Answer:
[tex]W=\frac{773}{4.45}=173.76 l b f[/tex]
Explanation:
[tex]W=\frac{G \cdot m_{e} \cdot m}{(R+h)^{2}}[/tex]
The law of gravitation
[tex]G=6.673\left(10^{-11}\right) m^{3} /\left(k g \cdot s^{2}\right)[/tex]
Universal gravitational constant [S.I. units]
[tex]m_{e}=5.976\left(10^{24}\right) k g[/tex]
Mass of Earth [S.I. units]
[tex]m=89 kg[/tex]
Mass of a man in a spacecraft [S.I. units]
[tex]R=6371 \mathrm{~km}[/tex]
Earth radius [km]
Distance between man and the earth's surface
[tex]h=261 \mathrm{~km} \quad[\mathrm{~km}][/tex]
ESULT [tex]W=\frac{6.673\left(10^{-11}\right) \cdot 5.976\left(10^{24}\right) \cdot 89}{\left(6371 \cdot 10^{3}+261 \cdot 10^{3}\right)^{2}}=773.22 \mathrm{~N}[/tex]
[tex]W=\frac{773}{4.45}=173.76 l b f[/tex]
In schematic diagrams, currents are indicated using arrows. What do the arrows indicate? a) the direction of motion of the electrons b) the direction of the current vector c) the direction of motion of the charge carriers d) the direction that positive charge carriers would move e) nothing; they are just a convenient drawing tool
Answer:
D
Explanation:
The direction that positive charge would move
12. Why do you think afternoons are much warmer than the mornings or evenings?
Answer:
Afternoons are much warmer than the mornings or evenings is explained below in complete detail.
Explanation:
Because in the afternoon the sun is on the head and the beams of sunshine arrive straight to the earth so we exhibit warmer and in daylight and evening the sun is sparse to the earth so we observe colder than an afternoon. However, in the evening, when the sun is weak, I don't sense or experienced loads of heat and sunburn on body parts that are viewing the sun directly.
an object accepts a pressure of 2:00 p.m. on the ground object is kept above it then the dead pressure will become Dash
Correct question is;
An object exerts a pressure of 2 pa on the ground if another object of the same weight is kept above it then the net pressure will become _____
Answer:
4 pa
Explanation:
We know that formula for pressure is;
P = Force/area
Where formula for force is;
F = mg
Thus;
P = mg/A
We are told P = 2 pa
Thus;
mg/A = 2
Now, when we add the same weight, our new Force is;
F_new = mg + mg = 2mg
Area remains the same A.
Thus, net pressure = F_new/A = 2mg/A
From earlier, we saw that mg/A = 2.
Thus;
net pressure = 2mg/A = 2 × 2 = 4 pa
A block of mass 3.20 kg is placed against a horizontal spring of constant k = 865 N/m and pushed so the spring compresses by 0.0650 m. HINT (a) What is the elastic potential energy of the block-spring system (in J)? (b) If the block is now released and the surface is frictionless, calculate the block's speed (in m/s) after leaving the spring. n/s
Answer:
a) The initial elastic potential energy of the block-spring system is 28.113 joules.
b) The final speed of the block is approximately 4.192 meters per second.
Explanation:
a) By applying Hooke's law and definition of work, we define the elastic potential energy ([tex]U_{g}[/tex]), measured in joules, by the following formula:
[tex]U_{g} = \frac{1}{2}\cdot k\cdot x^{2}[/tex] (1)
Where:
[tex]k[/tex] - Spring constant, measured in newtons per meter.
[tex]x[/tex] - Deformation of the spring, measured in meters.
If we know that [tex]k = 865\,\frac{N}{m}[/tex] and [tex]x = 0.065\,m[/tex], then the elastic potential energy is:
[tex]U_{g} = \frac{1}{2}\cdot \left(865\,\frac{N}{m} \right) \cdot (0.065\,m)[/tex]
[tex]U_{g} = 28.113\,J[/tex]
The initial elastic potential energy of the block-spring system is 28.113 joules.
b) According to the Principle of Energy Conservation, the initial elastic potential energy of the block-spring system becomes into translational kinetic energy, that is:
[tex]U_{g} = \frac{1}{2}\cdot m\cdot v^{2}[/tex] (2)
Where:
[tex]m[/tex] - Mass, measured in kilograms.
