A photographer uses his camera, whose lens has a 50 mm focal length, to focus on an object 2.5 m away. He then wants to take a picture of an object that is 50 cm away.

Part A
In which direction must the lens move to focus on this second object?

Part B
How far must the lens move to focus on this second object?

Answers

Answer 1

Answer:

0.004 m away from the film

Explanation:

u = Object distance

v = Image distance

f = Focal length = 50 mm

[tex]\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=\frac{1}{0.05}-\frac{1}{2.5}\\\Rightarrow \frac{1}{v}=\frac{98}{5} \\\Rightarrow v=\frac{5}{98}=0.051\ m[/tex]

The image distance is 0.051 m

When u = 50 cm

[tex]\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=\frac{1}{0.05}-\frac{1}{0.5}\\\Rightarrow \frac{1}{v}=18\\\Rightarrow v=\frac{1}{18}=0.055\ m[/tex]

The image distance is 0.055 m

The lens has moved 0.055-0.051 = 0.004 m away from the film

Answer 2
Final answer:

To focus on the closer object, the lens must move away from the object. By applying the lens formula, we calculate that the lens must move exactly 7mm away from the initial position.

Explanation:

This problem comes down to the concept of the lens formula in optical physics. The lens formula is 1/f = 1/v - 1/u where f is the focal length, v is the image distance, and u is the object distance.

Part A: To focus on the second object which is closer, the lens must move away from the object.

Part B: To calculate how far the lens must move, we need to apply the lens formula. Let's calculate the image distances for the two object distances.

For an object at 2.5 m, 1/v = 1/f + 1/u = 1/0.05 + 1/2.5 = 20 + 0.4 = 20.4. So, v = 0.049 m or 49 mm.

For an object at 0.5 m, 1/v = 1/f + 1/u = 1/0.05 - 1/0.5 = 20 - 2 = 18. So, v = 0.056 m or 56 mm.

Hence, the lens must move 56mm - 49mm = 7mm away from the initial position.

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Related Questions

An 80-kg quarterback jumps straight up in the air right before throwing a 0.43-kg football horizontally at 15 m/s . Suppose that the quarterback takes 0.30 s to return to the ground after throwing the ball. How far d will he move horizontally, assuming his speed is constant?

Answers

Answer:

0.0241875 m

Explanation:

[tex]m_1[/tex] = Mass of quarterback = 80 kg

[tex]m_2[/tex] = Mass of football = 0.43 kg

[tex]v_1[/tex] = Velocity of quarterback

[tex]v_2[/tex] = Velocity of football = 15 m/s

Time taken = 0.3 seconds

In this system as the linear momentum is conserved

[tex]m_1v_1+m_2v_2=0\\\Rightarrow v_1=-\frac{m_2v_2}{m_1}\\\Rightarrow v_1=-\frac{0.43\times 15}{80}\\\Rightarrow v_1=0.080625\ m/s[/tex]

Assuming this velocity is constant

[tex]Distance=Velocity\times Time\\\Rightarrow Distance=0.080625\times 0.3\\\Rightarrow Distance=0.0241875\ m[/tex]

The distance the quarterback will move in the horizontal direction is 0.0241875 m

Final answer:

The question is a physics problem regarding kinematics and requires calculating the horizontal distance an 80-kg quarterback moves while in the air after throwing a football. However, since the quarterback jumps vertically with zero initial horizontal speed and no horizontal force is acting after the throw, the resulting horizontal distance is zero.

Explanation:

The student is asking a question related to kinematics and the conservation of momentum in physics. Specifically, it concerns an 80-kg quarterback who jumps vertically and then throws a football horizontally. The core concept here is determining how far the quarterback will move horizontally during the time he is in the air, assuming a constant horizontal speed.

To calculate the distance (d) the quarterback moves horizontally, we can apply the equation for constant velocity motion: d = v * t. However, the information given does not directly provide the quarterback's horizontal speed after the throw; thus, we must assume that the horizontal speed of the quarterback is zero as the ball is thrown horizontally and no horizontal force on the quarterback has been mentioned.

Therefore, under the assumption that the quarterback's horizontal speed remains zero, the distance he moves horizontally d would also be zero, since he moves vertically up and down with no horizontal velocity component. If the problem had given a horizontal speed for the quarterback post-throw, we would use that given speed to calculate the distance using the equation.

Suppose that a simple pendulum consists of a small 81 g bob at the end of a cord of negligible mass. If the angle θ between the cord and the vertical is given by θ = (0.068 rad) cos[(9.2 rad/s) t + φ], what are (a) the pendulum's length and (b) its maximum kinetic energy?

Answers

Answer:

(a) 0.115 m

(b) 2.08 x 10^-5 J

Explanation:

mass of bob, m = 81 g = 0.081 kg

The equation of oscillation is given by

θ = 0.068 Cos {9.2 t + Ф}

Now by comparison

The angular velocity

ω = 9.2 rad/s

(a) [tex]\omega^{2} =\frac{g}{L}[/tex]

where, L be the length of the pendulum

[tex]L =\frac{g}{\omega ^{2}}[/tex]

[tex]L =\frac{9.8}{9.2 \times 9.2}[/tex]

L = 0.115 m

(b) A = L Sinθ

A = 0.115 x Sin 0.068

A = 7.8 x 10^-3 m

Maximum kinetic energy

K = 0.5 x mω²A²

K = 0.5 x 0.081 x 9.2 x 9.2 x 7.8 x 7.8 x 10^-6

K = 2.08 x 10^-5 J

An 15-cm-long bicycle crank arm, with a pedal at one end, is attached to a 21-cm-diameter sprocket, the toothed disk around which the chain moves. A cyclist riding this bike increases her pedaling rate from 62 rpm to 95 rpm in 12 s . A) What is the tangential acceleration of the pedal?

B)What length of chain passes over the top of the sprocket during this interval?

Answers

Final answer:

The tangential acceleration of the pedal is 0.259 m/s^2. The length of chain passing over the top of the sprocket depends on the number of teeth on the sprocket, which is not provided in the question.

Explanation:

A) The tangential acceleration of the pedal can be determined using the equation:

at = r × α

where at is the tangential acceleration, r is the radius of the pedal (half the length of the crank arm), and α is the angular acceleration. To find the angular acceleration, we can use the equation:

α = Δω / Δt

where Δω is the change in angular velocity and Δt is the change in time. Plugging in the given values:

α = (95 rpm - 62 rpm) * 2π / 60 s = (33 * 2π) / 60 s ≈ 3.459 rad/s2

Substituting this value and the radius into the first equation, we get:

at = 0.075 m * 3.459 rad/s2 = 0.259 m/s2

Therefore, the tangential acceleration of the pedal is 0.259 m/s2.

