Assume the three blocks (m1 = 1.0 kg, m2 = 2.0 kg, and m3 = 4.0 kg) portrayed in the figure below move on a frictionless surface and a force F = 34 N acts as shown on the 4.0-kg block. Answer parts a-c.

Assume The Three Blocks (m1 = 1.0 Kg, M2 = 2.0 Kg, And M3 = 4.0 Kg) Portrayed In The Figure Below Move

Answers

Answer 1

(a) The acceleration of the system is 8.5 m/s².

(b) The tension in the cord connecting the 4.0 kg and 1.0 kg blocks is 42.5 N.

(c) The force exerted by the 1.0 kg block on the 2.0 kg block is 59.5 N.

To solve this problem, we can use Newton's second law of motion (F = ma) and consider the forces acting on each block individually.

(a) Determine the acceleration given this system:

To find the acceleration (a) of the system, we can use the net force acting on the 4.0 kg block (m3). The only force acting on m3 is the applied force (F = 34 N).

F = m3 * a

34 N = 4.0 kg * a

Solving for a, we find:

a = 34 N / 4.0 kg

a = 8.5 m/s²

Therefore, the acceleration of the system is 8.5 m/s².

(b) Determine the tension in the cord connecting the 4.0-kg and the 1.0-kg blocks:

To find the tension in the cord (T), we can consider the forces acting on the 1.0 kg block (m1).

T - F = m1 * a

T - 34 N = 1.0 kg * 8.5 m/s²

T - 34 N = 8.5 N

T = 42.5 N

Therefore, the tension in the cord connecting the 4.0 kg and 1.0 kg blocks is 42.5 N.

(c) Determine the force exerted by the 1.0-kg block on the 2.0-kg block:

To find the force exerted by the 1.0 kg block (m1) on the 2.0 kg block (m2), we can consider the forces acting on the 2.0 kg block.

F - T = m2 * a

F - 42.5 N = 2.0 kg * 8.5 m/s²

F - 42.5 N = 17 N

F = 59.5 N

Therefore, the force exerted by the 1.0 kg block on the 2.0 kg block is 59.5 N.

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Related Questions

As a fish jumps vertically out of the water, assume that only two significant forces act on it: an upward force F exerted by the tail fin and the downward force due to gravity. A record Chinook salmon has a length of 1.50 m and a mass of 52.0 kg. If this fish is moving upward at 3.00 m/s as its head first breaks the surface and has an upward speed of 6.80 m/s after two-thirds of its length has left the surface, assume constant acceleration and determine the following. find a - the salmon's acceleration (answer in m/s^2 upward), find b - the magnitude of the force F during this interval (direction is N).

Answers

Answer:

To solve this problem, we need to use some principles of physics, specifically Newton's second law (F=ma) and the equations of motion. Here are the steps:

1. Calculate the acceleration (a)

We can use the equation of motion to find the acceleration:

v_f^2 = v_i^2 + 2a*d

where:

v_f = final velocity = 6.80 m/s

v_i = initial velocity = 3.00 m/s

d = distance = 2/3 of the length of the fish = 2/3 * 1.50 m = 1.00 m

a = acceleration (which we are trying to find)

Rearranging the equation to solve for a gives us:

a = (v_f^2 - v_i^2) / (2*d)

2. Calculate the magnitude of the force F

Once we have the acceleration, we can use Newton's second law (F=ma) to calculate the force. The net force acting on the fish as it jumps out of the water is the difference between the upward force F exerted by the tail fin and the downward force due to gravity (mg). The net force is also equal to the product of the mass of the fish and its acceleration (ma). Therefore, we have:

F - mg = ma

Rearranging this equation to solve for F gives us:

F = ma + mg

Now let's plug in the numbers and do the calculations.

First, let's find the acceleration:

a = (v_f^2 - v_i^2) / (2*d)

a = (6.80 m/s)^2 - (3.00 m/s)^2) / (2*1.00 m)

a = (46.24 m^2/s^2 - 9.00 m^2/s^2) / 2 m

a = 37.24 m^2/s^2 / 2 m

a = 18.62 m/s^2

The salmon's acceleration is 18.62 m/s^2 upward.

Next, let's find the force F. We know the mass of the fish is 52.0 kg, and the acceleration due to gravity is approximately 9.8 m/s^2. So,

F = ma + mg

F = (52.0 kg)(18.62 m/s^2) + (52.0 kg)(9.8 m/s^2)

F = 969.24 N + 509.6 N

F = 1478.84 N

So, the magnitude of the force F exerted by the salmon's tail fin during this interval is approximately 1479 N.

If a 9000kg water flows in a minute through a pipe of cross sectional area 0.3m², what is the speed of water in the pipe? ​

Answers

Answer:

5 m/s

Explanation:

We are given that 9000 kg of water flows through the pipe in 1 minute. Mass flow rate = mass/time

So, mass flow rate = 9000 kg / 1 minute = 150 kg/s

We know the cross sectional area of the pipe is 0.3 m2. From continuity equation, mass flow rate = density * area * velocity

So, 150 = 1000 * 0.3 * v (Density of water is approximately 1000 kg/m3)

Solving for v (velocity):

v = 150/(1000*0.3) = 5 m/s

Therefore, the speed of water in the pipe is 5 m/s.

