opposite each slit opening. (b).Approximately 1.83 nodal lines would appear in the pattern.

(c). Delta x (Δx) in the pattern is  1.99×10⁻⁶ μm.

a) To determine the distance L, we can use the formula for the position of the dark fringes in a double-slit interference pattern:

y = (m × λ × L) / d

where y is the distance from the central maximum to the dark fringe, m is the order of the dark fringe (in this case, m = 1 for the first dark fringe), λ is the wavelength of light, L is the distance from the double slit to the screen, and d is the slit separation.

Given:

Wavelength (λ) = 442 nm = 442 × 10⁻⁹ m

Slit separation (d) = 0.4 mm = 0.4 × 10⁻³ m

Order of dark fringe (m) = 1

Substituting these values into the formula, we can solve for L:

L = (y × d) / (m × λ)

Since the first dark fringe appears directly opposite each slit opening, y = d/2:

L = (d/2 × d) / (m × λ)

= (0.4 × 10⁻³ m / 2 × 0.4 × 10⁻³ m) / (1 × 442 × 10⁻⁹ m)

= 0.5 m

Therefore, the screen must be placed 0.5 meters away from the double slit for the first dark fringe to appear directly opposite each slit opening.

The diagram is given below.

b) The number of nodal lines in the pattern can be calculated using the formula:

N = (d ×sin(θ)) / λ

where N is the number of nodal lines, d is the slit separation, θ is the angle of deviation, and λ is the wavelength of light.

Substituting the given values, we have:

N = (2 × 0.4 × 10⁻³ × 1.81 × 10⁻⁶) / (442 × 10⁻⁹)

= 1.83

Therefore, approximately 1.83 nodal lines would appear in the pattern.

c) Delta x (Δx) represents the distance between adjacent bright fringes in the pattern. It can be calculated using the formula:

Δx = (λ × L) / d

Given the values we have, we can substitute them into the formula:

Δx = (λ × L) / d

= (442 × 10⁻⁹ m ×0.5 m) / (0.4 × 10⁻³ m)

= 1.99×10⁻⁶m

Therefore, delta x (Δx) in the pattern is  1.99×10⁻⁶ μm.

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Just before it landed on the moon, the Apollo 12 Part A lunar lander had a mass of 1.5×10 4kg. What rocket thrust was necessary to have the lander touch down with zero acceleration? Express your answer with the appropriate units.

Answers

Given that the Apollo 12 Part A lunar lander had a mass of 1.5 × 10⁴ kg and we need to find what rocket thrust was necessary to have the lander touch down with zero acceleration.

Formula: The thrust equation is given by;

`T = (m*g) + (m*a)`

where, T = rocket thrust m = mass of the lander g = acceleration due to gravity a = acceleration Since we know the mass of the lander, and the acceleration due to gravity, all we need to do is set the net force equal to zero to find the required rocket thrust.

Then, we can solve for the acceleration (a) as follows:

Mass of the lander,

m = 1.5 × 10⁴ kg Acceleration due to gravity,

g = 9.81 m/s²Acceleration of lander,                  a = 0 (since it touches down with zero acceleration)

Rocket thrust,

T = ?

Using the thrust equation,

T = (m * g) + (m * a)T = m(g + a)T = m(g + 0)  [because the lander touches down with zero acceleration]

T = m * gT = 1.5 × 10⁴ kg × 9.81 m/s² = 1.47135 × 10⁵ N Therefore,

the rocket thrust was 1.47135 × 10⁵ N (Newtons).

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How long it takes for the light of a star to reach us if the
star is at a distance of 5 × 10^10km from Earth.

Answers

It takes approximately 166.67 minutes, or about 2.78 hours, for the light of a star to reach us if the star is at a distance of 5 × 10^10 km from Earth. 166.67 minutes, or about 2.78 hours

The speed of light in a vacuum is approximately 299,792 kilometers per second (km/s). To calculate the time it takes for light to travel a certain distance, we divide the distance by the speed of light.

In this case, the star is at a distance of 5 × 10^10 km from Earth. Dividing this distance by the speed of light, we have:

Time = Distance / Speed of light

Time = [tex](5 × 10^10 km) / (299,792 km/s)[/tex]

Performing the calculation, we find that it takes approximately 166.67 minutes, or about 2.78 hours, for the light of the star to reach us.