B) The length of chain passing over the top of the sprocket during the given time interval can be calculated using the formula:

d = 2π * r * (ωf - ωi) * t / N

where d is the length of chain, r is the radius of the sprocket, ωi and ωf are the initial and final angular velocities respectively, t is the time interval, and N is the number of teeth on the sprocket. Plugging in the given values:

d = 2π * (21 cm / 2) * [(95 rpm - 62 rpm) * 2π / 60 s] * 12 s / N

The length of chain depends on the number of teeth on the sprocket. Since the number of teeth is not provided in the question, we cannot calculate the exact length of chain. However, we have all the necessary equation components to calculate it once the number of teeth is known.

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A motorcycle and a police car are moving toward one another. The police car emits sound with a frequency of 523 Hz and has a speed of 32.2 m > s. The motorcycle has a speed of 14.8 m > s. What frequency does the motorcyclist hear?

Answers

To solve this problem it is necessary to apply the concepts related to Dopler's Law. Dopler describes the change in frequency of a wave in relation to that of an observer who is in motion relative to the Source of the Wave.

It can be described as

[tex]f = \frac{c\pm v_r}{c\pm v_s}f_0[/tex]

c = Propagation speed of waves in the medium

[tex]v_r[/tex]= Speed of the receiver relative to the medium

[tex]v_s[/tex]= Speed of the source relative to the medium

[tex]f_0 =[/tex]Frequency emited by the source

The sign depends on whether the receiver or the source approach or move away from each other.

Our values are given by,

[tex]v_s = 32.2m/s \rightarrow[/tex] Velocity of car

[tex]v_r = 14.8 m/s \rightarrow[/tex] velocity of motor

[tex]c = 343m/s \rightarrow[/tex] Velocity of sound

[tex]f_0 = 523Hz \rightarrow[/tex]Frequency emited by the source

Replacing we have that

[tex]f = \frac{c + v_r}{c - v_s}f_0[/tex]

[tex]f = \frac{343 + 14.8}{343 - 32}(523)[/tex]

[tex]f = 601.7Hz[/tex]

Therefore the frequency that hear the motorcyclist is 601.7Hz

Final answer:

To determine the frequency heard by the motorcyclist, we apply the Doppler shift formula using the given speeds of the motorcycle and police car and the emitted frequency of the police car's siren. The calculated frequency is the observed frequency by the motorcyclist due to the relative motion of the two vehicles. The final answer is about 602.1 Hz

Explanation:

The subject of the question is the Doppler Effect, which is related to the change in frequency of sound waves due to the relative motion of the source and the observer. To calculate the frequency the motorcyclist hears, we use the following Doppler shift formula for sound:

f' = f(v + vo) / (v - vs)

Where:

f' is the observed frequency by the motorcyclist,

f is the emitted frequency by the police car (523 Hz),

v is the speed of sound in air, which can be assumed to be approximately 343 m/s at room temperature,

vo is the speed of the observer (motorcyclist) towards the source (14.8 m/s),

vs is the speed of the source (police car) towards the observer (32.2 m/s).

Plugging in the given values, the equation becomes:

f' = 523 Hz (343 m/s + 14.8 m/s) / (343 m/s - 32.2 m/s) = 602.089 Hz

After performing the calculations, the frequency heard by the motorcyclist can be determined. This is an application of the Doppler effect as studied in high school physics.

Determine the ratio of the flow rate through capillary tubes A and B (that is, Qa/Qb).

The length of A is twice that of B, and the radius of A is one-half that of B.

The pressure across both tubes is the same.

Answers

To solve this problem we can use the concepts related to the change of flow of a fluid within a tube, which is without a rubuleous movement and therefore has a laminar fluid.

It is sometimes called Poiseuille’s law for laminar flow, or simply Poiseuille’s law.

The mathematical equation that expresses this concept is

[tex]\dot{Q} = \frac{\pi r^4 (P_2-P_1)}{8\eta L}[/tex]

Where

P = Pressure at each point

r = Radius

[tex]\eta =[/tex] Viscosity

l = Length

Of all these variables we have so much that the change in pressure and viscosity remains constant so the ratio between the two flows would be

[tex]\frac{\dot{Q_A}}{\dot{Q_B}} = \frac{r_A^4}{r_B^4}\frac{L_B}{L_A}[/tex]

From the problem two terms are given

[tex]R_A = \frac{R_B}{2}[/tex]

[tex]L_A = 2L_B[/tex]

Replacing we have to

[tex]\frac{\dot{Q_A}}{\dot{Q_B}} = \frac{r_A^4}{r_B^4}\frac{L_B}{L_A}[/tex]

[tex]\frac{\dot{Q_A}}{\dot{Q_B}} = \frac{r_B^4}{16*r_B^4}\frac{L_B}{2*L_B}[/tex]

[tex]\frac{\dot{Q_A}}{\dot{Q_B}} = \frac{1}{32}[/tex]

Therefore the ratio of the flow rate through capillary tubes A and B is 1/32

If the pendulum is taken into the orbiting space station what will happen to the bob?
a. It will continue to oscillate in a vertical plane with the same period.
b. It will no longer oscillate because there is no gravity in space.
c. It will no longer oscillate because both the pendulum and the point to which it is attached are in free fall.
d. It will oscillate much faster with a period that approaches zero.

Answers

Answer:

Option C

Explanation:

The bob of the pendulum, when the pendulum is taken into the orbiting space station will be in free fall as there is no gravity in space and also the point to which it is attached.

The movement of the bob of the pendulum about its mean position is because of its weight due to which it oscillates and in the free fall, no force acts on it due to its weight since the body is not under the attraction force due to gravity thus it experiences weightlessness and does not oscillate.

Final answer:

On an orbiting space station with microgravity, a pendulum will not oscillate because the pendulum and its point of attachment are in free fall, absent of the normal gravitational forces.

Explanation:

If a pendulum is taken into an orbiting space station, its behavior will be altered due to the absence of an external gravitational force. This situation is best explained by option c. The pendulum will no longer oscillate because the pendulum and the point to which it is attached are in free fall, also known as microgravity. In an environment of microgravity, the forces that would ordinarily cause the pendulum bob to swing are absent, thus the pendulum will essentially float instead of oscillating back and forth.

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A 50-N crate is pulled up a 5-m inclined plane by a worker at constant velocity. If the plane is inclined at an angle of 37° to the horizontal and there exists a constant frictional force of 10 N between the crate and the surface, what is the force applied by the worker?

Answers

Answer:F=40.09 N

Explanation:

Given

weight of crate [tex]W=50 N[/tex]

Inclination [tex]\theta =37^{\circ}[/tex]

Frictional Force [tex]f=10 N[/tex]

as the crate is moving with constant velocity therefore net Force on crate is zero

[tex]F-50\sin (37)-f=0[/tex]

[tex]F=50\sin (37)+10[/tex]

[tex]F=30.09+10[/tex]

[tex]F=40.09 N[/tex]

A rigid cube (each side is 0.10 m) is filled with water and frozen solid. When water freezes its volume expands about 9%. How much pressure is exerted on the sides of the cube? Hint: Imagine trying to squeeze the block of ice back into the original cube.

