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12 days ago

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Discuss how the hardness or softness of the landing surface is related to the time required to stop the egg

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A car traveling at 70 mph70 mph down the interstate collides with a bug trying to cross the highway. Which of the following stat

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The force exerted by the car on the bug is identical to the force the bug applies back on the car.

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For lunch you and your friends decide to stop at the nearest deli and have a sandwich made fresh for you with 0.300 kg of Italia

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Answer:

Part a)

A = 0.0581 m

Part b)

T = 0.37 s

Explanation:

A slice is dropped onto the plate from a height of 0.250 m,

therefore the speed of the slice upon impact is calculated as

$v = \sqrt{2gh}$

We know that

$v = \sqrt{2(9.81)(0.250)}$

$v = 2.21 m/s$

Now applying the conservation of momentum:

$mv = (m + M)v_f$

$m = 0.300 kg$

$M = 0.400 kg$

From this equation, we find:

$0.300 (2.21) = (0.300 + 0.400) v_f$

$v_f = 0.95 m/s$

$0.400 (9.81) = 200 x_1$

When the slice rests on the plate, the new mean position can be expressed as

$x_1 = 0.01962 m$

$(0.300 + 0.400)9.81 = 200 x_2$

We also determine that the speed of SHM is represented as

$x_2 = 0.0343 m$

Here, we derive values from

$v = \omega\sqrt{A^2 - x^2}$

$\omega = \sqrt{\frac{k}{m + M}}$

$\omega = \sqrt{\frac{200}{0.300 + 0.400}}$

$\omega = 16.9 rad/s$

$a = x_2 - x_1 = 0.0343 - 0.01962 = 0.0147 m$

Using the previous formula gives:

$0.95 = 16.9\sqrt{A^2 - 0.0147^2}$

$A = 0.0581 m$

Part b)

The time period for the scale is computed as

$T = \frac{2\pi}{\omega}$

$T = \frac{2\pi}{16.9}$

$T = 0.37 s$

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3 months ago