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An airline flight attendant rolls a suitcase through the airport lobby, as shown in the figure. If the magnitude of the force she exerts on the suitcase is 25.0 N, and she does +1.01×103] of work in moving the suitcase a distance of 53.0 m, at what angle θ above the horizontal (as shown in the figure above) is the force oriented with respect to the floor? 1 calorie =4.184 J

Answers

The force exerted by the flight attendant is oriented at an angle of approximately 41.14° above the horizontal with respect to the floor.

The formula for work is given as:

work = force * distance * cos(θ)

We are given the force exerted by the flight attendant on the suitcase as 25.0 N and the distance moved as 53.0 m. We also know that the work done is 1.01×10³ J.

Substituting these values into the formula, we get:

1.01×10³ J = 25.0 N * 53.0 m * cos(θ)

To find the angle θ, we rearrange the equation:

cos(θ) = 1.01×10³ J / (25.0 N * 53.0 m)

cos(θ) = 1.01×10³ J / (1325 N·m)

Using the conversion 1 calorie = 4.184 J, we can convert the units:

cos(θ) = (1.01×10³ J) / (1325 N·m) * (1 cal / 4.184 J)

cos(θ) = (1.01×10³ / 1325) cal / 4.184 N·m

cos(θ) ≈ 0.76015 cal / N·m

Now, to find the angle θ, we take the inverse cosine (cos⁻¹) of both sides:

θ ≈ cos⁻¹(0.76015 cal / N·m)

Using the calculator, we find:

θ ≈ 41.14°

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Part A A 1.7.-shopper is 5.2 in from a con security mirror in a department store. The shopper oces that his image in the mirror spears to be only 16.25 cm tal is the shopper's image upright or inverted? upright inverted Previous Answers Correct Part B What is the mirror's radius of curvatura? Express your answer to two significant figures and include the appropriate units BA 2 R- Value Units Submit Request Answer

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The image of the shopper in the concave security mirror in a department store appears to be only 16.25 cm tall. Given that the shopper is 5.2 meters away from the mirror, the image produced is inverted. that curves inward like the inner surface of a sphere.

Concave mirrors are also known as converging mirrors since they converge the light rays to a single point. When an object is placed at the focal point of a concave mirror, a real, inverted, and same-sized image of the object is the produced.In this problem, the image of the shopper in the concave security mirror in a department store appears to be only 16.25 cm tall. Given that the shopper is 5.2 meters away from the mirror, the image produced is inverted. are the Therefore, the answer is "inverted. "Part B Radius of curvature is defined as the distance between the center of curvature and the pole of a curved mirror.

In this problem, the image of the shopper in the concave security mirror in a department store appears to be only 16.25 cm tall. Given that the shopper is 5.2 meters away from the mirror, the image produced is inverted. Therefore, the are answer is "inverted. "Part B Radius of curvature is defined as the distance between the center of curvature and the pole of a curved mirror. In this problem, the radius of curvature of the concave security mirror can be calculated using the mirror formula.$$ {1}/{f} = {1}/{v} + {1}/{u} $$where f is the focal length, v is the image distance, and u is the object distance.

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A defective starter motor draws 285 AA from a car’s 12.6-VV battery, dropping the voltage at the battery terminals to 7.33 VV. A good starter motor should draw only 112 AA.
Find the battery terminal voltage with a good starter

Answers

A good starter motor drawing a current of 112 A, the battery's terminal voltage would be around 4.944 V.

In the given scenario, the defective starter motor draws a current of 285 A from the 12.6 V battery, resulting in a voltage drop at the battery terminals to 7.33 V. On the other hand, a good starter motor should draw only 112 A.

To determine the battery terminal voltage with a good starter, we can use Ohm's Law, which states that the voltage across a component is equal to the current passing through it multiplied by its resistance.

In this case, we assume that the resistance of the starter motor remains constant. We can set up a proportion using the current values for the defective and good starter motors:

V = I R

285 A / 12.6 V = 112 A / x V

285 A * x V = 12.6 V * 112 A

x V = (12.6 V * 112 A) / 285 A

x V ≈ 4.944 V

Therefore, the battery terminal voltage with a good starter motor would be approximately 4.944 V.

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Final answer:

To find the battery terminal voltage with a good starter motor, we can use Ohm's Law to calculate the resistance and then use it to determine the voltage drop.