Answers

Answer:

P = 1.89 10⁸ N / m²

Explanation:

To solve this problem we can use the definition of bulk modules

       B = - P / (ΔV/V)

The negative sign is entered for the volume module to be positive, P is the pressure and ΔV/V is the volume change fraction

In this case the volume change is 9% this is

      ΔV / V 100 = 9%

      ΔV / V = ​​0.09

      P = B ΔV / V

The bulk modulus value is that of water since it is in a liquid state and then freezes

       B = 0.21 101¹⁰ N / m²

let's calculate

       P = 0.21 10¹⁰ 0.09

      P = 1.89 10⁸ N / m²

An applied force of 122 N is exerted horizontally on an 28 kg box of dogs in Yulin. The coefficient of kinetic friction between the floor and the box is 0.35. If the box is initially at rest, what is the final kinetic energy after it has been moved 9.46 m across the floor?

Answers

Answer:

The final kinetic energy is 242 J.

Explanation:

Hi there!

According to the work-energy theorem, the total work done on the box is equal to its change in kinetic energy:

W = ΔKE

Where:

W = work done on the box.

ΔKE = change in kinetic energy (final KE - initial KE).

The only forces that do work in this case are the applied force and the friction force because the box moves only horizontally.

The equation of work is the following:

W = F · s

Where:

F = force.

s = traveled distance.

Then, the work done by the applied force is:

W = 122 N · 9.46 m = 1.15 × 10³ J

To calculate the work done by friction, we have to find the friction force:

Fr = N · μ

Where:

Fr = friction force.

N = normal force.

μ = coefficient of kinetic friction.

The box does not have a net vertical acceleration. It means that the sum of the vertical forces acting on the box is zero:

∑Fy = 0

In this case, the only vertical forces are the weight of the box and the normal force. Then:

Weight + N = 0

N = - Weight

The weight of the box is calculated as follows:

Weight = m · g

Where:

m = mass of the box.

g = acceleration due to gravity.

Then:

-Weight = N = 28 kg · 9.8 m/s² = 274.4 N

Now, we can calculate the friction force:

Fr = N · μ

Fr = 274.4 N · 0.35 = 96 N

The work done by the friction force will be:

W = Fr · s

W = 96 N · 9.46 m = 908 J

Since the work done by friction opposes to the sense of movement, the work is negative.

Now, we can calculate the total work done on the box:

W total = W applied forece + W friction force

W = 1.15 × 10³ J - 908 J = 242 J

Applying the work-energy theorem:

W = final KE - initial KE

Since the box is initially at rest the initial kinetic energy is zero. Then:

W = final KE - 0

W = final KE

Final KE = 242 J

The final kinetic energy is 242 J.

A physics class conducting a research project on projectile motion constructs a device that can launch a cricket ball. The launching device is designed so that the ball can be launched at ground level with an initial velocity of 28 m/ s at an angle of 30° to the horizontal. Calculate the horizontal component of the velocity of the ball:a. initiallyb. after 1.0 sc. after 2.0 s

Answers

Answer:

Explanation:

initial velocity, u = 28 m/s

Angle of projection, θ = 30°

The acceleration in horizontal direction is zero, so the horizontal component of velocity is constant.

Horizontal component of velocity, u cos θ = 28 x Cos 30 = 24.25 m/s

At t = 2 sec, the horizontal component of velocity = 24.25 m/s

At t = 3 sec, the horizontal component of velocity = 24.25 m/s

In this exercise we have to use the knowledge about oblique launch to calculate the components of velocity for each case, so we have that:

For all times we will find a velocity equal to 24.25 m/s

organizing the information given in the statement we have that:

initial velocity, u = 28 m/sAngle of projection, θ = 30°The acceleration in horizontal direction is zerovelocity is constant.

Knowing that the component can be written as:

[tex]u cos \theta = 28 * Cos 30 = 24.25 m/s[/tex]

So as the formula does not depend on time we have that for any value it will have a constant velocity.

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Moe is carrying and supporting a large crate by exerting 1000 N. He is walking at constant velocity across a level, horizontal floor. If he covers a distance of 10 meters, how much work does he do on the crate? Use g = 10m/s2 if needed. Ignore friction between Moe and the crate.

Answers

Answer:

he did not work on the create

Explanation:

Honestly I have no idea it makes a lot of sense to use the work equation so that was my first answer but for some reason on UT Quest it said that was the wrong and moe not doing any work on the create was the right answer

Final answer:

The work done by Moe on the crate is 10,000 J.

Explanation:

The work done by Moe on the crate can be calculated using the formula:

Work = Force x Distance x Cosine(theta)

Since Moe is exerting a force of 1000 N and the distance covered is 10 meters, we can plug these values into the formula:

Work = 1000 N x 10 m x Cosine(0°)

The angle theta is 0° because Moe is exerting a force in the same direction as the displacement.

Cosine(0°) is equal to 1, so the work done by Moe on the crate is:

Work = 1000 N x 10 m x 1 = 10,000 J

Problem 12: A 20.0-m tall hollow aluminum flagpole is equivalent in strength to a solid cylinder 4.00 cm in diameter. A strong wind bends the pole much as a horizontal force of 1100 N exerted at the top would. How to the side does the top of the pole flex?

Answers

Answer:

The deformation in the pole due to force is 0.70 mm.

Explanation:

Given that,

Height = 20.0 m

Diameter = 4.00 cm

Force = 1100 N

We need to calculate the  area

Using formula of area

[tex]A=\pi\times r^2[/tex]

[tex]A=\pi\times(2.00\times10^{-2})^2[/tex]

[tex]A=0.00125\ m^2[/tex]

[tex]A=1.25\times10^{-3}\ m^2[/tex]

We need to calculate the deformation

Using formula of deformation

[tex]\Delta x=\dfrac{1}{s}(\dfrac{F}{A}\times L)[/tex]

Where, s = shear modulus

F = force

l = length

A = area

Put the value into the formula

[tex]\Delta x=\dfrac{1}{2.5\times10^{10}}\times(\dfrac{1100}{1.25\times10^{-3}}\times 20.0)[/tex]

[tex]\Delta x=0.000704\ m[/tex]

[tex]\Delta x=7.04\times10^{-4}\ m[/tex]

[tex]\DElta x=0.70\ mm[/tex]

Hence, The deformation in the pole due to force is 0.70 mm.

Light is incident along the normal to face AB of a glass prism of refractive index 1.54. Find αmax, the largest value the angle α can have without any light being refracted out of the prism at face AC if the prism is immersed in water (with index of refraction 1.33). Express your answer in degrees. Ignore any reflections from the surface BC.

Answers

To solve this problem it is necessary to use the concepts related to Snell's law.

