You wish to buy a motor that will be used to lift a 10-kg bundle of shingles from the ground to the roof of a house. The shingles are to have a 1.5-m/s2 upward acceleration at the start of the lift. The very light pulley on the motor has a radius of 0.17 m . Part A Determine the minimum torque that the motor must be able to provide. Express your answer with

Answers

Answer 1

Answer:

To lift a 10-kg bundle of shingles with an upward acceleration of 1.5 m/s², a motor with a pulley of radius 0.17 m must provide a torque of at least 19 N.m.

We wish to use a motor to lift a 10-kg (m) bund of shingles with an upward acceleration of 1.5 m/s² (a).

The resulting force (F = m.a) is the difference between the tension (T) of the pulley and the weight (w = m.g) of the shingles.

$T-m.g=m.a\nT = m.g+m.a = m(g+a) = 10 kg (9.8m/s^(2)+1.5m/s^(2) )=113 N$

where,

g: gravity

Then, we can calculate the minimum torque (τ) that the motor must apply using the following expression.

$\tau = r * T = 0.17m * 113N = 19N.m$

where,

r: radius of the pulley

To lift a 10-kg bundle of shingles with an upward acceleration of 1.5 m/s², a motor with a pulley of radius 0.17 m must provide a torque of at least 19 N.m.

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Answer 2

Answer:

Answer:

$\tau=19.21\ N-m$

Explanation:

It is given that,

Mass of bundle of shingles, m = 10 kg

Upward acceleration of the shingles, $a=1.5\ m/s^2$

The radius of the motor of the pulley, r = 0.17 m

Let T is the tension acting on the shingles when it is lifted up. It can be calculated as :

$T-mg=ma$

$T=m(g+a)$

$T=10* (9.8+1.5)$

T = 113 N

Let $\tau$ is the minimum torque that the motor must be able to provide. It is given by :

$\tau=r* T$

$\tau=0.17* 113$

$\tau=19.21\ N-m$

So, the minimum value of torque is 19.21 N-m. Hence, this is the required solution.

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Starting from rest, a basketball rolls from the top to the bottom of a hill, reaching a translational speed of 6.1 m/s. Ignore frictional losses. (a) What is the height of the hill? (b) Released from rest at the same height, a can of frozen juice rolls to the bottom of the same hill. What is the translational speed of the frozen juice can when it reaches the bottom?

When an external magnetic field is applied, what happens to the protons in a sample?A) All protons align with the field.
B) All protons align opposite to the field.
C) Some protons align with the field and some align opposite to it.
D) All protons assume a random orientation.

Answers

On account of external magnetic field, the protons will align with the magnetic field. Hence, option (a) is correct.

The given problem is based on the concept of magnetic field. The region where the magnetic force is experienced is known as magnetic field. Generally, the protons are the charged entities carrying the positive polarity and are one of the major constituents of modern atomic structure.

The origin of magnetic field occurs due to charged particles present in a specific space. And the magnetic field is due to the flowing of liquid metal in the outer core of the planet generates electric currents.
In the condition when an external field is applied, the majority of protons align to the field because these protons tend to act like small magnets under the effect of this external field.

Thus, we can conclude that on account of external magnetic field, the protons will align with the field.

Learn more about the magnetic field here:

brainly.com/question/14848188

Answer:

Some protons align with the field and some align opposite to it.

Explanation:

Majority align to the field because these protons tend to act like small magnets under the effect of this external field

A magnet of mass 0.20 kg is dropped from rest and falls vertically through a 35.0 cm copper tube. Eddy currents are induced, causing the copper to warm up. The speed of the magnet as it emerges from the tube is 1.10 m/s. How much heat energy is dissipated to the environment?

Answers

Answer:

0.58 J

Explanation:

We know that Total energy is conserved.

Initial Kinetic energy + Initial potential energy = final kinetic energy+ final potential energy + dissipated heat energy

Initial kinetic energy = 0 ( magnet is at rest initially)

Initial Potential energy = m g h = (0.20 kg)(9.81 m/s²)(0.35 m) = 0.69 J

Final kinetic energy = 0.5 m v² = 0.5 ×0.20 kg × 1.10 m/s = 0.11 J

Final potential energy = 0

∴ Dissipated heat energy = (0.69 -0.11) J = 0.58 J

A steel ball is dropped onto a thick piece of foam. The ball is released 2.5 meters above the foam. The foam compresses 3.0 cm as the ball comes to rest. What is the magnitude of the ball's acceleration as it comes to rest on the foam

Answers

Answer:

the magnitude of the ball's acceleration as it comes to rest on the foam is 817.5 m/s²

Explanation:

Given the data in the question;

initial velocity; u = 0 m/s

height; h = 2.5 m

we find the velocity of the ball just before it touches the foam.