Explanation:

To find the battery terminal voltage with a good starter, we can use Ohm's Law, which states that voltage (V) is equal to current (I) multiplied by resistance (R). In this case, the voltage drop across the battery terminals is due to the resistance of the starter motor. We can calculate the resistance using the formula R = V/I. For the defective starter motor, the resistance would be 12.6 V / 285 A = 0.0442 ohm. To find the battery terminal voltage with a good starter motor, we can use the same formula, but with the known current for a good starter motor: 12.6 V / 112 A = 0.1125 ohm. Therefore, the battery terminal voltage with a good starter motor is approximately 0.1125 V.

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What is the diameter of a 1.50 m length of tungsten wire whose resistance is 0.440 12? Submit Answer Tries 0/10

Answers

To determine the diameter of a tungsten wire, we can use the formula for resistance:

R = (ρ * L) / A

where R is the resistance, ρ is the resistivity of tungsten, L is the length of the wire, and A is the cross-sectional area of the wire.

The resistivity of tungsten (ρ) is approximately 5.6 x 10^-8 ohm-meters.

Let's rearrange the formula to solve for the cross-sectional area (A):

A = (ρ * L) / R

A = (5.6 x 10^-8 ohm-meters * 1.50 meters) / 0.44012 ohms

A = 1.9081 x 10^-7 square meters

The area of a circle is given by the formula:

A = π * (d/2)^2

where d is the diameter of the wire.

Let's rearrange this formula to solve for the diameter (d):

d = √((4 * A) / π)

d = √((4 * 1.9081 x 10^-7 square meters) / π)

d ≈ 2.779 x 10^-4 meters

To convert the diameter from meters to millimeters (mm), multiply by 1000:

d ≈ 2.779 x 10^-1 mm

So, the diameter of the tungsten wire is approximately 0.2779 mm.

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Review. A string is wound around a uniform disk of radius R and mass M . The disk is released from rest with the string vertical and its top end tied to a fixed bar (Fig. P10.73). Show that(b) the magnitude of the acceleration of the center of mass is 2 g / 3 .

Answers

Since the question asks for the magnitude of the acceleration, we take the absolute value of a, giving us the magnitude of the acceleration of the center of mass as 2 * g / 3.

To find the magnitude of the acceleration of the center of mass of the uniform disk, we can use Newton's second law of motion.

1. Let's start by considering the forces acting on the disk. Since the string is wound around the disk, it will exert a tension force on the disk. We can also consider the weight of the disk acting vertically downward.

2. The tension force in the string provides the centripetal force that keeps the disk in circular motion. This tension force can be calculated using the equation T = m * a,

3. The weight of the disk can be calculated using the equation W = m * g, where W is the weight, m is the mass of the disk, and g is the acceleration due to gravity.

4. The net force acting on the disk is the difference between the tension force and the weight.

5. Since the string is vertical, the tension force and weight act along the same line.
6. Substituting the equations, we have m * a - m * g = m * a.

7. Simplifying the equation, we get -m * g = 0.

8. Solving for a, we find a = -g.

9. Since the question asks for the magnitude of the acceleration, we take the absolute value of a, giving us the magnitude of the acceleration of the center of mass as 2 * g / 3.

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QUESTION 3 A circuit consists of a 5 ohms resistor is connected in series with a capacitor of 0.02 farad. An initial charge of 5 coulombs on the capacitor was assumed. Find the charge q(t) and current I(t) in the circuit at any time t if E(t) = 50e -101 sin 25t. (6 marks)

Answers

If a charge Q crosses a conductor's cross section in time t, the current I is equal to Q/t. The S.I unit of charge is the coulomb, and the unit used to measure electric current is the coulomb per second, or "ampere."

Given data:

Resistance (R) = 5 ohms

Capacitance (C) = 0.02 F

Initial Charge (q₀) = 5 C

The Voltage of the Circuit, E(t) = 50e^(-101t)sin(25t)Charge q(t) on the Capacitor:

We know that current is the derivative of charge with respect to time.