Snell's law establishes that reflection is subject to

[tex]n_1sin\theta_1 = n_2sin\theta_2[/tex]

Where,

[tex]\theta =[/tex] Angle between the normal surface at the point of contact

n = Indices of refraction for corresponding media

The total internal reflection would then be given by

[tex]n_1 sin\theta_1 = n_2sin\theta_2[/tex]

[tex](1.54) sin\theta_1 = (1.33)sin(90)[/tex]

[tex]sin\theta_1 = \frac{1.33}{1.54}[/tex]

[tex]\theta = sin^{-1}(\frac{1.33}{1.54})[/tex]

[tex]\theta = 59.72\°[/tex]

Therefore the [tex]\alpha_{max}[/tex] would be equal to

[tex]\alpha = 90\°-\theta[/tex]

[tex]\alpha = 90-59.72[/tex]

[tex]\alpha = 30.27\°[/tex]

Therefore the largest value of the angle α is 30.27°

The largest value the angle α can have without any light being refracted out of the prism at face AC is 30.27°.

How to calculate the angle?

From the information given, the total internal reflection would be:

1.54sinb = (1.33) sin90°

sin b = (1.33 sin90° / 1.54)

b = 59.72°

Therefore, the value of the angle will be:

= 90° - 59.72°

= 30.27°

In conclusion, the correct option is 30.27°.

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A rectangular coil of dimensions 5.40cm x 8.50cm consists of25 turns of wire. The coil carries a current of 15.0 mA.
a) Calculate the magnitude of its magnetic moment
b) Suppose a uniform magnetic field of magnitude of 0.350 T isapplied parallel to the plane of the loop. What is the magnitude ofthe torque acting on the loop?

Answers

Answer:

(a) Magnetic moment will be [tex]17.212\times 10^{-4}A-m^2[/tex]

(b) Torque will be [tex]6.024\times 10^{-4}N-m[/tex]

Explanation:

We have given dimension of the rectangular 5.4 cm × 8.5 cm

So area of the rectangular coil [tex]A=5.4\times 8.5=45.9cm^2=45.9\times 10^{-4}m^2[/tex]

Current is given as [tex]i=15mA=15\times 10^{-3}A[/tex]

Number of turns N = 25

(A) We know that magnetic moment is given by [tex]magnetic\ moment=NiA=25\times 45.9\times 10^{-4}\times 15\times 10^{-3}=17.212\times 10^{-4}A-m^2[/tex]

(b) Magnetic field is given as B = 0.350 T

We know that torque is given by [tex]\tau =BINA=0.350\times 15\times 10^{-3}\times 25\times 45.9\times 10^{-4}=6.024\times 10^{-4}N-m[/tex]

Final answer:

a) The magnitude of the magnetic moment is 0.0172 A·m². b) The magnitude of the torque on the loop is 0 N·m.

Explanation:

a) To calculate the magnitude of the magnetic moment, we can use the formula μ = nIA, where μ is the magnetic moment, n is the number of turns, I is the current, and A is the area of the coil. In this case, n = 25, I = 15.0 mA = 0.015 A, and A = (5.40 cm)(8.50 cm) = 45.9 cm² = 0.00459 m². Plugging these values into the formula, we get μ = (25)(0.015 A)(0.00459 m²) = 0.0172 A·m².

b) To calculate the magnitude of the torque, we can use the formula τ = μBsinθ, where τ is the torque, μ is the magnetic moment, B is the magnetic field strength, and θ is the angle between μ and B. In this case, μ = 0.0172 A·m² (calculated in part a),

B = 0.350 T, and since the magnetic field is applied parallel to the plane of the loop, θ = 0°.

Plugging these values into the formula, we get τ = (0.0172 A·m²)(0.350 T)(sin 0°)

= 0 N·m.

A horizontal vinyl record of mass 0.105 kg and radius 0.0757 m rotates freely about a vertical axis through its center with an angular speed of 5.80 rad/s and a rotational inertia of 5.18 x 10-4 kg·m2. Putty of mass 0.0213 kg drops vertically onto the record from above and sticks to the edge of the record. What is the angular speed of the record immediately afterwards?

Answers

Answer:

4.6939 rad/s

Explanation:

You have to use the conservation of angular momentum for both objects as an object is spinning and a collision happens. To properly solve, you have to know that the putty and the vinyl have the same angular speed after the putty lands on it and that the putty acts as a point mass so the formula of it's rotational inertia is mr^2.

What impulse occurs when an average force of 7.0 N is exerted on a cart for 2.5 s ?

Express your answer to two significant figures and include the appropriate units.

Answers

Answer:[tex]1.7\times 10[/tex]

Explanation:

Given

Force [tex]F=7 N[/tex]

time interval [tex]t=2.5 s[/tex]

Impulse is given by [tex]=Force \times time\ interval\ for\ applied\ force [/tex]

[tex]Impulse=7\times 2.5=17.5[/tex]

For two significant Figure

[tex]Impulse=1.7\times 10[/tex]

                       

A 5 kg ball takes 13.3 seconds for one revolution around the circle. What's the magnitude of the angular velocity of this motion?

Answers

Answer: 0.47 rad/sec

Explanation:

By definition, the angular velocity is the rate of change of the angle traveled with time, so we can state the following:

ω = ∆θ/ ∆t

Now, we are told that in 13.3 sec, the ball completes one revolution around the circle, which means that, by definition of angle, it has rotated 2 π rad (an arc of 2πr over the radius r), so we can find ω as follows:

ω = 2 π / 13.3 rad/sec = 0.47 rad/sec

Answer:

0.472rad/s

Explanation:

Angular velocity = 2πf where f = frequency and frequency is the number of revolution per second and 2π represent a cycle of revolution. The mass of the body was 5kg, the time taken to complete a cycle was 13.3 s.

Frequency = 1/period where period is the time it takes to complete a revolution.

F = 1/13.3 = 0.075hz

Angular velocity = 2* 3.142* 0.075 = 0.472rad/s

At a certain location, Earth's magnetic field of 34 µT is horizontal and directed due north. Suppose the net field is zero exactly 8.1 cm above a long, straight, horizontal wire that carries a constant current. What are (a) the magnitude of the current and (b) the angle between the current direction and due north?

Answers

Answer:

a) I = 13.77 A

b) 0 ° or to the East

Explanation:

Part a

The magnetic field by properties would be 0 at the radius on this case r =8.1 cm.Analyzing the situation the wirde would produce a magnetic field equals in magnitude to the magnetic field on Earth by with the inverse direction.

The formula for the magnetic field due to a wire with current is:

[tex] B = \frac{\mu_0 I}{2 \pi r} [/tex]

In order to have a value of 0 for the magnetic field at the radius then we need to have this balance

B (r=8.1) = B (Earth)

Replacing:

[tex] B = \frac{\mu_0 I}{2 \pi r)}= B_{Earth} [/tex]

Solving from I, from the last equation we got:

[tex] I = \frac{2 \pi r B_{earth}}{\mu_0} [/tex]

[tex] I=\frac{2 \pi 0.081 m (34 x 10^{-6} T)}{4 \pi x 10^{-7} Tm/A}[/tex] = 13.77 A

Part b

We can use the right hand rule for this case.

