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2 years ago

3- A ball of mass 0,50 kg falls and hits the floor at 10m/s,

Physics

1 answer:

Klio2033 [76]2 years ago

4 0

Answer:

3- A ball of mass 0,50 kg falls and hits the floor at 10m/s,

It rebounds at speed 8.0 m/s, as shown.

before collision

after collision

10 m/s

8 m/s

The collision between the ball and the floor lasts for 0.50 s.

What is the average force acting on the ball during the collision?

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Se lanza una piedra de 3.00 N verticalmente hacia arriba desde el suelo. Se observa que, cuando está 15.0 m sobre el suelo, viaj

BARSIC [14]

Answer:

(a). The speed at the moment of being thrown is 30.41 m/s.

(b). The maximum height is 47.18 m.

Explanation:

Given that,

Weight of stone = 3.00 N

Height = 15 m

Speed = 25.0 m/s

(a). We need to calculate the speed at the moment of being thrown

Using work energy theorem

$W=\dfrac{1}{2}m(v_{2}^2-v_{1}^2)$

$-mg\times d=\dfrac{1}{2}m(v_{2}^2-v_{1}^2)$

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$-9.8\times15=\dfrac{1}{2}\times(v_{2}^2-v_{1}^2)$

$-2\times9.8\times15=25^2-v_{1}^2$

$-v_{1}^2=-300-25^2$

$v_{1}=\sqrt{925}$

$v_{1}=30.41\ m/s$

(b). We need to calculate the maximum height

Using work energy theorem

$[tex]W=\dfrac{1}{2}mv_{2}^2-\dfrac{1}{2}mv_{1}^2$

$mg\times d=\dfrac{1}{2}mv_{2}^2-\dfrac{1}{2}mv_{1}^2$

Here, $\dfrac{1}{2}mv_{2}^2$ =0

$-(mg)\times d=\dfrac{1}{2}mv_{1}^2$

$d=\dfrac{v_{1}^2}{2g}$

Put the value into the formula

$d=\dfrac{30.41^2}{2\times9.8}$

$d=47.18\ m$

Hence, (a). The speed at the moment of being thrown is 30.41 m/s.

(b). The maximum height is 47.18 m.

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