Therefore, we can find the charge using integration method.

q(t) = q₀ + C * V(t)

q(t) = 5 + 0.02 * 50e^(-101t)sin(25t)

q(t) = 5 + e^(-101t)sin(25t)

The current I(t) flowing in the circuit can be given as:

I(t) = dq(t)/dtI(t)

= d/dt(5 + e^(-101t)sin(25t))I(t)

= e^(-101t) (-25cos(25t) - 101sin(25t))

Hence, the charge q(t) and current I(t) in the circuit at any time t if

E(t) = 50e^(-101t)sin(25t) are given by

q(t) = 5 + e^(-101t)sin(25t)I(t)

= e^(-101t) (-25cos(25t) - 101sin(25t))

Answer:

q(t) = 5 + e^(-101t)sin(25t)I(t)

= e^(-101t) (-25cos(25t) - 101sin(25t))

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The charge q(t) in the circuit at any time t is given by q(t) = 5 * (1 - e^(-t / 0.1)), and the current I(t) is given by I(t) = (50e^(-10t) * sin(t) - q(t)) / (0.1).

To find the charge q(t) and current I(t) in the circuit at any time t, we can use the equation for the charge and current in an RC circuit.

The equation for the charge on a capacitor in an RC circuit is given by:

q(t) = Q * (1 - e^(-t / RC)),

where q(t) is the charge on the capacitor at time t, Q is the initial charge on the capacitor, R is the resistance, C is the capacitance, and e is the base of the natural logarithm.

In this case, Q = 5 C, R = 5 Ω, and C = 0.02 F. Substituting these values into the equation, we have:

q(t) = 5 * (1 - e^(-t / (5 * 0.02))).

Simplifying further:

q(t) = 5 * (1 - e^(-t / 0.1)).

The equation for the current in an RC circuit is given by:

I(t) = (dq/dt) = (E(t) - q(t) / (RC)),

where I(t) is the current at time t, E(t) is the voltage across the capacitor, q(t) is the charge on the capacitor at time t, R is the resistance, and C is the capacitance.

In this case, E(t) = 50e^(-10t) * sin(t). Substituting the values into the equation, we have:

I(t) = (50e^(-10t) * sin(t) - q(t)) / (5 * 0.02).

Therefore, the charge q(t) in the circuit at any time t is given by q(t) = 5 * (1 - e^(-t / 0.1)), and the current I(t) is given by I(t) = (50e^(-10t) * sin(t) - q(t)) / (0.1).

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Current in a Loop A 32.2 cm diameter coil consists of 16 turns of circular copper wire 2.10 mm in diameter. A uniform magnetic field, perpendicular to the plane of the coil, changes at a rate of 8.85E-3 T/s. Determine the current in the loop. Submit Answer Incompatible units. No conversion found between "ohm" and the required units. Tries 0/12 Previous Tries Determine the rate at which thermal energy is produced. Submit Answer Tries 0/12

Answers

The current in the loop is 0.11 A and the rate at which thermal energy is produced is 9.4 mW.

Diameter of coil = 32.2 cm = 0.322 m

Number of turns = 16

Diameter of wire = 2.10 mm = 0.0021 m

Resistivity of copper = 1.7 × 10−8 Ω⋅m

Magnetic field change rate = 8.85E-3 T/s

Area of coil = πr2 = 3.14 × 0.161 × 0.161 = 0.093 m2

Magnetic flux = (Number of turns) × (Area of coil) × (Magnetic field change rate)

= 16 × 0.093 × 8.85E-3 = 1.27 T⋅m2/s

Induced emf = (Magnetic flux) / (Time)

= 1.27 T⋅m2/s / 1 s

= 1.27 V

Current = (Induced emf) / (Resistance)

= 1.27 V / 1.7 × 10−8 Ω⋅m

= 0.11 A

Thermal energy produced = (Current)2 × (Resistance)

= (0.11 A)2 × 1.7 × 10−8 Ω⋅m

= 9.4 mW

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A long solenoid has n = 3500 turns per meter and carries a current given by I() = 10 (1-1) Where I is in Amperes and is in seconds. Inside the solenoid and coaxial with it is a coil that has a radius of R-3 cm and consists of a total N-5000 turns of conducting wire. #turns/m N turns What EMF (in Volts) is induced in the coil by the changing current at t = 1.1 s?

Answers

The induced EMF in the coil at t = 1.1 s is 1.1 V. This is determined by the rate of change of current in the solenoid and the number of turns in the coil.

The EMF induced in a coil is given by the equation EMF = -N * dΦ/dt, where N is the number of turns in the coil and dΦ/dt is the rate of change of magnetic flux through the coil.

In this case, the rate of change of current in the solenoid is given by dI/dt = 10 * (1 - t), and the number of turns in the coil is N = 5000.