[tex]v[/tex] - Final speed, measured in meters per second.
Then, the final speed is cleared:
[tex]v = \sqrt{\frac{2\cdot U_{g}}{m} }[/tex]
If we know that [tex]U_{g} = 28.113\,J[/tex] and [tex]m = 3.20\,kg[/tex], then the final speed of the block is:
[tex]v = \sqrt{\frac{2\cdot (28.113\,J)}{3.20\,kg} }[/tex]
[tex]v \approx 4.192\,\frac{m}{s}[/tex]
The final speed of the block is approximately 4.192 meters per second.
Calculate the work done by an applied force of 76.0 N on a crate for the following. (Include the sign of the value in your answer.)
a. The force is exerted horizontally while pushing the crate 5.20 m.
b. The force is exerted at an angle of 41.0
Answer:
a) 400.4Joules
b) 262.69Joules
Explanation:
Work is said to be done if the force applied to an object cause the object to move through a distance
Workdone = Force × Distance
Given
Force = 76N
Distance= 5.2m
Work done = 77 × 5.2
Work done = 400.4Joules
b) If the force is exerted at an angle of 41°
Work done = Fdsin theta
Work done = 77(5.2)sin41
Work done = 400.4sin41
Work done = 262.69Joules
According to the FITT Principle you should exercise how many days ?
In a test run, a certain car accelerates uniformly from zero to 20.4 m/s in 2.60 s.
Required:
a. What is the magnitude of the cars acceleration?
b. How long does it take the car to change speed from 10.0 m/s to 20 m/s.
c. Will doubling the time always double the change in speed? why?
Answer:
(a) The acceleration is 7.85 m/s²
(b) It takes the car to change speed from 10.0 m / s to 20 m / s in a time of 1.27 seconds.
(c) Doubling the time will double the change in velocity if the acceleration is kept constant.
Explanation:
(a) Acceleration is the physical quantity that measures the rate of change of velocity with time. That is, acceleration relates changes in speed with the time in which they occur, that is, it measures how fast the changes in speed are.
The average acceleration is calculated using the following expression:
[tex]a=\frac{vf-vi}{t}[/tex]
where a is the acceleration, vf is the final velocity, vi is the initial velocity and t is the time.
In this case:
vf= 20.4 m/svi=0 m/st= 2.60 sReplacing:
[tex]a=\frac{20.4 \frac{m}{s} - 0\frac{m}{s} }{2.60 s}[/tex]
a= 7.85 m/s²
The acceleration is 7.85 m/s²
(b) In this case you know:
a= 7.85 m/s²vf= 20 m/svi= 10 m/sReplacing:
[tex]7.85 \frac{m}{s^{2} } =\frac{20 \frac{m}{s} - 10\frac{m}{s} }{t}[/tex]
and solving you get:
[tex]t=\frac{20 \frac{m}{s} - 10\frac{m}{s} }{7.85 \frac{m}{s^{2} } }[/tex]
t=1.27 s
It takes the car to change speed from 10.0 m / s to 20 m / s in a time of 1.27 seconds.
(c) Being:
[tex]a=\frac{vf-vi}{t}[/tex]
Then:
a*t= vf - vi
vf - vi represents the change in velocity. You can see that, if a (acceleration) is constant, then (vf - vi) is directly proportional to the time t: therefore, if t doubles, the change in velocity doubles as well.
In other words, doubling the time will double the change in velocity if the acceleration is kept constant.
A steel wire in a piano has a length of 0.540 m and a mass of 4.800 ✕ 10−3 kg. To what tension must this wire be stretched so that the fundamental vibration corresponds to middle C (fC = 261.6 Hz on the chromatic musical scale)?
