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3 years ago

7

A student throws a 130 g snowball at 6.5 m/s at the side of the schoolhouse, where it hits and sticks. What is the magnitude of

the average force on the wall if the duration of the collision is 0.18 s?

1 answer:

Alex73 [517]3 years ago

8 0

Answer:

4.7 N

Explanation:

130 g = 0.13 kg

The momentum of the snowball when it's thrown at the wall is

$p = mv = 0.13*6.5 = 0.845 kgm/s$

Which is also the impulse. From here we can calculate the magnitude of the average force F knowing the duration of the collision is 0.18 s

$p = F\Delta t$

$F*0.18 = 0.845$

$F = 0.845 / 0.18 = 4.7 N$

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11 months ago

A. Calculate the electric potential energy stored in a capacitor that stores <img src="https://tex.z-dn.net/?f=3.40%20x%2010%5E%

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3 years ago

Read 2 more answers

The cylindrical tub of a dryer in a laundromat rotates counterclockwise about a horizontal axis at 41.5 rev/min as it dries the

frozen [14]

Answer:

$\theta = 49.81^0$

Explanation:

Given that:

$\omega = 41.5 \ rev/min\\\\\omega = 41.5 *\frac{1}{60}* 2 \pi\\\\\omega = 4.45 \ rad/s\\\\\\diameter = 0.748 m$

If we let the piece of the close lose contact at ∠θ;

Then ; from force balance;

we have:

$\\\\mg sin \theta = \frac{mv^2}{r}\\\\sin \theta = \frac{2v^2}{dg}\\\\\theta = sin^{-1} (\frac{2v^2}{dg})$

where;

$v = \frac{\omega d}{2}\\\\v = \frac{4.45 *0.748}{2}\\\\v = 1.6643\\\\v^2 = 2.77$

Again:

$\theta = sin^{-1}(\frac{2v^2}{dg})\\\\\theta = sin^{-1}( \frac{2*2.77}{0.74*9.8})\\\\\theta = 49.81^0$

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2 years ago