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77julia77 [94]
3 years ago
6

A rock is thrown vertically upward from some height above the ground. It rises to some maximum height and falls back to the grou

nd. What one of the following statements is true if air resistance is neglected? The acceleration of the rock is zero when it is at the highest point. The speed of the rock is negative while it falls toward the ground. As the rock rises, its acceleration vector points upward. At the highest point the velocity is zero, the acceleration is directed downward. The velocity and acceleration of the rock always point in the same direction.

Physics
1 answer:
True [87]3 years ago
6 0

Answer:

At the highest point the velocity is zero, the acceleration is directed downward.

Explanation:

This is a free-fall problem, in the case of something being thrown or dropped, the acceleration is equal to -gravity, so -9.80m/s^2. So, the acceleration is never 0 here.

I attached an image from my lecture today, I find it to be helpful. You can see that because of gravity the acceleration is pulled downwards.

At the highest point the velocity is 0, but it's changing direction and that's why there's still an acceleration there.

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7 0
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#1 Not sure where to start. This is for AP Physics!
yaroslaw [1]

First,

\rho=\dfrac mV

where \rho is density, m is mass, and V is volume. We can compute the volume of the roll:

2.7\,\dfrac{\mathrm g}{\mathrm{cm}^3}=\dfrac{1275\,\mathrm g}V

\implies V\approx472.22\,\mathrm{cm}^3\approx4.72\,\mathrm m^3

When the roll is unfurled, the aluminum will be a rectangular box (a very thin one), so its volume will be the product of the given area and its thickness x. Note that we're assuming the given area is not the actual total surface area of the aluminum box, but just the area of the largest face (i.e. the area of one side of the unrolled sheet of aluminum).

So we have

V=Ax

where A is the given area, so

4.72\,\mathrm m^3=\left(18.5\,\mathrm m^2\right)x

\implies x\approx0.255\,\mathrm m=255\,\mathrm{mm}

If we're taking significant digits into account, the volume we found would have been V=470\,\mathrm m^3, in turn making the thickness x=250\,\mathrm{mm}.

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Answer:

b) 1.67×10^7 m/s

Explanation:

The solution is attached in the attachment section

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