A student drops a ball from a slope. He notices the ball rolling down as shown in the image.

Physics

1 answer:

zavuch27 [327]2 years ago

5 0

Answer:

You are correct, it is (3) the ball is in motion as it rolls down the slope.

Explanation:

For number 1 it would make sense because how would the student be in motion if he is just standing there throwing the ball down the slope, not himself.

For number 2 still wouldn't make so much sense since we are talking about the ball being in motion not the student.

For number 3 it seems the most reasonable because the ball is in motion as it is rolling down the slope.

For number 4 it doesn't make sense for a ball to lose its shape just by rolling, also even if you're like: "oh, but what if its leaking air or has a hole??" well easy then. It STILL wouldn't matter because in the sentence it never specifies it has a "hole" or is "leaking air" therefore this answer is wrong.

YOUR ANSWER IS number 3.

You might be interested in

What substance is used by plants as a source of food

harkovskaia [24]

Answer:

dart

Explanation:

dart and sun and water so that the plant be okay

7 0

3 years ago

A friend insists that electric current is the same as electrical energy. Work with a partner to develop a rebuttal

kotykmax [81]

Answer: The electrons flowing through the wire are referred to as a quantity of electricity, and the flow of electricity is referred to as “an electric current.”

Explanation: Hope it Helps have a blessed day

4 0

3 years ago

child slides down a snow‑covered slope on a sled. At the top of the slope, her mother gives her a push to start her off with a s

Strike441 [17]

Answer:

θ = 13.7º

Explanation:

According to the work-energy theorem, the change in the kinetic energy of the combined mass of the child and the sled, is equal to the total work done on the object by external forces.
The external forces capable to do work on the combination of child +sled, are the friction force (opposing to the displacement), and the component of the weight parallel to the slide.
As this last work is just equal to the change in the gravitational potential energy (with opposite sign) , we can write the following equation:

$\Delta K + \Delta U = W_{nc} (1)$

ΔK, is the change in kinetic energy, as follows:

$\Delta K = \frac{1}{2}* m* (v_{f} ^{2} - v_{0} ^{2}) (2)$

ΔU, is the change in the gravitational potential energy.
If we choose as our zero reference level, the bottom of the slope, the change in gravitational potential energy will be as follows:

$\Delta U = 0 - m*g*h = -m*g*d* sin\theta (3)$

Finally, the work done for non-conservative forces, is the work done by the friction force, along the slope, as follows:

$W_{nc} = F_{f} * d * cos 180\º \\\\ = 0.2*m*g*d* cos 180\º = -0.2*m*g*d (4)$

Replacing (2), (3), and (4) in (1), simplifying common terms, and rearranging, we have:

$\frac{1}{2}* (v_{f} ^{2} - v_{0} ^{2}) = g*d* sin\theta -0.2*g*d$

Replacing by the givens and the knowns, we can solve for sin θ, as follows: $\frac{1}{2}*( (4.30 m/s) ^{2} - (0.75 m/s)^{2}) = 9.8 m/s2*25.5m* sin\theta -0.2*9.8m/s2*25.5m\\ \\ 8.56 (m/s)2 = 250(m/s)2* sin \theta -50 (m/s)2\\ \\ sin \theta = \frac{58.6 (m/s)2}{250 (m/s)2} = 0.236$ ⇒ θ = sin⁻¹ (0.236) = 13.7º