Answer:
T = 708.81 N
Explanation:
Given that,
Length of a steel wire in a piano, l =0.54 m
Mass, [tex]m=4.8\times 10^{-3}\ kg[/tex]
We need to find the tension must this wire be stretched so that the fundamental vibration corresponds to middle C, fc = 261.6 Hz
The equation for fundamental frequency is given by :
[tex]f=\dfrac{1}{2l}\times \sqrt{\dfrac{T}{\mu}} \\\\f=\dfrac{1}{2l}\times \sqrt{\dfrac{T}{\dfrac{m}{l}}} \\\\261.6=\dfrac{1}{2\times 0.54}\times \sqrt{\dfrac{T}{\dfrac{4.8\times 10^{-3}}{0.54}}} \\\\261.6\times 2\times 0.54=\sqrt{\dfrac{T}{\dfrac{4.8\times 10^{-3}}{0.54}}}\\\\282.528=\sqrt{\dfrac{T}{0.00888}} \\\\(282.528)^2=\dfrac{T}{0.00888}\\\\T=708.81\ N[/tex]
So, the required tension in the wire is 708.81 N.
PLEASE HELP!!
To conclude, how would you explain
electromagnetic induction to a 5 year old?
words to include: magnets, electricity, and
change.
Answer:
fgdyauqushxhxxgxggxgxgxgsgsysysy
sysydyydydysysysysyxhxyt56fys7w727ueudu
A tennis ball is hit with an initial velocity of 15 m/s at an angle of
40°. What is the horizontal distance of the ball after 2 seconds?
(round to the nearest meter)
Answer:
Horizontal acceleration is 0 in a projectile motion
Explanation:
first we have to find the horizontal velocity
15cos40 = 11.49ms^-1
and then use the following equation to find the distance
[tex]s = ut + \frac{1}{2} a {t}^{2} \\ s = 11.49 \times 2 \\s = 22.98m[/tex]
Given:
Initial velocity = 15 m/sAngle = 40°According to the question,
→ [tex]15 \ Cos 40^{\circ} = 11.49 \ m/s^{-1}[/tex]
By using the relation, we get
→ [tex]s = ut +\frac{1}{2}at^2[/tex]
By substituting the values, we get
[tex]= 11.49\times 2[/tex]
[tex]= 22.98 \ m[/tex]
Thus the response above is correct.
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PLEASE ANSWER ASAP BEFORE MY TEACHER AND MY MOM KILLES ME PLEASE ASAP
The first person with the right answer gets to be a brainlest
In the attachment there is a density column where there is colour
Question: tell me why is the red at the bottom of the density column if it is the least dense
The most dense thing inside the column is the little star. It's pink, and it's resting on the bottom of the cylinder.
The red part under the star is the solid base that the cylinder is standing in. It's probably wood or plastic, it's not free to move, and it's not involved in the experiment.
Answer: That is not meant to be red, it‘s the bottom of the beaker. The star is at the very bottom of the beaker. it’s just the base of the beaker.
A sports car is accelerated from 0
car?
100 km per hour in 3 s. What is the acceleration of the?
A. 0.1g
100
B. 0.3 g
3s 1 1m 11h
160, 160m
C. 0.9 g
D. 3 g
23. [1 mark]
I have a cylinder shaped glass container that is 15 cm tall. It holds 100 cm^3 of nitrogen gas at 1 atm pressure (101.3 kPa) and 22 degrees C. I place a rubber stopper in the top so that no gas can escape. If a 40 N force is required to force the rubber stopper off the top of the flask, what temperature can I heat the nitrogen to with a Bunsen burner before the rubber stopper pops off
Answer:
T₂ = 469.73 K = 196.73 °C
Explanation:
First we will find the surface area of rubber stop:
[tex]Area = A= \frac{Volume}{Length} \\\\A = \frac{100\ cm^3}{15\ cm}\\\\A = 6.67\ cm^2 = 6.67 \ x\ 10^{-4}\ m^2[/tex]
Now, we will find the final pressure required to remove the rubber stop:
[tex]Final\ Pressure\ = P_{2} = \frac{Force}{Area}+Atmospheric Pressure \\\\P_{2} = \frac{40\ N}{6.67\ x\ 10^{-4}\ m^2} + 101.3 KPa\\\\ P_{2} = 60000\ Pa + 101.3 KPa = 60\ KPa + 101.3 KPa\\\\P_{2} = 161.3\ KPa[/tex]
Now, we use equation of state:
[tex]\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}[/tex]
for constant volume due to rigid cylinder:
[tex]\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}\\\\T_{2} = \frac{P_{2} T_{1}}{P_{1}}[/tex]
where,
P₁ = initial pressure = 101.3 KPa
P₂ = final pressure = 161.3 KPa
T₁ = Initial Temperature = 22°C = 295 K
T₂ = Final Temperature = ?
