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3 years ago

Calculate the workdone to stretch an elastic string by 40cm if a force of 10 newton produces an extension of 4cm in it

Physics

1 answer:

notka56 [123]3 years ago

5 0

Answer:

Workdone = 20 Joules

Explanation:

Given the following data;

Force = 10N

Extension, e = 4cm to meters = 4/100 = 0.04 meters

Workdone extension = 40cm to meters = 40/100 = 0.4 meters

To find the work done;

First of all, we would find the spring constant using the formula;

Force = spring constant * extension

10 = spring constant * 0.04

Spring constant = 10/0.04

Spring constant = 250 N/m

Next, we find the work done;

Workdone = ½ke²

Where;

k is the spring constant.

e is the extension.

Substituting into the formula, we have;

Workdone = ½ * 250 * 0.4²

Workdone = 125 * 0.16

Workdone = 20 Joules

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A Ping-Pong ball moving East at a speed of 4 m/s collides with a stationary bowling ball. The Ping-Pong ball bounces back to the

In-s [12.5K]

Answer:

They experience the same magnitude impulse

Explanation:

We have a ping-pong ball colliding with a stationary bowling ball. According to the law of conservation of momentum, we have that the total momentum before and after the collision must be conserved:

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where

$p_p$ is the initial momentum of the ping-poll ball

$p_b$ is the initial momentum of the bowling ball (which is zero, since the ball is stationary)

$p'_p$ is the final momentum of the ping-poll ball

$p'_f$ is the final momentum of the bowling ball

We can re-arrange the equation as follows

$p_p - p'_p = p_b'-p_b$

$-\Delta p_p = \Delta p_b$

which means

$|\Delta p_p | = |\Delta p_b|$ (1)

so the magnitude of the change in momentum of the ping-pong ball is equal to the magnitude of the change in momentum of the bowling ball.

However, we also know that the magnitude of the impulse on an object is equal to the change of momentum of the object:

$I=\Delta p$ (2)

Therefore, (1)+(2) tells us that the ping-pong ball and the bowling ball experiences the same magnitude impulse:

$|I_p| = |I_b|$

4 0

4 years ago

If raindrops are falling vertically at 8.30 m/s, what angle from the vertical do they make for a person jogging at 2.07 m/s?

dmitriy555 [2]

Answer:

∆ = 14°

Explanation:

Let the angle be ∆

Opposite = 2.07 m/s

Adjacent = 8.3 m/s

Tan ∆ = opposite / adjacent

Tan ∆ = 2.07 / 8.3

Tan ∆ = 0.2494

∆ = ArcTan (0.2494)

∆ = 14°

The angle from vertical the raindrops make for a person jogging is 14°

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