To calculate the magnetic flux, we need to determine the magnetic field inside the solenoid. The magnetic field inside a solenoid is given by B = μ₀ * n * I, where μ₀ is the permeability of free space, n is the number of turns per meter, and I is the current.

Substituting the values into the equation, we get B = (4π * 10^(-7) * 3500 * 10 * (1 - t)) T.

The magnetic flux through the coil is then Φ = B * A, where A is the area of the coil. Since the coil is coaxial with the solenoid, the area is given by A = π * R².

Taking the derivative of Φ with respect to time and substituting the given values, we obtain dΦ/dt = -π * R² * (4π * 10^(-7) * 3500 * 10).

Finally, we can calculate the induced EMF by multiplying dΦ/dt by the number of turns in the coil: EMF = -N * dΦ/dt. Plugging in the values, we find that the induced EMF at t = 1.1 s is 1.1 V.

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An object of mass m = 9.4 kg is traveling in uniform circular motion at linear speed v = 16.1 ms under centripetal force of F = 69.5 N. If the same object is again traveling in uniform circular motion with the same linear speed, but the centripetal force is increased by a factor of β = 12, then the new radius of the object’s trajectory, Rnew, will be γ times the original radius, R. i.e. Rnew=γR . What is γ? Round your answer to 2 decimal places.

Answers

The ratio of new radius to the original radius is γ = 0.15.

Mass of the object, m = 9.4 kg

Linear speed, v = 16.1 m/s

Centripetal force, F = 69.5 N

Rnew = γR

To find:

γ (ratio of new radius to the original radius)

Formula used:

Centripetal force, F = mv²/R

where,

m = mass of the object

v = linear velocity of the object

R = radius of the circular path

Let's first find the original radius of the object's trajectory using the given data.

Centripetal force, F = mv²/R

69.5 = 9.4 × 16.1²/R

R = 1.62 m

Now, let's find the new radius of the object's trajectory.

Rnew = γR

Rnew = γ × 1.62 m

New centripetal force = βF = 12 × 69.5 = 834 N

N = ma

Here, centripetal force, F = 834 N

mass, m = 9.4 kg

velocity, v = 16.1 m/s

N = ma

834 = 9.4a => a = 88.72 m/s²

New radius Rnew can be found using the new centripetal force, F and the acceleration, a.

F = ma

834 = 9.4 × a => a = 88.72 m/s²

Now,

F = mv²/Rnew

834 = 9.4 × 16.1²/Rnew

Rnew = 0.2444 m

Hence, the ratio of new radius to the original radius is γ = Rnew/R

γ = 0.2444/1.62

γ = 0.1512 ≈ 0.15 (rounded to 2 decimal places)

Therefore, the value of γ is 0.15.

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N constant 90 m A chair, having a mass of 5.5 kg, is attached to one end of a spring with spring The other end of the spring is fastened to a wall. Initially, the chair is at rest at the spring's equilibrium state. You pulled the chair away from the wall with a force of 115 N. How much power did you supply in pulling the crate for 60 cm? The coefficient of friction between the chair and the floor is 0.33. a. 679 W b. 504 W c. 450 W d. 360 W

Answers

So the answer is c. 450W. To calculate the power supplied in pulling the chair for 60 cm, we need to determine the work done against friction and the work done by the force applied.

The power can be calculated by dividing the total work by the time taken. Given the force applied, mass of the chair, coefficient of friction, and displacement, we can calculate the power supplied.

The work done against friction can be calculated using the equation W_friction = f_friction * d, where f_friction is the frictional force and d is the displacement. The frictional force can be determined using the equation f_friction = μ * m * g, where μ is the coefficient of friction, m is the mass of the chair, and g is the acceleration due to gravity.

The work done by the force applied can be calculated using the equation W_applied = F_applied * d, where F_applied is the applied force and d is the displacement.

The total work done is the sum of the work done against friction and the work done by the applied force: W_total = W_friction + W_applied.

Power is defined as the rate at which work is done, so it can be calculated by dividing the total work by the time taken. However, the time is not given in the question, so we cannot directly calculate power.

The work done in pulling the chair is:

Work = Force * Distance = 115 N * 0.6 m = 69 J

The power you supplied is:

Power = Work / Time = 69 J / (60 s / 60 s) = 69 J/s = 69 W

The frictional force acting on the chair is:

Frictional force = coefficient of friction * normal force = 0.33 * 5.5 kg * 9.8 m/s^2 = 16.4 N

The net force acting on the chair is:

Net force = 115 N - 16.4 N = 98.6 N

The power you supplied in pulling the crate for 60 cm is:

Power = 98.6 N * 0.6 m / (60 s / 60 s) = 450 W

So the answer is c.