The magnetic field of the wire would point to the South, because the magnetic field of the earth given points to the North. Based on this the current need's to flow from West to East in order to create a magnetic field pointing to the south, because the current would be perpendicular to the magnetic field created.

Consider the amideanion.What is the central atom? Enter its chemical symbol.How many lone pairs are around the central atom?What is the ideal angle between the nitrogen-hydrogen bonds?Compared to the ideal angle, you would expect the actual angle between the nitrogen-hydrogen bonds to be ...

Answers

Answer:

a) The central atom of the amide ion is nitrogen

b) (NH2)-

c) There are two lone pairs around the nitrogen atom

d) The ideal angle between nitrogen hydrogen bonds is 109.5° C, typical of a tetrahedral electron density arrangement

e) I would expect the nitrogen-hydrogen bond angles in the amide ion to be less than 109.5 ° because unbonded pairs repel bonded pairs much more than other bonded pairs do. So the bonds in this ion would be pushed closer together than normal

Sometime around 2022, astronomers at the European Southern Observatory hope to begin using the E-ELT(European Extremely Large Telescope), which is planned to have a primary mirror 42 m in diameter. Let us assume that the light it focuses has a wavelength of 600 nm. (1 light-year = 9.461×1015 m) Note: Jupiter's Diameter dj=1.43×108 m 1) What is the most distant Jupiter-sized planet the telescope could resolve, assuming it operates at the diffraction limit? (Express your answer to two significant figures.)

Answers

Answer:

[tex]8.2\times 10^{15}\ m[/tex]

Explanation:

[tex]\lambda[/tex] = Wavelength = 600 nm

d = Diameter of mirror = 42 m

D = Distance of object

x = Diameter of Jupiter = [tex]1.43\times 10^8\ m[/tex]

Angular resoulution is given by

[tex]\Delta\theta=1.22\frac{\lambda}{d}\\\Rightarrow \Delta\theta=1.22\frac{600\times 10^{-9}}{42}\\\Rightarrow \Delta\theta=1.74286\times 10^{-8}\ rad[/tex]

We also have the relation

[tex]\Delta\theta\approx=\frac{x}{D}\\\Rightarrow D\approx\frac{x}{\Delta\theta}\\\Rightarrow D\approx\frac{1.43\times 10^8}{1.74286\times 10^{-8}}\\\Rightarrow D\approx 8.2049\times 10^{15}\ m[/tex]

The most distant Jupiter-sized planet the telescope could resolve is [tex]8.2\times 10^{15}\ m[/tex]

The auditory canal behaves like a resonant tube to aid in hearing. One end terminated at the eardrum, while the other opens to the outside. Typically, the canal is about 2.4 cm long. At which frequency would it resonate in its first harmonic?
4.2 kHz
3.6 kHz
2.9 kHz
5.7 kHz

Answers

Answer:

3.6 kHz

Explanation:

The pipes behave like a closed pipe . The end open is the end of the air canal outside the ear and the closed end is the eardrum.

The first harmonic will be as seen in the figure attached.

The length of the first harmonic will be λ/4.

λ/4=2.4 cm

λ=2.4 * 4=9.6 cm 0.096 m

Speed of Sound- 344 m/s(in air)

velocity(v) * Time Period(T) = Wavelength (λ)

Also, Time Period(T)= \frac{\textrm{1}}{\textrm{Frequency(f)}}

\frac{\textrm{Velocity}}{\textrm{Wavelength}}=\frac{\textrm{1}}{\textrm{Time Period}} =Frequency

Plugging in the values into the equation,

Frequency = [tex]\frac{344}{0.096}[/tex] Hz

                  = 3583.3 Hz≈3600 Hz= 3.6 kHz

Frequency= 3.6 kHz

Answer:

3.6 kHz

Explanation:

The auditory canal is a closed pipe because it has one closed end, the end terminated at the eardrum.

The length of the first harmonic of a closed pipe is given as;

L = λ/4 --------------------  (i)

where L = Length

and λ = wavelength

2.4 = λ/4

λ = 2.4 x 4 = 9.6 cm

Also, v = fλ  ------------------ (ii)

where v = speed of sound in air = 344 m/s

f = frequency of wave in Hertz

f = v/λ ------------------ (iii)

convert 9.6 cm to m = 0.096 m

substitute for λ and v in (iii)

[tex]f = \frac{344}{0.096} = 3583.33[/tex]

3583.33 Hz = 3600 Hz = 3.6 kHz

Three masses (3 kg, 5 kg, and 7 kg) are located in the xy-plane at the origin, (2.3 m, 0), and (0, 1.5 m), respectively.
A) Determine the center of mass of the system.
B) A 6 kg mass is added to the system. This four-mass system has a new center of mass at the origin. What is the location of the 6 kg mass?

Answers

Answer:

a) C.M [tex]=(\bar x, \bar y)=(0.767,0.7)m[/tex]

b) [tex](x_4,y_4)=(-1.917,-1.75)m[/tex]

Explanation:

The center of mass "represent the unique point in an object or system which can be used to describe the system's response to external forces and torques"

The center of mass on a two dimensional plane is defined with the following formulas:

[tex]\bar x =\frac{\sum_{i=1}^N m_i x_i}{M}[/tex]

[tex]\bar y =\frac{\sum_{i=1}^N m_i y_i}{M}[/tex]

Where M represent the sum of all the masses on the system.

And the center of mass C.M [tex]=(\bar x, \bar y)[/tex]

Part a

[tex]m_1= 3 kg, m_2=5kg,m_3=7kg[/tex] represent the masses.

[tex](x_1,y_1)=(0,0),(x_2,y_2)=(2.3,0),(x_3,y_3)=(0,1.5)[/tex] represent the coordinates for the masses with the units on meters.

So we have everything in order to find the center of mass, if we begin with the x coordinate we have:

[tex]\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)}{3kg+5kg+7kg}=0.767m[/tex]

[tex]\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)}{3kg+5kg+7kg}=0.7m[/tex]

C.M [tex]=(\bar x, \bar y)=(0.767,0.7)m[/tex]

Part b

For this case we have an additional mass [tex]m_4=6kg[/tex] and we know that the resulting new center of mass it at the origin C.M [tex]=(\bar x, \bar y)=(0,0)m[/tex] and we want to find the location for this new particle. Let the coordinates for this new particle given by (a,b)

[tex]\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)}{3kg+5kg+7kg+6kg}=0m[/tex]

If we solve for a we got:

[tex](3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)=0[/tex]

[tex]a=-\frac{(5kg*2.3m)}{6kg}=-1.917m[/tex]

[tex]\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)}{3kg+5kg+7kg+6kg}=0m[/tex]

[tex](3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)=0[/tex]

And solving for b we got:

[tex]b=-\frac{(7kg*1.5m)}{6kg}=-1.75m[/tex]

So the coordinates for this new particle are:

[tex](x_4,y_4)=(-1.917,-1.75)m[/tex]

The specifications for a ceiling fan you have just bought indicate that the total moment of inertia of its blades is 0.23 kg · m2 and they experience a net torque of 2.8 N · m.