using the equation of motion;

v² = u² + 2gh

we know that acceleration due gravity g = 9.81 m/s²

so we substitute

v² = ( 0 )² + ( 2 × 9.81 × 2.5 )

v² = 49.05

v = √49.05

v = 7.00357 m/s

Now as the ball touches the foam

final velocity v₀ = 0 m/s

compresses S = 3 cm = 0.03 m

v₀² = v² + 2as

we substitute

( 0 )² = 49.05 + 0.06a

0.06a = -49.05

a = -49.05 / 0.06

a = -817.5 m/s²

Therefore, the magnitude of the ball's acceleration as it comes to rest on the foam is 817.5 m/s²

Two cars, one with mass mi = 1200 kg is traveling north, and the other a large car with mass m2= 3000 kg is traveling East. If they collide while each car is traveling with a speed v = 5.0 m/s, what is the car's final speed and direction (vector notation can be given as well). Assume the collision is perfectly inelastic.

Answers

Answer:

Both cars travel at < 10 , 4 > m/s

Explanation:

Conservation of Linear Momentum

The total momentum of a system of particles of masses m1 and m2 traveling at velocities v1 and v2 (vectors) is given by

$\vec p_1=m_1\vec v_1+m_2\vec v_2$

When the particles collide, their velocities change to v1' and v2' while their masses remain unaltered. The total momentum in the final condition is

$\vec p_2=m_1\vec v'_1+m_2\vec v'_2$

We know the collision is perfectly inelastic, which means both cars stick together at a common final velocity v'. Thus

$\vec p_2=m_1\vec v'+m_2\vec v'=(m_1+m_2)\vec v'$

Both total momentums are equal:

$m_1\vec v_1+m_2\vec v_2=(m_1+m_2)\vec v'$

Solving for v'

$\displaystyle v'=(m_1\vec v_1+m_2\vec v_2)/(m_1+m_2)$

The data obtained from the question is

$m_1=1200\ kg$

$m_2=3000\ kg$

The first car travels north which means its velocity has only y-component

$\vec v_1=<0,5>$

The second car travels east, only x-component of the velocity is present

$\vec v_2=<5,0>$

Plugging in the values

$\displaystyle v'=(1200<0,5>+3000<5,0>)/(1200+3000)$

$\displaystyle v'=(<0,6000>+<15000,0>)/(1500)$

$\displaystyle v'=(<15000,6000>)/(1500)=<10,4>\ m/s$

The magnitude of the velocity is

$|v'|=√(10^2+4^2)=10.77\ m/s$

And the angle

$\displaystyle tan\alpha=(4)/(10)=0.4$

$\alpha=21.8^o$

Parallel Plates Consider a very large conducting plate at potential V0 suspended a distance d above a very large grounded plane. Find the potential between the plates. The plates are large enough so that they may be considered to be infinite. This means that one can neglect fringing fields.

Answers

Answer:

$V = (V_0x)/(d)$

Explanation:

Since the field lines are parallel and the electric field is uniform between two parallel plates, a test charge would experience the same force of attraction or repulsion no matter where it is located in the field,

I attached an image that could help to understand the representation of the field. The formula used to calculate it is given by,

$E= -(\Delta V)/(x)$ (1)

If we want to consider the change in Voltage with respect to the position then it would be,

$E= -(dV)/(dx)$

According to the information provided, the potential is $V_0$ and there is a distance d, therefore

$E= -(V_0)/(d)$ (2)

Taking equation (1) we can clear V, to what we have,

$(dV)/(dx) = -E$

$dV = -Edx$

Integrating,

$V= - \int Edx$

Substituting (2)

$V = -\int (V_0)/(d) dx$

$V = (V_0x)/(d)$

Where x is the height from the grounded plate.

Each mass in the figure is 3 kg. Find the magnitude and direction of the net gravitational force on mass A due to the other masses.A. 2.45 × 10–7 N toward B
B. 3.75 × 10–7 N toward C
C. 2.00 × 10–7 N toward D
D. 1.15 × 10–7 N toward D

Answers

The magnitude and direction of the net gravitationalforce on mass A due to the other masses is 1.15 × 10⁻⁷ N toward D.

The gravitational force between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. In this case, all of the masses are equal to 3 kg, and the distance between mass A and mass D is 3 m.

The gravitational force between mass A and mass D is therefore:

F = G * m_A * m_D / r²

= 6.674 × 10⁻¹¹ N m² / kg² * 3 kg * 3 kg / 3 m²

= 1.15 × 10⁻⁷ N

The direction of the gravitational force is towards mass D.

Therefore, the net gravitational force on mass A due to the other masses is 1.15 × 10⁻⁷ N toward D.

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Answer:

THE ANSER IS B

Explanation:

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