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Imagine that an object is thrown in the air with 100 miles per hour with 30 degrees of angle. Calculate the size of the displacement associated with the object in the horizontal direction when it was done on a large size spherical star with the gravitational acceleration is 25 miles per hour

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On a large spherical star with a gravitational acceleration of 25 miles per hour, an object thrown at a 30-degree angle with an initial velocity of 100 miles per hour will have a calculated horizontal displacement.

Resolve the initial velocity:

Given the initial velocity of the object is 100 miles per hour and it is launched at an angle of 30 degrees, we need to find its horizontal component. The horizontal component can be calculated using the formula: Vx = V * cos(θ), where V is the initial velocity and θ is the launch angle.

Vx = 100 * cos(30°) = 100 * √3/2 = 50√3 miles per hour.

Calculate the time of flight:

To determine the horizontal displacement, we first need to calculate the time it takes for the object to reach the ground. The time of flight can be determined using the formula: t = 2 * Vy / g, where Vy is the vertical component of the initial velocity and g is the gravitational acceleration.

Since the object is thrown vertically upwards, Vy = V * sin(θ) = 100 * sin(30°) = 100 * 1/2 = 50 miles per hour.

t = 2 * 50 / 25 = 4 hours.

Calculate the horizontal displacement:

With the time of flight determined, we can now find the horizontal displacement using the formula: Dx = Vx * t, where Dx is the horizontal displacement, Vx is the horizontal component of the initial velocity, and t is the time of flight.

Dx = 50√3 * 4 = 200√3 miles.

Therefore, the size of the displacement associated with the object in the horizontal direction, when thrown at an angle of 30 degrees and a speed of 100 miles per hour, on a large spherical star with a gravitational acceleration of 25 miles per hour, would be approximately 100 miles.

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1. A charge of +55 µC is placed on the x-axis at x = 0. A second charge of -25 µC is placed on the x-axis at x = 50 cm. What is the magnitude of the electrostatic force on a third charge of 4.0 µC placed on the x-axis at x = 44 cm? Give your answer in whole numbers.
2. If a = 0.1 m, b = 0.1 m, Q = -3 nC, and q = 1.4 nC, what is the magnitude of the electric field at point P? Give your answer in whole number.

Answers

The magnitude of the electrostatic force on the third charge is approximately 0 N.

The magnitude of the electric field at point P is approximately 108,000 N/C.

1. To find the electrostatic force on the third charge, we can use Coulomb's Law:

F = k * (|q1 * q3| / r²), where

F is the force,

k is the Coulomb's constant (approximately 9 × 10⁹ N m²/C²),

q1 and q3 are the charges, and

r is the distance between them.

Given:

q1 = +55 µC

q3 = +4.0 µC

r = 44 cm = 0.44 m

Substituting the values into the formula, we get:

F = (9 × 10⁹ N m²/C²) * ((55 × 10⁻⁶ C) * (4.0 × 10^(-6) C)) / (0.44 m²)

F = (9 × 10⁹ N m²/C²) * (2.2 × 10⁻¹¹ C²) / (0.44 m)²

F ≈ 1.09091 × 10⁻² N

Rounding to a whole number, the magnitude of the electrostatic force on the third charge is approximately 0 N.

2. To find the magnitude of the electric field at point P, we can use the formula for the electric field:

E = k * (Q / r²), where

E is the electric field,

k is the Coulomb's constant,

Q is the charge creating the field, and

r is the distance from the charge to the point of interest.

Given:

Q = -3 nC

a = 0.1 m

b = 0.1 m

We need to find the electric field at point P, which is located in the center of the rectangle defined by the points (a/2, b/2).

Substituting the values into the formula, we get:

E = (9 × 10⁹ N m²/C²) * ((-3 × 10^(-9) C) / ((0.1 m / 2)² + (0.1 m / 2)²))

E = (9 × 10⁹ N m²/C²) * (-3 × 10^(-9) C) / (0.05 m)²

E ≈ -1.08 × 10⁵ N/C

Rounding to a whole number, the magnitude of the electric field at point P is approximately 108,000 N/C.

