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3 years ago

A paper clip that has a mass of 1.5 grams is thrown into the air and initially has a kinetic energy

Physics

1 answer:

Vsevolod [243]3 years ago

3 0

Answer:

$v = 4.2 \ m/s$

Explanation:

Given data:

Mass of the paper clip, $m = 1.5 \ g = 0.0015 \ kg$

Kinetic energy, $K = 0.013 \ \rm J$

Let the velocity of the paper clip when it is thrown be <em>v</em>.

Thus,

$K = \frac{1}{2}mv^{2}$

$0.013 = 0.5 \times 0.0015 \times v^{2}$

$\Rightarrow \ v = 4.16 \ m/s$

$v = 4.2 \ m/s.$ (rounding to nearest tenth)

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A rock is thrown from a bridge at an angle 30∘ below horizontal.immediately after the rock is released, is the magnitude of its

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3 years ago

What is the solution to the problem expressed to the correct number of significant figures?

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Answer:

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The least precise number is the number with the lowest significant numbers:

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3 years ago

A ball of mass 0.160 kg is dropped from a height of 2.25 m. When it hits the ground it compresses 0.087 m.

Studentka2010 [4]

A) 6.64 m/s downward

B) 0.026 s

C) -40.9 N

Explanation:

We can solve this problem by using the law of conservation of energy.

In fact, since the total mechanical energy of the ball must be conserved, this means that the initial gravitational potential energy of the ball before the fall is entirely converted into kinetic energy just before it reaches the floor.

So we can write:

$PE=KE\\mgh = \frac{1}{2}mv^2$

where

m = 0.160 kg is the mass of the ball

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

h = 2.25 m is the initial height of the ball

v is the final velocity of the ball before hitting the ground

Solving for v, we find:

$v=\sqrt{2gh}=\sqrt{2(9.8)(2.25)}=6.64 m/s$

And the direction of the velocity is downward.

The motion of the ballduring the collision is a uniformly accelerated motion (= with constant acceleration), so the time of impact can be found by using a suvat equation:

$s=(\frac{u+v}{2})t$