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Vika [28.1K]
3 years ago
7

A box accidentally drops from a truck traveling at a speed of 20.0 m/s and slides along the ground for a distance of 30.0 m Calc

ulate a) The acceleration of the box (assuming it is constant while sliding)
Physics
1 answer:
Hoochie [10]3 years ago
6 0

Answer:

a= - 6.667 m/s²

Explanation:

Given that

The initial speed of the box ,u= 20 m/s

The final speed of the box ,v=  0 m/s

The distance cover by box ,s= 30 m

Lets take the acceleration of the box = a

We know that

v²= u ² + 2 a s

Now by putting the values in the above equation we get

0²=20² + 2 a x 30

a=- \dfrac{20^2}{2\times 30} \ m/s^2

a= - 6.667 m/s²

Negative sign indicates that velocity and acceleration are in opposite direction.

Therefore the acceleration of the box will be  - 6.667 m/s² .

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Answer:

Hi

before I answer a question I think very deeply and try my best, hope it helps...

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3 years ago
A man is standing on the edge of a 20.0 m high cliff. He throws a rock horizontally with an initial velocity of 10.0 m/s.
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Answer:

<em>a. The rock takes 2.02 seconds to hit the ground</em>

<em>b. The rock lands at 20,2 m from the base of the cliff</em>

Explanation:

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The time taken by the object to hit the ground is calculated by:

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\displaystyle d=v.t

The man is standing on the edge of the h=20 m cliff and throws a rock with a horizontal speed of v=10 m/s.

a,

The time taken by the rock to reach the ground is:

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\displaystyle t=\sqrt{4.0816}

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b.

The range is calculated now:

\displaystyle d=10\cdot 2.02

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