(a) What is the angular acceleration of the blades in rad/s2? rad/s2

(b) When the blades rotate at 205 rpm, what is the rotational kinetic energy, in joules? J

Answers

Answer:(a)12.17 rad/s

Explanation:

Given

Moment of Inertia [tex]I=0.23 kg.m^2[/tex]

Torque [tex]T=2.8 N-m[/tex]

(a)Torque is given by Product of Moment of inertia and angular acceleration

[tex]T=I\cdot \alpha [/tex]

[tex]2.8=0.23\cdot \alpha [/tex]

[tex]\alpha =\frac{2.8}{0.23}=12.17 rad/s[/tex]

(b)RPM of blades [tex]N=205 rpm [/tex]

angular velocity [tex]\omega =\frac{2\pi N}{60}[/tex]

[tex]\omega =\frac{2\pi 205}{60}=21.47 rad/s[/tex]

Rotational Kinetic Energy [tex]=\frac{I\omega ^2}{2}[/tex]

[tex]=\frac{0.23\times (21.47)^2}{2}=53.01 J[/tex]

Final answer:

The angular acceleration of the blades is approximately 12.1739 rad/s^2. When the blades rotate at 205 rpm, the rotational kinetic energy is approximately 0.0948 J.

Explanation:

(a) To find the angular acceleration of the blades, we can use the formula:

torque = moment of inertia × angular acceleration

Plugging in the given values:

torque = 2.8 N · m

moment of inertia = 0.23 kg · m2

Rearranging the formula, we get:

angular acceleration = torque / moment of inertia

Substituting the values:

angular acceleration = 2.8 N · m / 0.23 kg · m2

Solving for angular acceleration gives us:

angular acceleration ≈ 12.1739 rad/s2

(b) To find the rotational kinetic energy, we can use the formula:

rotational kinetic energy = ½ × moment of inertia × (angular velocity)2

Plugging in the given values:

moment of inertia = 0.23 kg · m2

angular velocity = 205 rpm = 205 revolutions / 60 seconds = 3.4167 rev/s

Rearranging the formula, we get:

rotational kinetic energy = ½ × moment of inertia × (angular velocity)2

Substituting the values:

rotational kinetic energy = ½ × 0.23 kg · m2 × (3.4167 rev/s)2

Solving for rotational kinetic energy gives us:

rotational kinetic energy ≈ 0.0948 J

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A heating coil inside an electric kettle delivers 2.2 kW of electrical power to the water in the kettle. How long will it take to raise the temperature of 0.50 kg of water from 30.0° C to 72.0° C

Answers

Answer:

0.04s

Explanation:

Specific heat of water c = 4.186J/g.C

The heat energy it would take to heat up 0.5 kg of water from 30C to 72 C is

[tex]E = mc\Delta T = 0.5 * 4.186 * (72 - 30) = 87.9 J[/tex]

If the power of the electric kettle is 2.2kW (or 2200W) and suppose the work efficiency is 100%, then the time it takes to transfer that power is

[tex] t = \frac{E}{P} = \frac{87.9}{2200} = 0.04 s[/tex]

A 19 g bullet is fired into the bob of a ballistic pendulum of mass 1.3 kg. When the bob is at its maximum height, the strings make an angle of 60° with the vertical. The length of the pendulum is 2.3 m. Find the speed of the bullet.

Answers

Answer:

217.43298 m/s

Explanation:

[tex]m_1[/tex] = Mass of bullet = 19 g

[tex]m_2[/tex] = Mass of bob = 1.3 kg

L = Length of pendulum = 2.3 m

[tex]\theta[/tex] = Angle of deflection = 60°

u = Velocity of bullet

Combined velocity of bullet and bob is given by

[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2aL(1-cos\theta)+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times (1-cos60)+0^2}\\\Rightarrow v=3.13209\ m/s[/tex]

As the momentum is conserved

[tex]m_1u=(m_1+m_2)v\\\Rightarrow u=\frac{(m_1+m_2)v}{m_1}\\\Rightarrow v=\frac{(0.019+1.3)\times 3.13209}{0.019}\\\Rightarrow v=217.43298\ m/s[/tex]

The speed of the bullet is 217.43298 m/s

Final answer:

The problem involves conservation of momentum and energy principles. Initially, bullet's momentum equals the final momentum of the system. The bullet's speed can be found by solving these equations, using the provided values.

Explanation:

This problem can be solved using principles from both conservation of momentum and conservation of energy. To find the speed of the bullet, we need to consider two scenarios: the before and after the bullet is fired into the bob. Initial momentum is the mass of the bullet multiplied by its velocity and final momentum is the combined mass of the bullet and bob at their highest point. Assuming there's no external force acting, we can have:

m_bullet * v_bullet = (m_bullet + m_bob) * v_final.

The final velocity here is the vertical component of the velocity when the pendulum reach its highest point. This can be calculated by:

v_final = sqrt(2*gravity*height).

The height can be calculated using trigonometry:

height = length - length * cos(60).

Filling all the given values into the equations will give the speed of the bullet.

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A glass flask whose volume is 1000 cm3 at a temperature of 0.300 ∘C is completely filled with mercury at the same temperature. When the flask and mercury are warmed together to a temperature of 52.0 ∘C , a volume of 8.25 cm3 of mercury overflows the flask.

Answers

Final answer:

When the glass flask and mercury are warmed together, the mercury expands and overflows the flask. To calculate the initial volume of the mercury, use the equation: Volume of mercury = Volume of flask + Volume overflowed.

Explanation:

When the glass flask and mercury are warmed together, both substances expand due to the increase in temperature. As a result, some of the mercury overflows the flask. To calculate the initial volume of the mercury, we can use the equation:

Volume of mercury = Volume of flask + Volume overflowed

So, the initial volume of mercury is 1000 cm³ + 8.25 cm³ = 1008.25 cm³.

The colors of the stars in the sky range from red to blue. Assuming that the color indicates the frequency at which the star radiates the maximum amount of electromagnetic energy, estimate the surface temperature of red, yellow, white, and blue stars.
A) Estimate the surface temperature of red star. Assume that red color corresponds to wavelengths of approximately 650 nm.B) Estimate the surface temperature of yellow star. Assume that yellow color corresponds to wavelengths of approximately 570 nm.C) Estimate the surface temperature of blue star. Assume that blue color corresponds to wavelengths of approximately 470 nmD) Estimate the surface temperature of white star. Since green is in the middle of the visible spectrum, choose 520 nm wavelength for white star.