A tennis player moves in a straight-line path as shown in the figure below. Find her average velocity in the following time intervals. Find (a) - 0 to 1.0 s, find (b) 0 to 4s, find (c) - 1.0 s to 5.0 s, find (d) - 0 to 5.0 s.

Answers

(a) The average velocity of the tennis player at 0 to 1 s is  4 m/s.

(b) The average velocity of the tennis player at 0 to 4 s is -0.5 m/s.

(c) The average velocity of the tennis player at 1 to 5 s is  1 m/s.

(d) The average velocity of the tennis player at 0 to 5 s is 0.8 m/s.

What is the average velocity of the tennis player?

The average velocity of the tennis player at the given time, is calculated by applying the formula for average velocity as follows;

average velocity = total displacement / total time

(a) The average velocity at 0 to 1 s;

average vel. = (4 m - 0 m ) / (1 s ) = 4 m/s

(b) The average velocity at 0 to 4 s;

average vel. = (-2 - 0 )m / 4 s = -0.5 m/s

(c) The average velocity at 1 to 5 s;

average vel. = (4 - 0 )m / (5 - 1) s =  1 m/s

(d) The average velocity at 0 to 5 s;

average vel. = (4 - 0 )m / (5 - 0) s =  0.8 m/s

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Suppose that you're facing a straight current-carrying conductor, and the current is flowing toward you. The lines of magnetic force at any point in the magnetic field will act in
Question 17 options:

A)

a clockwise direction.

B)

a counterclockwise direction.

C)

the direction opposite to the current.

D)

the same direction as the current.

Answers

Suppose that you're facing a straight current-carrying conductor, and the current is flowing toward you. The lines of magnetic force at any point in the magnetic field will act in option c)  the direction opposite to the current.

Lenz's law is the law that governs the direction of magnetic force.According to Lenz's law, magnetic fields induced by an electric current have a polarity such that the current's magnetic field opposes any change in current flow. Based on this law, the induced current must produce a magnetic field that opposes the current that produced it.

If the current is flowing towards us, the induced magnetic field must flow in the opposite direction to the current. Therefore, the direction of the lines of magnetic force at any point in the magnetic field will act in the direction opposite to the current.Hence, the correct option is C) the direction opposite to the current.

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If you were trying to build a soundproof room, which of the following materials would you choose to absorb the most sound, based on the coefficient of absorption for each material?
Question 19 options:

A)

Concrete

B)

Wood

C)

Carpet

D)

Heavy curtains

Answers

Answer:

C) Carpet

Explanation:

If you were trying to build a soundproof room, the material that would absorb the most sound would be carpet. Carpet has a high coefficient of absorption, which means that it is effective in reducing sound transmission. Concrete and wood are hard surfaces that reflect sound, making them poor choices for sound absorption. Heavy curtains may help to reduce sound transmission, but they are not as effective as carpet. So, if you want to build a soundproof room, you should consider using carpet as a primary material for sound absorption.

With a force of 200 N a body is lifted 20 meters in 20 seconds. Calculate the weight of this body. Use the formula for distance as a function of acceleration with initial velocity equal to zero.

Answers

The weight of the body is approximately 19600 N when a body is lifted with a force of 200 N about  20 meters in 20 seconds.

We know that distance = initial velocity*time  + (1/2) * acceleration * [tex]time^2[/tex]

Given that, the initial velocity equal to zero

distance= (1/2) * acceleration * [tex]time^2[/tex]

Also given-

distance = 20 meters

time = 20 seconds

Rearranging the formula-

20 = (1/2) * acceleration * [tex](20^2)[/tex]

20 = (1/2)   * acceleration * 400

40 = acceleration * 400

acceleration = 40/400 = 0.1 metre/[tex]sec^{2}[/tex]

Force= mass*acceleration

mass= force/acceleration

=200/0.1= 2000 Kg

But this acceleration is due to applied force but weight only involves gravitational force only.  

Since weight is defined as the force acting on an object due to gravity,

weight = mass * acceleration due to gravity

= 2000 * 9.8 (acceleration due to gravity)

= 19600 N

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what type of force

a child on a sled slides down the hill

Answers

Answer:

Gravity.

Explanation:

Gravity causes the child on a sled to slide down the hill.

Hope this helps!

Fig above shows a wave traveling through a medium. Use the fig to answer the questions below.

A.)What is the amplitude of the wave ? Include correct units.
B.)Use the graph to determine the time of one wave. Use it to find the frequency.
C.)If the speed of the wave is 25 m/s, what is the wavelength of the wave ? Show data listing, equation , substitution leading to the answer for full credit.

Answers

(a) The amplitude of the wave is 0.2 m.

(b) The period of the wave is  4 s.

(c) The wavelength of the wave is 100 m.

What is the amplitude of the wave?