Therefore,
[tex]T_{2} = \frac{(161.3\ KPa)(295\ K)}{101.3\ KPa}[/tex]
T₂ = 469.73 K = 196.73 °C
To what temperature will 9400J of heat raise 3.5kg of water that is initially at 18.0?C? The specific heat of water is 4186 J/kg?C?.
Answer:
[tex]T_f=18.641^{\circ} C[/tex]
Explanation:
Given that,
Heat raised, Q = 9400 J
Mass, m = 3.5 kg
Initial temperature of water, [tex]T_i=18^{\circ} C[/tex]
The specific heat of water, [tex]c=4186\ J/kg^{\circ} C[/tex]
We need to find the final temperature of water. The heat required to raise the temperature is given by :
[tex]Q=mc\Delta T\\\\Q=mc(T_f-T_i)\\\\\dfrac{Q}{mc}=(T_f-T_i)\\\\T_f=\dfrac{Q}{mc}+T_i\\\\T_f=\dfrac{9400}{3.5\times 4186}+18\\\\=18.641^{\circ} C[/tex]
So, the final temperature is [tex]18.641^{\circ} C[/tex].
______ is the limiting factor that can keep some people from exercising if they can’t afford to purchase certain types of exercise equipment or pay for activities. Money Understanding Motivation Location Family behaviors
Answer:
I think it is Location.
Explanation:
If im wrong im sorry
Answer:
Money
Explanation:
______ is the limiting factor that can keep some people from exercising if they (can’t afford) to purchase certain types of exercise equipment or pay for activities.
Says they cant afford it, therefore it would obviously be money
A 3.1 kg block initially at rest is pulled to the right along a horizontal frictionless surface by a constant horizontal force of 16.6 N. Find the speed after it has moved 3.8m.
We know, acceleration is given by :
a = F/m
a = 16.6/3.1 m/s²
a = 5.35 m/s²
Now, we need to find the speed after it has moved 3.8 m.
By equation of motion :
[tex]v^2 -u^2 = 2as\\\\v=\sqrt{2as}\\\\v = \sqrt{2\times 5.35 \times 3.8 }\ m/s\\\\v = 6.38\ m/s[/tex]
Therefore, the speed after it has moved 3.8 m is 6.38 m/s.
. Momentum is ____________ proportional to an objects mass and its velocity, meaning that if either the mass or the velocity increases, the momentum will _______________.
Answer:
1) directly
2) increase or double
Explanation:
Momentum is directly proportional to an objects mass and its velocity, meaning that if either the mass or the velocity increases, the momentum will double/increase.
A standard inverting op-amp circuit has an R1 of 10 kΩ and an Rf of 220 kΩ. If the offset current is 100 nA the output offset voltage due to this current is ________.
Answer:
The value is [tex]V_{os} = 0.001 \ V[/tex]
Explanation:
From the question we are told that
The circuit resistance is [tex]R_1 = 10 \ k \Omega[/tex]
The feedback resistance is [tex]R_f = 220 \ k \Omega[/tex]
The offset current is [tex]I_{os } = 100 \ nA = 100 * 1)^{-9} \ A[/tex]
Generally the offset voltage is mathematically reparented as
[tex]V_{os} = R_f * I_{os}[/tex]
=> [tex]V_{os} = 10 *10^{3}* 100 *10^{-9}[/tex]
=> [tex]V_{os} = 0.001 \ V[/tex]
Under which of the following conditions is the magnitude of the average velocity of a particle moving in one dimension smaller than the average speed over some time interval?