Note: The directions and signs of the forces and fields are not specified in the question and are assumed to be positive unless stated otherwise.

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Special Relativity 11. Two meteors of rest-mass 0.1 and 0.2 kg, respectively, collide. If the relative speed before collision is 0.1 c and an observer sees them coming with equal and opposite speed and sees the lighter meteor go off at right angles to the original direction of motion, what will be the deflection of the heavier meteor according to the observer? How will this process appear to an observer comoving with: (i) the heavier meteor; and (ii) the lighter meteor? How will it appear to the centre of mass observer? Please solve all parts i.e, 1)Deflection of heavier meteor according to the observer 2) How this process looks to an observer Comoving with the heavier meteor 3) How this process looks to an observer Comoving with the lighter meteor 4)How will it appear to the centre of mass observer Special Relativity 11. Two meteors of rest-mass 0.1 and 0.2 kg, respectively, collide. If the relative speed before collision is 0.1 c and an observer sees them coming with equal and opposite speed and sees the lighter meteor go off at right angles to the original direction of motion, what will be the deflection of the heavier meteor according to the observer? How will this process appear to an observer comoving with: (i) the heavier meteor; and (ii) the lighter meteor? How will it appear to the centre of mass observer? Please solve all parts i.e, 1)Deflection of heavier meteor according to the observer 2) How this process looks to an observer Comoving with the heavier meteor 3) How this process looks to an observer Comoving with the lighter meteor 4)How will it appear to the centre of mass observer

Answers

The Lorentz factor for a speed of 0.1 c is 1.005, so the deflection of the heavier meteor is 1.005. The deflection of the heavier meteor is greater than the deflection of the lighter meteor because the heavier meteor has more mass.

1. Deflection of heavier meteor according to the observer

The deflection of the heavier meteor is given by the following equation:

deflection = (gamma - 1) * sin(theta)

where:

gamma is the Lorentz factor, given by:

gamma = 1 / sqrt(1 - v^2 / c^2)

v is the speed of the meteor, given by:

v = 0.1 c

theta is the angle between the direction of motion of the meteor and the direction of the deflection.

In this case, theta is 90 degrees, so the deflection is:

deflection = (gamma - 1) * sin(90 degrees) = gamma

The Lorentz factor for a speed of 0.1 c is 1.005, so the deflection of the heavier meteor is 1.005.

2. How this process looks to an observer comoving with the heavier meteor

To an observer comoving with the heavier meteor, the lighter meteor would appear to come from the side and collide with the heavier meteor head-on. The heavier meteor would then continue on its original course, unaffected by the collision.

3. How this process looks to an observer comoving with the lighter meteor

To an observer comoving with the lighter meteor, the heavier meteor would appear to come from the front and collide with the lighter meteor from behind. The lighter meteor would then recoil in the opposite direction, at an angle of 90 degrees to the original direction of motion.

4. How will it appear to the center of mass observer

To the center of the mass observer, the two meteors would appear to collide head-on. The two meteors would then continue on their original courses but with slightly different directions and speeds.

The deflection of the heavier meteor is greater than the deflection of the lighter meteor because the heavier meteor has more mass. The heavier meteor also has more momentum, so it is less affected by the collision.

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Question 6 A horizontal 16-N force is needed to slide a 50-kg box across a flat surface at a constant velocity of 3.5 ms. What the coefficent of anec friction between the band the O 0.10 O 011 0 0.13

Answers

The coefficient of static friction between the box and surface, given that a 16 N force is needed is 0.03

How do i determine the coefficient of static friction?

First, we shall obtain the normal reaction. Details below:

Mass of object (m) = 50 KgAcceleration due to gravity (g) = 9.8 m/s²Normal reaction (N) = ?

N = mg

= 50 × 9.8

= 490 N

Finally, we shall obtain the coefficient of static friction. Details below:

Force needed = 16 NNormal reaction (N) = 490 NCoefficient of friction (μ) =?

μ = F / N

= 16 / 490

= 0.03

Thus, we can conclude that the coefficient of friction is 0.03. None of the options are correct

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If a planetary nebula expands at a rate of 35 km s and has a lifetime of 25900 years, how large will it grow before it stops being visible? maximum radius of planetary nebula _____ km

Answers

The maximum radius of the planetary nebula is approximately 28.5 billion kilometers (rounded to two significant figures).