(a) The amplitude of the wave is the maximum displacement of the wave.

amplitude of the wave = 0.2 m

(b) The period of the wave is the time taken for the wave to make one complete cycle.

period of the wave = 5.5 s - 1.5 s = 4 s

(c) The wavelength of the wave is calculated as follows;

λ = v / f

where;

v is the speed of the wavef is the frequency of the wave

f = 1/t = 1 / 4s = 0.25 Hz

λ = ( 25 m/s ) / 0.25 Hz

λ = 100 m

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Two forces are applied to a car in an effort to move it, as shown in the figure below. (Let
F1 = 445 N and F2 = 368 N. Assume up and to the right are in the positive directions.) Find part a and b.

Answers

a. The resultant vector of the forces F1 and F2 applied to the car has a magnitude of approximately 801.86 N and is directed approximately 19.85 degrees to the right (forward direction).

b. With a mass of 3000 kg, the car experiences an acceleration of approximately 0.2673 m/s² when subjected to the given forces, neglecting friction.

To find the resultant vector of the two forces, we can use vector addition. The given forces are F1 = 445 N and F2 = 368 N. Let's resolve these forces into their horizontal and vertical components.

For F1:

The angle between the normal and F1 is 10 degrees. We can find the horizontal and vertical components using trigonometry.

Horizontal component of F1 = F1 * cos(10 degrees)

                          = 445 N * cos(10 degrees)

                          ≈ 438.37 N

Vertical component of F1 = F1 * sin(10 degrees)

                        = 445 N * sin(10 degrees)

                        ≈ 77.06 N

For F2:

The angle between the normal and F2 is 30 degrees. Again, we can use trigonometry to find the components.

Horizontal component of F2 = F2 * cos(30 degrees)

                          = 368 N * cos(30 degrees)

                          ≈ 318.64 N

Vertical component of F2 = F2 * sin(30 degrees)

                        = 368 N * sin(30 degrees)

                        ≈ 184 N

Now, we can add the horizontal and vertical components separately to find the resultant vector.

Horizontal component of the resultant vector = Horizontal component of F1 + Horizontal component of F2

                                           ≈ 438.37 N + 318.64 N

                                           ≈ 757.01 N

Vertical component of the resultant vector = Vertical component of F1 + Vertical component of F2

                                         ≈ 77.06 N + 184 N

                                         ≈ 261.06 N

To find the magnitude of the resultant vector, we can use the Pythagorean theorem:

Magnitude of the resultant vector = sqrt((Horizontal component)^2 + (Vertical component)^2)

                                = [tex]\sqrt{((757.01 N)^2 + (261.06 N)^2)}[/tex]

                                ≈ 801.86 N

The direction of the resultant vector can be found using trigonometry:

Direction = arctan(Vertical component / Horizontal component)

         = arctan(261.06 N / 757.01 N)

         ≈ 19.85 degrees to the right (forward direction)

So, the resultant vector of the two forces has a magnitude of approximately 801.86 N and a direction of approximately 19.85 degrees to the right (forward direction).

Now, let's calculate the acceleration of the car using Newton's second law: F = ma.

Given that the mass of the car is 3000 kg, we can rearrange the equation to solve for acceleration:

Acceleration = Force / Mass

Using the magnitude of the resultant vector (801.86 N), we have:

Acceleration = 801.86 N / 3000 kg

            ≈ 0.2673 m/s²

Therefore, the car has an acceleration of approximately 0.2673 m/s² in the direction of the resultant vector, assuming there is no friction present.

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What happens when a substance undergoes a physical change

Answers

Answer: the material involved in the change is structurally the same before and after the change. Types of some physical changes are texture, shape, temperature, and a change in the state of matter. A change in the texture of a substance is a change in the way it feels

Explanation:

Which statement best describes the refraction of light as it moves from air to glass?

A. Light bends due to the difference in the speed of light in air and glass.

B. Although the light bends, its speed remains the same as before.

C. Although the light changes speed, it continues in the same direction as before.

D. Light undergoes diffraction due to the difference in the speed of light in air and glass.

Answers

A. Light bends due to the difference in the speed of light in air and glass.

Explain the function of power supply, readout, peripheral, microcomputer, transducer and processor​

Answers

The function of the power supply is to provide electrical energy to the device or system that needs it. The power supply converts the incoming voltage from the power source into a form that is usable by the device, such as DC voltage.

The readout is a device or component that displays data or information to the user. The readout could be a simple LED display or a complex graphical display.

A peripheral is a device or component that connects to a computer or other electronic device to provide additional functionality. Examples of peripherals include printers, scanners, and external hard drives.

A microcomputer is a type of computer that is designed to fit on a single microchip. Microcomputers are found in a wide range of devices, including smart phones, tablets, and embedded systems.

A transducer is a device that converts one form of energy to another. In electronics, transducers are commonly used to convert electrical energy into mechanical energy, or vice versa.

The processor is the central component of a computer or electronic device. The processor is responsible for executing instructions and controlling the other components of the system. The performance and capabilities of a device are largely determined by the speed and power of the processor.

A rifle is fired straight up, and the bullet leaves the rifle with an initial velocity
magnitude of 630 m/s. After 5.00 s, the velocity is 581 m/s. At what rate is the bullet
decelerated?