Answer:
A particle moves in the +x direction and then reverses the direction of its velocity
Explanation:
This is illustrated in that when a particle moves in a straight route with no alterations in direction, this will lead to displacement and distance being equal at any point in time during the movement. Thereby the quantity of average speed and average velocity will equal.
On the other hand, should the particle reverses direction, the distance traveled will be greater than it's displacement, thereby, the average speed will be greater than the average velocity.
If a person throws a wiffle ball and a tennis ball with the same speed, which object would travel with more kinetic
energy? Justify your selection
Answer:
The tennis ball will have more kinetic energy
Explanation:
It will have more kinetic energy because the tennis ball is more heavier than a wiffle ball. Furthermore, objects that are heavy will have more kinetic energy than objects that are light. Therefore, the tennis ball is the correct answer.
The tennis ball will have more kinetic energy as compared to the wiffle ball, because tennis ball is more heavier.
The energy possessed by any object due to virtue of its motion is known as kinetic energy. The standard expression for the kinetic energy is given as,
[tex]KE = \dfrac{1}{2}mv^{2}[/tex]
Here,
m is the mass of balls.
v is the speed of balls.
Clearly the kinetic energy of any object is dependent on the mass and velocity. So with same velocity, the heavier object will have more kinetic energy. Since, the tennis ball is more heavier as compared to the wiffle ball, therefore kinetic energy of tennis ball will be more.
Thus, we can conclude that the tennis ball will have more kinetic energy as compared to the wiffle ball, because tennis ball is more heavier.
Learn more about the kinetic energy here:
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Older televisions display a picture using a device called a cathode ray tube, where electrons are emitted at high speed and collide with a phosphorescent surface, causing light to be emitted. The paths of the electrons are altered by magnetic fields. Consider one such electron that is emitted with an initial velocity of 1.85 107 m/s in the horizontal direction when magnetic forces deflect the electron with a vertically upward acceleration of 5.45 1015 m/s2. The phosphorescent screen is a horizontal distance of 5.6 cm away from the point where the electron is emitted. (a) How much time does the electron take to travel from the emission point to the screen? (b) How far does the electron travel vertically before it hits the screen?
Answer:
a) t = 3.027 10⁻⁹ s , b) y = 2.25 10⁻² m
Explanation:
We can solve this problem using the kinematic relations
a) as on the x-axis there is no relationship
vₓ = x / t
t = x / vₓ
We reduce the magnitudes to the SI system
x = 5.6 cm (1m / 100 vm) = 0.056 m
we calculate
t = 0.056 / 1.85 10⁷
t = 3.027 10⁻⁹ s
b) the time is the same for the two movements, on the y axis
y = v₀t + ½ a t²
as the beam leaves horizontal there is no initial vertical velocity
y = ½ a t²
let's calculate
y = ½ 5.45 10¹⁵ (3.027 10⁻⁹)²
y = 2.25 10⁻² m
A uniform rod of length L, mass M, is suspended by two thin strings. Which of the following statements is true
regarding the tensions in the strings? Please explain
(a) T2 = 0.8 T1
(b) None of the above
(c) T2 = T1
(d) Not enough information to determine
(e) T2 = 0.6 T1
(f) T2 = 2.5 T1
Answer:
(d) not enough info
Explanation:
because it doesn't specify where the strings are attached
if it was the two ends of the rod then T1 would be equal to T2
Which of the following is not a metric base unit? *
Meter
Liter
Inch
Gram
Answer:
Liter
Explanation:
BERE
Which describes the positions on a horizontal number line?
0
O All points to the left of one are positive.
O All points to the right of one are positive.
O All points to the left of zero are negative.
O All points to the right of zero are negative.
