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3 years ago

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How does collision time affect impact forces

1 answer:

Licemer1 [7]3 years ago

5 0

Answer:

A large increase in collision time can mean decrease in impact force. This is essentially what an air bag in your car does. It increases the time over which it stops you and decreases the force. ... In this case, the piano would have a smaller change in momentum and a smaller impact force.

Explanation:

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Two parallel plates of area 2.34*10-3 M2 have 7.07*10-7C of charge placed on them. A6.62*10-5C charge q1 is placed between the p

ira [324]

Answer:

The force exerted on the $q_1$ is $F = 2.25*10^{3} \ N$

Explanation:

From the question we are told that

The area is $A = 2.34*10^{-3} \ m^2$

The magnitude of charge placed on them is $q = 7.07 * 10^{-7} C$

The charge placed between the plate is $q_1 = 6.62 *10^{-5} C$

The electric field generated around the plate is mathematically represented as

$E = \frac{q}{A \epsilon_o}$

Substituting values

$E = \frac{7.07*10^{-7}}{2.34*10^{-3} * 8.85 *10^{-12}}$

$E = 34*10^{6} \ V/m$

The force exerted the charge $q_1$ is mathematically represented as

$F = q_1 * E$

Substituting values

$F = 6.62 *10^{-5} * 34*10^{6}$

$F = 2.25*10^{3} \ N$

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3 years ago

Two particles execute simple harmonic motion of the same amplitude and frequency along close parallel lines. They pass each othe

deff fn [24]

Answer:

$\theta_2 - \theta_1 = 156.93 degree$

Explanation:

As we know that the displacement of the particle from the mean position is 1/5 times of its amplitude

so we have

$y = A sin\omega t$

$y = \frac{A}{5}$

so now we have

$\frac{A}{5} = A sin\omega t$

now we have

$\theta_1 = 11.53 degree$

so the phase other particle in opposite direction is given as

$\theta_2 = 180 - 11.53 = 168.46 degree$

so we have phase difference given as

$\theta_2 - \theta_1 = 168.46 - 11.53$

$\theta_2 - \theta_1 = 156.93 degree$

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3 years ago

Is O2 classified as a compound?

dybincka [34]

Answer: NO.

Explanation: Oxygen is not a compound. It has only one element in it.

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3 years ago

A quarterback throws a football 40 yards in 4 seconds.what is the average speed the football

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The answer is 10 yards per second

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4 years ago

Read 2 more answers

A 5cm tall object is placed 4cm in front of a converging lens that has a focal length of 8cm. Where is the image located in ____

OverLord2011 [107]

Answer:

a. -8 cm

Explanation:

$d_{o}$ = distance of the object = 4 cm

$d_{i}$ = distance of the image = ?

$f$ = focal length of the converging lens = 8 cm

using the lens equation

$\frac{1}{d_{o}} + \frac{1}{d_{i}} = \frac{1}{f}$

$\frac{1}{4} + \frac{1}{d_{i}} = \frac{1}{8}$

$d_{i}$ = - 8 cm

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3 years ago