Answers

Explanation:

acceleration definition = change in velocity / change in time =

(630 - 581)  m/s  /  5 s = 49 / 5 = 9.8 m/s^2  was the deceleration

Question 1 of 10
What is the slope of the line plotted below?
B. 2
5
10
C. 1
O A. 0.5
о
9
OD. -0.5
5

Answers

The answer is B. 2


Explanation

Masses m and 2m are joined by a light inextensible string which runs without slipping over a uniform circular pulley of mass 2m and radius a. Using the angular position of the pulley as generalized coordinate, write down the Lagrangian function and Lagrange's equation. Find the acceleration of the masses.​

Answers

The acceleration of the mass 2m is - (8/5) a θ´´.

We have two masses m and 2m connected by a string without slipping over a uniform circular pulley of mass 2m and radius a. We have to find the acceleration of the masses and write down the Lagrangian function and Lagrange's equation. The angular position of the pulley as generalized coordinate is used. Lagrangian function

L = T – VL = Kinetic energy - Potential energy

The kinetic energy is the sum of the kinetic energies of the two masses and the pulley. The potential energy is the sum of the potential energies of the two masses. The potential energy of the pulley can be ignored since it is fixed. Let θ be the angular position of the pulley, x be the distance fallen by the mass m and y be the distance fallen by the mass 2m.Kinetic energy of mass m (K1)K1 = ½ mv²where v = (dx/dt) is the velocity of mass mKinetic energy of mass 2m

(K2)K2 = ½ (2m) (dy/dt)²where (dy/dt) is the velocity of mass 2mKinetic energy of pulley (K3)The pulley is rolling without slipping, so the velocity of the point at the edge of the pulley is given byv = R(θ´)where R = a is the radius of the pulley. Hence, the kinetic energy of the pulley is

K3 = ½ I (θ´)²where I = (2/5) M R² = (2/5) (2m) a² is the moment of inertia of the pulleyPotential energy of mass m (V1)V1 = mgywhere g is the acceleration due to gravityPotential energy of mass 2m (V2)V2 = 2mgxThe Lagrangian function isL = K1 + K2 + K3 - V1 - V2L = ½ m(dx/dt)² + ½ (2m) (dy/dt)² + ½ (2/5) (2m) a² (θ´)² - mgy - 2mgxL = ½ m(dx/dt)² + ½ (2m) (dy/dt)² + ½ (4/5) ma² (θ´)² - mgy - 2mgxLagrange's

equationLet's find the equation of motion of the mass m using Lagrange's equation. The Lagrangian function depends on three variables, so we need three equations of motion.Lagrange's equation isd/dt (δL/δ(dx/dt)) - δL/δx = 0The first term gives usd/dt (δL/δ(dx/dt)) = m(dx/dt) + (4/5) ma² (θ´)(d/dt)(θ´) = m(dx²/dt²) + (4/5) ma² θ´´The second term gives usδL/δx = -d/dx (mgy) = 0The third term gives usδL/δ(θ) = d/dt (δL/δ(θ´))δL/δ(θ) = d/dt [(4/5) ma² (θ´)] = (4/5) ma² θ´´

The equation of motion ism(dx²/dt²) + (4/5) ma² θ´´ = 0We can solve this equation to find the acceleration of the mass m.The acceleration of the mass mThe acceleration of the mass m is given bya = dx²/dt² = - (4/5) a θ´´Therefore, the acceleration of the mass m is - (4/5) a θ´´.The equation of motion of the pulley is obtained in

the same way as above but we need to use the moment of inertia I of the pulley in the Lagrangian. We get(4/5) ma² θ´´ + 2mgRθ´² = 0Dividing by (4/5) ma², we getθ´´ + (5/8gR) θ´² = 0The acceleration of the mass 2m is given by the same expression as above but with m replaced by 2m.

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Consider the figure below. (a) Find the tension in each cable supporting the 524-N cat burglar. (Assume the angle of the inclined cable is 34.0°.) (b) Suppose the horizontal cable were reattached higher up on the wall. Would the tension in the inclined cable increase, decrease, or stay the same?

Answers

(a) The tension in the inclined cable (T1) and horizontal cable (T2) supporting the cat burglar is equal. The tension in the vertical cable (T3) is 524 N.

(b) If the horizontal cable is reattached higher up, the tension in the inclined cable (T1) would increase.

(a) To find the tension in each cable supporting the 524-N cat burglar, we'll consider the forces acting on the system. Let's denote the tension in the inclined cable as T1, the tension in the horizontal cable as T2, and the tension in the vertical cable as T3. The angle between the inclined cable and the vertical cable is given as θ.

In the vertical direction, the tension in the vertical cable T3 balances the weight of the cat burglar:

T3 - 524 N = 0

T3 = 524 N

In the horizontal direction, the tension in the inclined cable T1 can be expressed as:

T1 * cos(θ) = T2

Now, we need to determine the value of θ to calculate T1 and T2. Let's assume that θ is the given angle of θ = 0.

Substituting the angle and rearranging the equation, we have:

T1 = T2 / cos(θ)

T1 = T2 / cos(0)

T1 = T2 / 1

T1 = T2

So, the tension in the inclined cable (T1) is equal to the tension in the horizontal cable (T2).

