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Step2247 [10]
3 years ago
12

Ans

Physics
1 answer:
Tresset [83]3 years ago
5 0
Probably the last one
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A box moving to the right experiences four forces, as shown.
Free_Kalibri [48]

Answer:b

Explanation:g

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What is the role of the neutral wire
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5 0
3 years ago
Help with physical science please
alex41 [277]

1. Elastic potential energy (D. EEl)

In this situation, the spring is compressed with the toy on top of it. The toy is stationary, so it does not have kinetic energy. However, the spring is compressed, so it does have elastic potential energy, given by:

E_{EL}=\frac{1}{2}kx^2

where k is the spring constant and x is the compression of the spring.

2. Gravitational potential energy (C. Eg)

In this situation, the spring has been released, so it returns to its natural position, so its elastic potential energy is zero. The toy is also stationary, since it is at its top position, where its velocity is zero, so its kinetic energy is also zero. However, the toy is now at a certain height h above the spring, so it has gravitational potential energy given by:

E_g = mgh

where m is the mass of the toy and g is the gravitational acceleration.

3. Gravitational potential and kinetic energy (A. Eg and EK)

In this situation, the toy is falling: so, it is moving with a certain speed v, so it has kinetic energy given by

E_k = \frac{1}{2}mv^2

Also, since it is at a certain height above the spring, it still has some gravitational potential energy, as in the previous point.

4. Gravitational potential energy (C. Eg)

The jumper is standing on the bridge, so it has gravitational potential energy given by its height h above the ground:

E_g=mgh

where m is the mass of the jumper.

5. This exercise has the same text of the previous one.

8 0
3 years ago
Jake drags a sled of mass 25 kg across the snow. The sled is attached to a rope that pulls with a force of 50 N at an angle of 3
madam [21]

Answer:

218.5 N

Explanation:

In order for the sled to be in equilibrium along the vertical direction, the forces acting along this direction must be balanced. So the equilibrium equation is:

N+F sin \theta - mg =0

where

N is the normal force

F = 50 N is the force that pulls the sled

\theta is the angle between the force and the horizontal, so

F sin \theta is the component of F acting along the vertical direction

(mg) is the weight of the sled, with

m = 25 kg being the mass of the sled

g = 9.8 m/s^2 is the acceleration due to gravity

Solving the formula for N, we find

N=mg-F sin \theta = (25 kg)(9.8 m/s^2)-(50 N)sin 32^{\circ}=218.5 N

7 0
3 years ago
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