Mark this and return
Save and Exit
Next
Submit
Answer:
All points to the left of zero are negative
Explanation:
Answer:
C
Explanation:
on edge
Mental processes refers to
overt actions and reactions.
only animal behavior.
internal, covert processes.
outward behavior.
An alpha particle has twice the charge of a beta particle. Why does the former deflect less than the latter when passing between electrically charged plates, assuming they both have the same speed?
*PLEASE HELP*
WILL MARK BRAINLIEST
Light passing through a double slit with
separation d = 6.42x10-5 m creates its
first maximum (m=1) at an angle of
0.396 deg. What is the wavelength of
the light, IN NANOMETERS?
(Hint: The answer will be between 400 and
700.) (Unit = nm)
Answer:
443.8 nm
Explanation:
mλ = yd/D = d(y/D)
m = 1
d = 6.42 x 10-5 m
θ = 0.396°
y/D = tanθ = tan(0.396°) ≅ 6.912 x 10-3
λ = (6.42 x 10-5 m) tan(0.396°) = (6.42 x 10-5 m)(6.912 x 10-3) ≅ 4.438 x 10-7 m = 443.8 nm
I JUST WANT TO MAKE CLEAR THAT d=6.42*10^-5 NOT d = 6.42x10-5 right?
What is the net force on a truck if the force of friction is 31 N and the force of the engine is 79 N?
48 N backword
79 N forward
48 N forward
110 N forward
Answer:
Fr = 48 [N] forward.
Explanation:
Suppose the movement is on the X axis, in this way we have the force of the engine that produces the movement to the right, while the force produced by the brake causes the vehicle to decrease its speed in this way the sign must be negative.
∑F = Fr
[tex]F_{engine}-F_{brake} =F_{r}\\F_{r}=79-31\\F_{r}=48[N][/tex]
The movement remains forward, since the force produced by the movement is greater than the braking force.
Kelli drew a diagram to compare cast and imprint fossils. Which label belongs in the area marked Y? Involves minerals replacing remains of organism Takes millions of years to form Gives evidence of organism's activity Involves mold filling with minerals and sediment
Answer:
D. Involves mold filling with minerals and sediment
Explanation:
The given part y is in the inter region of the circle labelled cast and imprints. Thus, it belongs to both. Cast fossils are a type of imprint fossils which are filled with minerals from sediments and hardens to form the fossils. Thus, option d is correct.
What are fossils?Fossils are decomposed plants or animals or their imprints indicating their actions such as footprints, shells etc. There are broadly two types of fossils namely body fossils and trace fossils.
Body fossils are decomposed body parts of animals or plants such as bones, tree branches, skull etc. Trace fossils are imprints of the animals or plants such as footprints, shells, caves, skin etc.
Imprint fossils are the impression of any activities or behaviour of living things. shell, leaf, caves etc are imprint fossils. Cast fossils are formed by filling of imprint fossils with minerals reached from water and sediments and hardened to fossils. Therefore, the given area is cast fossil and option d is correct.
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A soccer ball with a mass of 0.427 kg approaches a player horizontally with a speed of 16.0 m/s. The player kicks the ball with her foot, which causes the ball to move in the opposite direction with a speed of 21.7 m/s.
Required:
a. What magnitude of impulse (in kg · m/s) is delivered to the ball by the player?
b. What is the direction of the impulse delivered to the ball by the player?
c. If the player's foot is in contact with the ball for 0.0600 s, what is the magnitude of the average force (in N) exerted on the player's foot by the ball?
Explanation:
Given that,
Mass of a soccer ball, m = 0.427 kg
Initial speed, u = 16 m/s
Final speed, v = -21.7 m/s (negative as it is in opposite direction)
(a) Impulse = change in momentum
J = 0.427 × (-21.7 -16)
J = -16.097 kg-m/s
(b) Impulse delivered to the ball by the player is in opposite direction.
(c) Time of contant, t = 0.0600 s
Average force,
J = force×time
[tex]F=\dfrac{J}{t}\\\\F=\dfrac{16.097}{0.0600 }\\\\=268.28\ N[/tex]
Hence, this is the required solution.