Therefore, the tension in each cable is as follows:

T1 (inclined cable) = T2 (horizontal cable)

T1 = T2

T3 (vertical cable) = 524 N

(b) If the horizontal cable were reattached higher up on the wall, the tension in the inclined cable (T1) would increase.

The correct answer is option A.  

This is because reattaching the horizontal cable at a higher point on the wall would increase the horizontal component of the tension, resulting in a larger tension in the inclined cable. The tension in the vertical cable (T3) would remain the same as it is independent of the position of the horizontal cable.

In summary, the tension in the inclined cable (T1) and the horizontal cable (T2) are equal, and their value depends on the angle θ. The tension in the vertical cable (T3) is 524 N. If the horizontal cable were reattached higher up on the wall, the tension in the inclined cable would increase, while the tension in the vertical cable would remain the same.

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Select the correct answer.
In which situation is maximum work considered to be done by a force?
A.
The angle between the force and displacement is 180°.
B.
The angle between the force and displacement is 90°.
C.
The angle between the force and displacement is 60°.
D.
The angle between the force and displacement is 45°.
E.
The angle between the force and displacement is 0°.

Answers

Option A. The angle between the force and displacement is 180°, the maximum work is considered to be done by the force.

Work is defined as the product of the force applied to an object and the displacement of the object in the direction of the force. Mathematically, work (W) is given by the equation:

W = F * d * cos(theta)

Where

F = magnitude of the force

d = magnitude of the displacement

theta = angle between the force and displacement vectors.

In order to maximize the work done by a force, we need to maximize the value of the cosine of the angle theta. The cosine function reaches its maximum value of 1 when the angle theta is 0° or 180°.

When the angle between the force and displacement is 0° (option E), the force and displacement vectors are perfectly aligned in the same direction. In this case, the work done is maximized. Therefore, the correct answer is option A.

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Planets sweep out
close to the sun, it travels a
areas in
time in their orbits around the sun, but the distance they move varies. When the planet is
when it is closer to the sun.
✓distance as the area stays the same. So, the planet moves

Answers

In planetary motion, as a planet orbits the Sun, it sweeps out equal areas in equal time intervals. This principle, known as Kepler's Second Law of Planetary Motion, describes the behavior of planets as they move in elliptical orbits around the Sun.

When a planet is closer to the Sun, it experiences a stronger gravitational pull, which causes it to move faster. As a result, the planet covers a larger distance in a given amount of time compared to when it is farther from the Sun. This compensates for the smaller distance, ensuring that the area swept out by the planet remains the same.

To illustrate this, imagine a line connecting the Sun and the planet, called the radius vector. As the planet moves along its orbit, the radius vector sweeps out a wedge-shaped area. The rate at which this area is swept out is constant. When the planet is closer to the Sun, it moves faster, covering more distance along its orbit in a given time. Consequently, the narrower end of the wedge is compensated by the planet's higher speed, resulting in an equal area to that covered when it is farther from the Sun.

This phenomenon is a consequence of the conservation of angular momentum in the gravitational field of the Sun and allows for a consistent distribution of the planet's orbital motion.

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An object of mass M = 14.0 kg is attached to a cord that is wrapped around a wheel of radius r = 12.0 cm (see figure). The acceleration of the object down the frictionless incline is measured to be a = 2.00 m/s2 and the incline makes an angle = 37.0° with the horizontal. Assume the axle of the wheel to be frictionless. Answer parts a-c.

Answers

a.  the tension in the rope is  91.5 N.

b.   the moment of inertia of the wheel is  0.1008 kg⋅m².

c.  the angular speed of the wheel 2.30 s after it begins rotating is  38.34 rad/s.

How do we calculate?

(a)

The tension in the rope can be found by considering the forces acting on the object.

ma = mg*sin(θ) - T

(14.0 kg)(2.00 m/s²)

= (14.0 kg)(9.8 m/s²)*sin(37°) - T

T = (14.0 kg)(9.8 m/s²)*sin(37°) - (14.0 kg)(2.00 m/s²)

T =  91.5 N

(b)

The moment of inertia of a wheel:

I = (1/2)MR²

I = (1/2)(14.0 kg)(0.12 m)²

I = 0.1008 kg⋅m²

(c)

The angular acceleration of the wheel:

α = a/R

α = angular acceleration,

a = linear acceleration of the object,

R =  radius of the wheel.

α = (2.00 m/s²)/(0.12 m)

α = 16.67 rad/s²

The angular speed (ω) of the wheel after time t is :

ω = ω₀ + αt

ω = 0 + (16.67 rad/s²)(2.30 s)

ω = 38.34 rad/s

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D 4.8
This is a harder question based on the Law of Conservation of Momentum. Take the time to work
your way through it. Start with a diagram.
A 400 kg bomb sitting at rest on a table explodes into three pieces. A 150 kg piece moves off to the
east with a velocity of 150 m s². A 100 kg piece moves off with a velocity of 200 m s at a direction of
south 60° west. What is the velocity of the third piece?

It is possible

Answers

The velocity of the third piece is v₃ = -12500 kg·m/s / m₃

How do we calculate?

The law of conservation of momentum states that the total momentum before the explosion is equal to the total momentum after the explosion.

velocity of the third piece =  v₃.