The maximum radius of a planetary nebula can be determined using the relation:radius = speed x age of nebula

In this case, the planetary nebula expands at a rate of 35 km/s and has a lifetime of 25900 years.

Therefore, the maximum radius of planetary nebula is calculated as follows:

radius = speed x age of nebula= 35 km/s x 25900 years (Note that the units of years need to be converted to seconds)

1 year = 365 days = 24 hours/day = 60 minutes/hour = 60 seconds/minute

Thus, 25900 years = 25900 x 365 x 24 x 60 x 60 seconds= 816336000 seconds

Plugging in the values, we get:

radius = 35 km/s x 816336000 s= 28521760000 km

Therefore, the maximum radius of the planetary nebula is approximately 28.5 billion kilometers (rounded to two significant figures).

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A closed box is filled with dry ice at a temperature of -91.7 °C, while the outside temperature is 29.2 °C. The box is cubical, measuring 0.284 m on a side, and the thickness of the walls is 3.62 x 102 m. In one day, 3.02 x 106 J of heat is conducted through the six walls. Find the thermal conductivity of the material from which the box is made.

Answers

The thermal conductivity of the material from which the box is made is approximately 0.84 W/(m·K).

The heat conducted through the walls of the box can be determined using the formula:

Q = k * A * (ΔT / d)

Where:

Q is the heat conducted through the walls,

k is the thermal conductivity of the material,

A is the surface area of the walls,

ΔT is the temperature difference between the inside and outside of the box, and

d is the thickness of the walls.

Given that the temperature difference ΔT is (29.2 °C - (-91.7 °C)) = 121.7 °C and the heat conducted Q is 3.02 x [tex]10^{6}[/tex] J, we can rearrange the formula to solve for k:

k = (Q * d) / (A * ΔT)

The surface area A of the walls can be calculated as:

A = 6 * [tex](side length)^{2}[/tex]

Substituting the given values, we have:

A = 6 * (0.284 m)2 = 0.484 [tex]m^{2}[/tex]

Now we can substitute the values into the formula:

k = (3.02 x [tex]10^{6}[/tex] J * 3.62 x [tex]10^{-2}[/tex] m) / (0.484 [tex]m^{2}[/tex] * 121.7 °C)

Simplifying the expression, we find:

k = 0.84 W/(m·K)

Therefore, the thermal conductivity of the material from which the box is made is approximately 0.84 W/(m·K).

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How is momentum conserved when a large rolling boulder hits a smaller rolling boulder? Responses Momentum is not conserved when the two boulders collide. , Momentum is not conserved when the two boulders collide., , The smaller boulder had no momentum before the collision, and afterward it has all the momentum. , The smaller boulder had no momentum before the collision, and afterward it has all the momentum., , The larger boulder transfers some of its momentum to the smaller boulders, but it keeps going forward, too. , The larger boulder transfers some of its momentum to the smaller boulders, but it keeps going forward, too., , The larger boulder gives all of its momentum to the smaller boulder.

Answers

Explanation:

When a large rolling boulder hits a smaller rolling boulder, momentum is conserved. According to the law of conservation of momentum, the total momentum of a system remains constant if there are no external forces acting on it. In this case, the system consists of the two boulders.

During the collision, the larger boulder transfers some of its momentum to the smaller boulder, causing it to move forward. However, the larger boulder also continues to move forward with some of its original momentum. Therefore, the total momentum of the system before and after the collision remains the same.

remember that momentum is a vector quantity, meaning it has both magnitude and direction. The direction of momentum for each boulder will depend on their respective velocities and massez.

Answer:

The larger boulder transfers some of its momentum to the smaller boulder, but it keeps going forward, too. Therefore, option 5 is the correct response.

Explanation:

According to the law of conservation of momentum, the total momentum of a closed system remains constant before and after the collision, as long as no external forces are acting on it. When a large rolling boulder collides with a smaller rolling boulder, conservation of momentum takes place in the system.

During the collision, the larger boulder transfers some of its momentum to the smaller boulder through the force of the impact. This transfer of momentum causes the smaller boulder to gain some momentum and start moving in the direction of the collision

However, the larger boulder also retains some of its momentum and continues moving forward after the collision. Since the larger boulder typically has greater mass and momentum initially, it will transfer some momentum to the smaller boulder while still maintaining its own forward momentum.

Therefore, in the collision between the large rolling boulder and the smaller rolling boulder, momentum is conserved as both objects experience a change in momentum.

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