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2 years ago

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Earth's gravity acts upon objects with a steady force of __________. A. 8. 9 meters per second B. 9. 8 meters per minute C. 8. 9

meters per minute squared D. 9. 8 meters per second squared Please select the best answer from the choices provided. A B C D.

1 answer:

Mila [183]2 years ago

8 0

Answer:

9.8 meters per second squared

Explanation:

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A hollow cylinder of mass 2.00 kgkg, inner radius 0.100 mm, and outer radius 0.200 mm is free to rotate without friction around

kipiarov [429]

Answer: 2.86 m

Explanation:

To solve this question, we will use the law of conservation of kinetic and potential energy, which is given by the equation,

ΔPE(i) + ΔKE(i) = ΔPE(f) + ΔKE(f)

In this question, it is safe to say there is no kinetic energy in the initial state, and neither is there potential energy in the end, so we have

mgh + 0 = 0 + KE(f)

To calculate the final kinetic energy, we must consider the energy contributed by the Inertia, so that we then have

mgh = 1/2mv² + 1/2Iw²

To get the inertia of the bodies, we use the formula

I = [m(R1² + R2²) / 2]

I = [2(0.2² + 0.1²) / 2]

I = 0.04 + 0.01

I = 0.05 kgm²

Also, the angular velocity is given by

w = v / R2

w = 4 / (1/5)

w = 20 rad/s

If we then substitute these values in the equation we have,

0.5 * 9.8 * h = (1/2 * 0.5 * 4²) + (1/2 * 0.05 * 20²)

4.9h = 4 + 10

4.9h = 14

h = 14 / 4.9

h = 2.86 m

8 0

3 years ago

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What happens to the magnitude of the force of gravitation between two objects if: 1. distance between the objects is tripled? 2.

Ludmilka [50]

<span>2. mass of both objects doubled?
Hope it helped!</span>

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4 years ago

Read 2 more answers

A spring gun is made by compressing a spring in a tube and then latching the spring at the compressed position. A 4.97-g pellet

dimaraw [331]

Answer:

$v = 2.8898 \frac{m}{s}$

Explanation:

This is a problem easily solve using energy conservation. As there are no non-conservative forces, we know that the energy is conserved.

When the spring is compressed downward, the spring has elastic potential energy. When the spring is relaxed, there is no elastic potential energy, but the pellet will have gained gravitational potential energy and kinetic energy. Lets see what are the terms for each of this.

<h3>Elastic potential energy</h3>

We know that a spring following Hooke's Law has a elastic potential energy:

$E_{ep} = \frac{1}{2} k (\Delta x)^2$

where $\Delta x$ is the displacement from the relaxed length and k is the spring's constant.

To obtain the spring's constant, we know that Hooke's law states that the force made by the spring is :

$\vec{F} = - k \Delta \vec{x}$

as we need 9.12 N to compress 4.60 cm, this means:

$k = \frac{9.12 \ N}{4.6 \ 10^{-2} \ m}$

$k = 198.26 \ \frac{ N}{m}$

So, the elastic energy of the compressed spring is:

$E_{ep} = \frac{1}{2} 198.26 \ \frac{ N}{m} (4.6 \ 10^{-2} \ m)^2$

$E_{ep} = 0.209759 \ Joules$

And when the spring is relaxed, the elastic potential energy will be zero.

<h3>Gravitational potential energy</h3>

To see how much gravitational potential energy will the pellet win, we can use

$\Delta E_{gp} = m g \Delta h$

where m is the mass of the pellet, g is the acceleration due to gravity and \Delta h is the difference in height.

Taking all this together, the gravitational potential energy when the spring is relaxed will be:

$\Delta E_{gp} = 4.97 \ 10^{-3} kg \ 9.8 \frac{m}{s^2} 4.6 \ 10^{-2} m$

$\Delta E_{gp} = 0.00224 \ Joules$

<h3>Kinetic Energy</h3>

We know that the kinetic energy for a mass m moving at speed v is:

$E_k = \frac{1}{2} m v^2$

so, for the pellet will be

$E_k = \frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2$

<h3>All together</h3>

By conservation of energy, we know:

$E_{ep} = \Delta E_{gp} + E_k$

$0.209759 \ Joules = 0.00224 \ Joules + \frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2$

So

$\frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2 = 0.209759 \ Joules - 0.00224 \ Joules$

$\frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2 = 0.207519 \ Joules$

$v = \sqrt{ \frac{ 0.207519 \ Joules}{ \frac{1}{2} \ 4.97 \ 10^{-3} kg } }$

$v = 2.8898 \frac{m}{s}$

7 0

3 years ago

how much work does tension do on the laptop as it moves 2.0 mm ? express your answer with the appropriate units.

8090 [49]

The work done by tension force of 14N applied on the laptop by a rope as it moves 2.0 mm up the slope is 0.028 J

W = F d cos θ

W = Work done

F = Force

d = Displacement

θ = Angle between force and displacement vector

F = 14 N

d = 2 mm = 0.002 m

θ = 0

W = 14 * 0.002 * 1

W = 0.028 J

Work done is the change in energy of an object. So if an object moves a certain distance, work is done on the object. If the force and displacement are perpendicular to each other there is no work done on the object.

Therefore, the work done by tension on the laptop is 0.028 J

To know more about work done

brainly.com/question/12834956

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8 0

1 year ago

A slinky is stretched by two forces. When the forces are removed, the slinky returns to its original length. The slinky has been

e-lub [12.9K]

Answer: <u>elastically</u> deformed or <u>non-permanently</u> deformed

Explanation:

According to classical mechanics, there are two types of deformations:

-Plastic deformation (also called irreversible or permanent deformation), in which the material does not return to its original form after removing the applied force, therefore it is said that the material was permanently deformed.

This is because the material undergoes irreversible thermodynamic changes while it is subjected to the applied forces.

-Elastic deformation (also called reversible or non-permanent deformation), in which the material returns to its original shape after removing the applied force that caused the deformation.

In this case t<u>he material also undergoes thermodynamic changes, but these are reversible, causing an increase in its internal energy by transforming it into elastic potential energy.</u>

<u />

Therefore, the situation described in the question is related to elastic deformation.

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3 years ago