The total initial momentum before the explosion = 0

The total final momentum after the explosion= 0

Initial momentum = 0 kg·m/s (since the bomb is at rest)

Final momentum = m₁v₁ + m₂v₂ + m₃v₃

m₁ = mass of the first piece = 150 kg

v₁ = velocity of the first piece = 150 m/s (to the east)

m₂ = mass of the second piece = 100 kg

v₂ = velocity of the second piece = 200 m/s (south 60° west)

m₃ = mass of the third piece = unknown

v₃ = velocity of the third piece = unknown

0 = (150 kg)(150 m/s) + (100 kg)(200 m/s)(cos(60°)) + (m₃)(v₃)

final momentum = 0 and hence  v₃ is found as :

0 = 22500 kg·m/s - 10000 kg·m/s + (m₃)(v₃)

-12500 kg·m/s = (m₃)(v₃)

v₃ = -12500 kg·m/s / m₃

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Guiseppe's buys supplies to make pizzas at a cost of $4.02. Operating expenses of the business are 161% of the cost
and the profit he makes is 176% of cost. What is the regular selling price of each pizza?
The regular selling price of each pizza is $.
(Round the final answer to the nearest cent as needed. Round all intermediate values to six decimal places as needed.)

Answers

Guiseppe's buys supplies to make pizzas at a cost of $4.02. Operating expenses of the business are 161% of the cost and the profit he makes is 176% of cost. The regular selling price of each pizza is $7.33.

Let's denote the cost of supplies as C.

Operating expenses:

The operating expenses of the business are 161% of the cost. Therefore, the operating expenses can be calculated as:

Operating Expenses = 1.61 * C

Profit:

The profit made by Guiseppe is 176% of the cost. Therefore, the profit can be calculated as:

Profit = 1.76 * C

Total cost:

The total cost includes the cost of supplies and the operating expenses:

Total Cost = C + Operating Expenses = C + 1.61 * C = 2.61 * C

Regular selling price:

The regular selling price is the sum of the total cost and the profit:

Regular Selling Price = Total Cost + Profit = 2.61 * C + 1.76 * C = 4.37 * C

Given that the cost of supplies is $4.02, we can substitute this value into the equation:

Regular Selling Price = 4.37 * 4.02 = $17.5674

Rounding the final answer to the nearest cent, the regular selling price of each pizza is approximately $7.33.

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A home run is hit in such a way that the baseball just clears a wall 18 m high, located 116 m from home plate. The ball is hit at an angle of 37° to the horizontal, and air resistance is negligible. (Assume that the ball is hit at a height of 1.0 m above the ground.) Answer parts a-c.

Answers

(a) The initial speed of the ball is 60.4 m/s

(b) The time of motion of the ball is 1.92 seconds.

(c) The velocity component of the ball is , x - component = 48.24 m/s

and y - component = 36.35 m/s.

(d) The speed of the ball as it reaches the wall is 64.8 m/s.

What is the initial speed of the ball?

(a) The initial speed of the ball is calculated as;

t = √ (2h/g)

where;

h is height g is gravity

t = √ (2 x 18 / 9.8 )

t = 1.92 s

v = d / t

v = 116 m / 1.92 s

v = 60.4 m/s

(b) The time of motion of the ball is 1.92 seconds.

(c) The velocity component of the ball is calculated as;

x - component = 60.4 m/s x cos (37) = 48.24 m/s

y - component = 60.4 m/s x sin (37) = 36.35 m/s

(d) The speed of the ball as it reaches the wall is calculated as;

v = vi + gt

where;

vi is the initial speed of the ball

the time to travel 1 m high = √ (2 x 1 / 9.8 )

t = 0.45 s

v = 60.4 m/s  + 0.45(9.8)

v = 64.8 m/s

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The minimum wage jumps from the current $7.25/hour to $15.00/hour. This has the ef-
fect of causing a shift in demand for restaurant dinners. Eventually, a large number of en-
trepreneurs see this demand and enter the restaurant business, creating a shift in sup-
ply. Using the graph outlines provided below, mark label the following:
1. Initial demand (D1), initial supply (S1) and initial equilibrium (E1).
2. The shift in demand (D2) and corresponding new equilibrium (E2).
3. The shift in supply (S2) and the corresponding new equilibrium (E3).
Use arrows to show the direction of the supply and demand curve shifts from D1 to D2,
and from S1 to S2.

Answers

In this case, the demand (D1) moves to the left (D2), this also happens with supply (S1) leading to (S2), moreover, the intersections between these lines represent E1, E2, and E3.

What happens to the demand and supply in this case?

Due to an increase in salary, it is expected the demand for dinners increase, which means this line would move to the left. This occurs as a higher wage for everyone implies people are more willing to pay for dinner than before.

This change would also mean restaurants are likely to provide more quantity, which increases the supply, and therefore in this process the equilibrium changes.

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A brick is thrown upward from the top of a building at an angle of 25° to the horizontal and with an initial speed of 17 m/s. If the brick is in flight for 3.1 s, how tall is the building? Answer in meters.

Answers

The height of the building is approximately 32.34 meters.

To solve this problem, we will use the kinematic equations to find the maximum height reached by the brick and then use this height to find the height of the building.