Answers

Answer:

a) 4458K b) 5048K, c) 6166K, d) 5573K

Explanation:

The temperature of the stars and many very hot objects can be estimated using the Wien displacement law

    [tex]\lambda_{max}[/tex] T = 2,898 10⁻³ [m K]

    T = 2,898 10⁻³ / [tex]\lambda_{max}[/tex]

a) indicate that the wavelength is

    Lam = 650 nm (1 m / 109 nm) = 650 10⁻⁹ m

    Lam = 6.50 10⁻⁷ m

    T = 2,898 10⁻³ / 6.50 10⁻⁷

    T = 4,458 10³ K

    T = 4458K

b) lam = 570 nm = 5.70 10⁻⁷ m

    T = 2,898 10⁻³ / 5.70 10⁻⁷

    T = 5084K

c) lam = 470 nm = 4.70 10⁻⁷ m

    T = 2,898 10⁻³ / 4.7 10⁻⁷

    T = 6166K

d) lam = 520 nm = 5.20 10⁻⁷ m

    T = 2,898 10⁻³ / 5.20 10⁻⁷

    T = 5573K

A ) the surface temperature of red star is about 4500 K

B ) the surface temperature of yellow star is about 5100 K

C ) the surface temperature of blue star is about 6200 K

D ) the surface temperature of white star is about 5600 K

[tex]\texttt{ }[/tex]

Further explanation

Let's recall the Wien's Displacement Law as follows:

[tex]\boxed {\lambda_{max}\ T = 2.898 \times 10^{-3} \texttt{ m.K}}[/tex]

where:

λ_max = the wavelength of the maximum radiation energy ( m )

T = surface temperature of the star ( K )

Let us now tackle the problem!

[tex]\texttt{ }[/tex]

Given:

wavelength of red light = λ_r = 650 nm = 650 × 10⁻⁹ m

wavelength of yellow light =  λ_y = 570 nm = 570 × 10⁻⁹ m

wavelength of blue light =  λ_b = 470 nm = 470 × 10⁻⁹ m

wavelength of white light =  λ_w = 520 nm = 520 × 10⁻⁹ m

Asked:

A ) the surface temperature of red star = T_r = ?

B ) the surface temperature of yellow star = T_y = ?

C ) the surface temperature of blue star = T_b = ?

D ) the surface temperature of white star = T_w = ?

Solution:

Part A) :

[tex]T_r = ( 2.898 \times 10^{-3} ) \div \lambda_r[/tex]

[tex]T_r = ( 2.898 \times 10^{-3} ) \div ( 650 \times 10^{-9} )[/tex]

[tex]\boxed {T_r \approx 4500 \texttt{ K} }[/tex]

[tex]\texttt{ }[/tex]

Part B) :

[tex]T_y = ( 2.898 \times 10^{-3} ) \div \lambda_y[/tex]

[tex]T_y = ( 2.898 \times 10^{-3} ) \div ( 570 \times 10^{-9} )[/tex]

[tex]\boxed {T_y \approx 5100 \texttt{ K} }[/tex]

[tex]\texttt{ }[/tex]

Part C) :

[tex]T_b = ( 2.898 \times 10^{-3} ) \div \lambda_b[/tex]

[tex]T_b = ( 2.898 \times 10^{-3} ) \div ( 470 \times 10^{-9} )[/tex]

[tex]\boxed {T_b \approx 6200 \texttt{ K} }[/tex]

[tex]\texttt{ }[/tex]

Part D) :

[tex]T_w = ( 2.898 \times 10^{-3} ) \div \lambda_w[/tex]

[tex]T_w = ( 2.898 \times 10^{-3} ) \div ( 520 \times 10^{-9} )[/tex]

[tex]\boxed {T_w \approx 5600 \texttt{ K} }[/tex]

[tex]\texttt{ }[/tex]

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[tex]\texttt{ }[/tex]

Answer details

Grade: High School

Subject: Mathematics

Chapter: Energy

A toy cannon uses a spring to project a 5.35-g soft rubber ball. The spring is originally compressed by 5.08 cm and has a force constant of 8.07 N/m. When the cannon is fired, the ball moves 16.0 cm through the horizontal barrel of the cannon, and the barrel exerts a constant friction force of 0.033 0 N on the ball.
(a) With what speed does the projectile leave the barrel of the cannon? m/s
(b) At what point does the ball have maximum speed? cm (from its original position)
(c) What is this maximum speed? m/s

Answers

Answer:

a) the velocity is v=1.385 m/s

b) the ball has its maximum speed at 4.68 cm away from its compressed position

c)  the maximum speed is 1.78 m/s

Explanation:

if we do an energy balance over the ball, the potencial energy given by the compressed spring is converted into kinetic energy and loss of energy due to friction, therefore

we can formulate this considering that the work of the friction force is equal to to the energy loss of the ball

W fr = - ΔE = - ΔU - ΔK = Ui - Uf + Kf - Ki

therefore

Ui + Ki = Uf + Kf + W fr  

where U represents potencial energy of the compressed spring , K is the kinetic energy W fr is the work done by the friction force. i represents inicial state, and f final state.

since

U= 1/2 k x² , K= 1/2 m v²  , W fr = F*L

X= compression length , L= horizontal distance covered

therefore

Ui + Ki = Uf + Kf + W fr

1/2 k xi² + 1/2 m vi² = 1/2 k x² + 1/2 m vf² + F*L

a) choosing our inicial state as the compressed state , the initial kinetic energy is Ki=0 and in the final state the ball is no longer pushed by the spring thus Uf=0

1/2 k X² + 0 = 0 + 1/2 m v² + F*L

1/2 m v² = 1/2 k X² - F*L

v = √[(k/m)x² -(2F/m)*L] = √[(8.07N/m/5.35*10^-3 Kg)*(-0.0508m)² -(2*0.033N/5.35*10^-3 Kg)*(0.16 m)] = 1.385 m/s

b) in any point x , and since L= d-(X-x) , d = distance where is no pushed by the spring.

1/2 k X² + 0 = 1/2 k x² + 1/2 m v² + F*[d-(X+x)]

1/2 m v² =1/2 k X²-1/2 k x² - F*[d-(X-x)] = (1/2 k X²+ F*X) - 1/2k x² - F*x + F*d

taking the derivative

dKf/dx = -kx - F = 0 → x = -F/k = -0.033N/8.07 N/m = -4.089*10^-3 m = -0.4cm

at x m = -0.4 cm the velocity is maximum

therefore is 5.08 cm-0.4 cm=4.68 cm away from the compressed position

c) the maximum speed is

1/2 m v max² = (1/2 k X²+ F*X) - 1/2k x m² - F*(x m) + 0

v =√[ (k/m) (X²-xm²) + (2F/m)(X-xm) ] = √[(8.07N/m/5.35*10^-3 Kg)*[(-0.0508m)² - (-0.004m)²] + (2*0.033N/5.35*10^-3 Kg)*(-0.0508m-(-0.004m)] = 1.78 m/s

In a downtown office building, you notice each of the four sections of a rotating door has a mass of 75 kg. What is the width, in meters, of each section of the door if a force of 56 N applied to the outer edge of a section produces an angular acceleration of 0.420 rad/s2?