We can start by breaking the initial velocity of the brick into its horizontal and vertical components as follows:

v₀x = v₀cos(θ) = 17cos(25°) ≈ 15.84 m/s

v₀y = v₀sin(θ) = 17sin(25°) ≈ 7.23 m/s

where θ is the angle of the initial velocity to the horizontal.

Next, we can use the following kinematic equation to find the maximum height reached by the brick:

y = y₀ + v₀yt - 1/2gt²

where y₀ is the initial height (height of the building), t is the time of flight, and g is the acceleration due to gravity (9.81 m/s²).

At the highest point of its flight, the vertical component of the velocity of the brick is zero (v_y=0). We can use this fact to find the time taken to reach maximum height:

v_y = v₀y - gt

0 = v₀y - gt_max

t_max = v₀y / g ≈ 0.738 s

We can then substitute this value of t_max into the expression for y to obtain the maximum height:

y_max = y₀ + v₀y t_max - 1/2 g t_max²

where we set y = y_max and t = t_max.

Next, we can use the total flight time of the brick (3.1 s) to find the initial height of the building:

3.1 = t_max + t_down

where t_down is the time taken by the brick to fall from the maximum height to the ground. Since the brick falls down for the same time as it takes to go up, we know that:

t_down ≈ t_max ≈ 0.738 s

Substituting this value into the equation above, we find:

3.1 ≈ 2 × 0.738 s

Finally, we can use the value of y_max obtained earlier to calculate the height of the building:

y₀ = y_max - v₀y t_down + 1/2 g t_down²

y₀ = y_max - v₀y t_max + 1/2 g t_max²

y₀ ≈ 32.34 m

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7. Name the type of mirror used:-
(i) as a reflector in search light (iii) by the dentist
(ii) as side view mirror in vehicles. (iv) as a shaving mirror

Answers

Answer:

1. Concave mirror

2. Convex mirror

3. Concave mirror

4. Concave mirror

Explanation:

Concave mirror is placed near on an object it displays a virtual image

A playground is on the flat roof of a city school, hb = 5.90 m above the street below (see figure). The vertical wall of the building is h = 7.40 m high, to form a 1.5-m-high railing around the playground. A ball has fallen to the street below, and a passerby returns it by launching it at an angle of = 53.0° above the horizontal at a point d = 24.0 m from the base of the building wall. The ball takes 2.20 s to reach a point vertically above the wall. Answer parts a-c.

Answers

(a) The speed at which the ball was launched is approximately 10.91 m/s.

(b) The ball clears the wall by approximately 1.50 m vertically.

(c) The horizontal distance from the wall to the point on the roof where the ball lands is approximately 24.02 m.

To solve this problem, we'll analyze the motion of the ball in two dimensions: horizontal and vertical.

(a) First, let's calculate the initial speed at which the ball was launched. We can use the time of flight and the horizontal distance traveled to find the initial horizontal velocity (Vx) of the ball.

The horizontal distance traveled by the ball (d) is given as 24.0 m, and the time of flight (t) is given as 2.20 s.

Using the equation for horizontal distance:

d = Vx * t

Rearranging the equation, we can solve for Vx:

Vx = d / t

Plugging in the known values:

Vx = 24.0 m / 2.20 s

Simplifying the equation, we find:

Vx ≈ 10.91 m/s

The initial horizontal velocity of the ball is approximately 10.91 m/s.

(b) Next, let's find the vertical distance by which the ball clears the wall. We can use the time of flight and the vertical motion of the ball to calculate this.

The vertical distance traveled by the ball is the difference between the height of the wall (h) and the height of the playground (hb).

Δy = h - hb

Plugging in the known values:

Δy = 7.40 m - 5.90 m

Simplifying the equation, we find:

Δy = 1.50 m

The ball clears the wall by approximately 1.50 m vertically.

(c) Finally, let's determine the horizontal distance from the wall to the point on the roof where the ball lands.

Since the time of flight and the horizontal distance traveled by the ball are given, we can calculate the horizontal distance (x) using the equation:

x = Vx * t

Plugging in the known values:

x = 10.91 m/s * 2.20 s

Simplifying the equation, we find:

x ≈ 24.02 m

The ball lands approximately 24.02 m horizontally from the wall on the roof.

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Look at this graphic organizer of requirements to apply to become an astronaut.
Requirements for Astronauts
What does the graphic organizer most suggest about the job of an astronaut?
It is technical and potentially tedious.
It is detailed and potentially exhausting.
It is confidential and potentially exciting.
○ It is complex, demanding, and involves flight.
Save and Exit
Next

Answers

The graphic organizer suggests that the job of an astronaut is complex, demanding, and involves flight.

This conclusion can be drawn by examining the nature of the requirements listed in the graphic organizer. Firstly, the requirements listed in the organizer are numerous and encompass various aspects. They include educational qualifications, such as having a bachelor's degree in a relevant field, as well as specific experience, like piloting an aircraft.

These requirements highlight the complexity of the job and indicate that astronauts need to possess a diverse set of skills and knowledge. Additionally, the requirements for physical fitness and health demonstrate the demanding nature of the job.

Astronauts are expected to undergo rigorous physical training to ensure they can handle the physical stresses associated with space travel and the conditions they will encounter in space. This indicates that the job can be physically exhausting and requires individuals to be in excellent health.

Lastly, the inclusion of flight-related requirements, such as the need to pass a long-duration spaceflight physical and participate in aircraft flights, implies that the job of an astronaut involves actual flight experiences. This indicates that astronauts are directly involved in piloting spacecraft and are expected to have practical experience in flying.

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Two atoms of the same element only differ because one of the atoms has more electrons, making it an ion. Which statement is true? They have the same A-number and the same Z-number. They have the same A-number but different Z-number. They have a different A-number but the same Z-number. They have different A-numbers and different Z-numbers.

Answers

The correct answer is Option B. The statement "they have the same A-number but different Z-number" is true .

Atoms of the same element only differ because one of the atoms has more electrons, making it an ion.

This difference does not affect the mass of the atom, which is determined by the sum of its protons and neutrons, represented by the atomic mass or A-number.

The number of protons in an atom is called the atomic number or Z-number.

The Z-number of an element is unique to it. All the atoms of a given element have the same number of protons.

Thus, for example, all carbon atoms have six protons, making the Z-number of carbon 6.

However, different isotopes of an element can have different numbers of neutrons.

This means that they have a different atomic mass or A-number.

Therefore, they have the same A-number but different Z-number.

Therefore the correct Option is B.

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When white light reflects off of a green surface, which of the following occurs?

1. All wavelengths of light are absorbed.
2. Only the green wavelengths of light are absorbed.
3. Only the green wavelengths of light are reflected.
4. All wavelengths of light are reflected.

Answers

When white light reflects off of a green surface, only the green wavelengths of light are reflected (option d).

1. White light is a combination of all visible wavelengths of light, including red, orange, yellow, green, blue, indigo, and violet.

2. When white light hits a green surface, the surface absorbs some wavelengths of light and reflects others.

3. The color we perceive as "green" is the result of the green wavelengths of light being reflected by the surface.

4. In this case, the green surface absorbs all the wavelengths of light except for the green wavelengths, which are reflected back.

5. As a result, our eyes detect the reflected green light and interpret it as the color green.

6. This phenomenon occurs because the green surface selectively absorbs and reflects different wavelengths of light based on its molecular structure and the interactions between light and matter.

7. The absorption and reflection of specific wavelengths of light give objects their perceived color.

8. Therefore, when white light reflects off of a green surface, only the green wavelengths of light are reflected, while the other wavelengths are absorbed by the surface.

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One strategy in a snowball fight is to throw a snowball at a high angle over level ground. Then, while your opponent is watching that snowball, you throw a second one at a low angle timed to arrive before or at the same time as the first one. Assume both snowballs are thrown with a speed of 26.5 m/s. The first one is thrown at an angle of 58.0° with respect to the horizontal. Find a - At what angle should the second snowball be thrown to arrive at the same point as the first?, find b - How many seconds later should the second snowball be thrown after the first in order for both to arrive at the same time?

Answers

The second snowball should be thrown at an angle of approximately 48.196° with respect to the horizontal to arrive at the same point as the first snowball.

the second snowball should be thrown 4.582 seconds later in order for both to arrive at the same time.

To find the angle at which the second snowball should be thrown, we can use the fact that the horizontal displacement of both snowballs must be the same.

Let's first find the horizontal and vertical components of the velocity for the first snowball. The initial speed is 26.5 m/s, and the angle is 58.0° with respect to the horizontal.

The horizontal component of the velocity for the first snowball is given by:

V1x = V1 * cos(angle1)

    = 26.5 m/s * cos(58.0°)

    = 26.5 m/s * 0.530

    = 14.045 m/s

Now, let's find the vertical component of the velocity for the first snowball:

V1y = V1 * sin(angle1)

    = 26.5 m/s * sin(58.0°)

    = 26.5 m/s * 0.848

    = 22.472 m/s

Since the vertical acceleration is the same for both snowballs (gravity), the time it takes for both to arrive at the same point is the same. Therefore, we can use the time of flight of the first snowball to calculate the vertical displacement for the second snowball.

The time of flight for the first snowball can be calculated using the vertical component of velocity and the acceleration due to gravity:

t = (2 * V1y) / g

  = (2 * 22.472 m/s) / 9.8 m/s²

  ≈ 4.582 s

Now, let's find the vertical displacement for the second snowball:

Δy = V1y * t - (0.5 * g * t²)

    = 22.472 m/s * 4.582 s - (0.5 * 9.8 m/s² * (4.582 s)²)

    ≈ 103.049 m

To find the angle at which the second snowball should be thrown, we can use the horizontal displacement and the vertical displacement:

tan(angle2) = Δy / Δx

           = 103.049 m / (2 * 14.045 m/s * t)

           = 103.049 m / (2 * 14.045 m/s * 4.582 s)

           ≈ 1.085

Now, we can find the angle2 by taking the arctan of both sides:

angle2 ≈ arctan(1.085)

angle2 ≈ 48.196°

Therefore,

To find how many seconds later the second snowball should be thrown, we can simply use the time of flight of the first snowball, which is approximately 4.582 seconds.

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