Answers

Answer:

each door has a width of 2.666 meters

Explanation:

from Newton's second law applied to rotational motion:

ζ = I α

where ζ= torque , I = moment of inertia , α = angular acceleration

the moment of inertia for a flat plate around its central axis is

I = 1/12 m a² , where m= mass, a= total width = 2L

therefore the moment of inercia for a flat plate with length 2L ( 2 doors, one in each side of the central axis) is

I1 = 1/12 m (2L)² = 1/3 m L²

if we have 4 doors , that is 2 flat plates with length of 2L perpendicular to each other:

I = Ix + Iy = 2*I1 = 2/3 m L²

thus

ζ = I α

4* F * L = 2/3 * (4*m) L² * α

L = 3/2* F/ ( m*α) = 3/2* 56 N / ( 75 Kg * 0.420 rad/seg²) = 2.666 m

You are lying in bed and want to shut your bedroom door. You have a bouncy "superball" and a blob of clay, both with the same mass. Which one would be more effective to throw at your door to close it?

a. The superball.
b. Neither will work.
c. The blob of clay.
d. Both the same.

Answers

Answer:

Choice A. The bouncy "superball" will exert approximately twice as much force than the blob of clay when tossed out at the same speed.

Explanation:

The momentum of an object is equal to the product of its mass and its velocity. The momentum of the system is conserved even if the collision is inelastic.

When the bouncy ball or the blob of clay hits the door, their momentum changes. Since momentum is conserved, the door would move in the opposite direction such that the total momentum stays at zero.

Let the mass of each object be [tex]m[/tex]. Suppose that both objects hit the door at a speed of [tex]v[/tex] in the same direction.

The bouncy ball will likely bounce back at about the same speed (in the opposite direction, away from the door.) The magnitude of the change in velocity will be at most [tex]2v[/tex]. The magnitude of the change in momentum will be at most [tex]2 \,m\cdot v[/tex].However, when the blob of clay hits the door, its shape would change (and it would potentially stick to the door.) In other words, its velocity becomes almost [tex]0[/tex] after the collision. The magnitude of the change in its momentum will only be around [tex]m \cdot v[/tex], which is only about [tex]\dfrac{1}{2}[/tex] the momentum change due to the bouncy ball.

By Newton's Second Law, the net force on the door is proportion to the rate of change in its momentum. Assume that the two objects are in contact with the door for the same amount of time, the bouncy ball would exert about twice as much force on the door as the clay would. Hence it is "more effective" for closing the door.

Final answer:

The clay is more effective than the superball for shutting the door because upon impact the clay sticks to the door and transfers all of its momentum to it, while the superball bounces back and doesn't transfer all of its momentum.

Explanation:

The most effective object to throw at your door to shut it would be c. The blob of clay. This is because of the principle of conservation of momentum. When the clay hits the door, it sticks to it, thus transferring all of its momentum to the door. This momentum is what causes the door to move and potentially close. On the other hand, when the superball hits the door, it bounces back, meaning not all of its momentum is transferred to the door. Hence, the clay, by virtue of its ability to stick and transfer its full momentum, would be more effective.

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Based on this experiment, the student can conclude that all plants need water and sunlight to grow because only the plant that received water and sunlight grew and stayed healthy.B. Yes. Based on this experiment, the student can conclude that all plants need water and sunlight to grow because all three plants received water, sunlight, or both, and all three grew at the beginning of the experiment.C. No. Based on this experiment, the student observed that plants can grow without water and sunlight.D. No. Based on this experiment, the student can only conclude that this type of bean plant needs water and sunlight to grow. Express the sum of the polynomial 5x^2+6x17 and the square of the binomial (x+6) as a polynomial in standard form. How many bar lines are there in this staff including the final bar line What were the causes and results of the thirty years war Which of the following changes would bring the U.S. capital stock, currently below the Golden Rule level, closer to the steady-state, consumption-maximizing level? - increasing the population growth rate - increasing the saving rate - increasing the rate of capital depreciation - increasing the rate of technological progress Tolerable misstatement is used to A. Determine sample size Select the sample Evaluate results Yes Yes No B. Determine sample size Select the sample Evaluate results No Yes No C. Determine sample size Select the sample Evaluate results No No Yes D. Determine sample size Select the sample Evaluate results Yes No Yes Please help ASAP From Caribbean Climate Change.Weather events of this degree also cause expensive damage to infrastructure. In 2013, tropical storms in St. Vincent and the Grenadines resulted in widespread flooding and landslides. Bridges, roads, and houses were destroyed.Based on the context clues in the passage, what does the word infrastructure mean?A) beaches, forests, and mountainsB) buildings, bridges, and roadsC) cars, trucks, and trainsD) food, water, and clothing Our own experiences with the state through encounters with institutions such as the police or airport customs lend credence to the idea of the state as a collection of institutions and structures, but as political anthropologists, we can understand the state as: (+4) + (-13) = solve. How did Galileo use his knowledge in music to understanding the movement of a pendulum? Suppose that a chicken farm uses a nearby stream to dispose of the wastes released by its chickens. These wastes flow downstream into a lake that has become thick with algae and polluted due to the minerals in the waste matter. The local office of a nonprofit environmental organization collects enough donations to stop the farm's pollution. Which of the following types of private solutions to the externality of pollution has occurred in this case? a. contracts b. Moral codes and social sanctions c. Integration of different types of businesses through merger or acquisition d. Charities Its important to note that sometimes private solutions to externalities do not work. For example, this occurs when one party repeatedly holds out for a better deal. This describes the problem of (transaction cost, breakdown in bargaining, property held in common) Select all of the following that are ordered pairs of the given function.f(x) = 3- 2(-2, -1)(-1, 5)(0,3)(1, 0)(2, -1) One number has a prime factorization of 23.32, and another number has a prime factorization of 22.3. Which of thefollowing expressions would equal the greatest common factor of these two numbers?25.3523.3322.322.3 What is most likely the authors reason for mentioning the State of the Union in the first line of the address? 1st line: Ladies and gentlemen, Id planned to speak to you tonight to report on the state of the Union, but the events of earlier today have led me to change those plans. Question 5 options: A) to make it clear that he just came up with these remarks on the spot B) to tie in the Challenger explosion with the rest of his planned remarks for the State of the Union C) to emphasize the historic nature of the tragedy that occurred D) to vent his frustration at the change of plans help me out pleaseee The circle with center F is divided into sectors. In circle F , EB is a diameter. The radius of circle F is 3 units If the blood vessels of the skin dilate, the person appearsA.) CyanoticB.) FlushedC.) Jaundiced D.) Pale and washed out Ashlin, age 24, started an extreme diet last month. She has already lost 20 pounds, and due to restricting her food intake, Ashlin rarely feels hungry. Ashlin's eating behavior has triggered new _____, allowing her to function with fewer daily calories.