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3 years ago

How does collision time affect impact forces

Physics

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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1. The volume of a given mass of gas is 20cm when its

Pavel [41]

Answer:

Explanation:

The way to show a cubed substance is either like this³ or like this x^3. The small three is found at the bottom toolbar at the bottom of the question space marked by the Ω symbol.

100 mmHg

Givens

V1 = 20 cm^3

V2 = 80 cm^3

P1 = 400 mmHg

P2 = ?

Formula

V1 * P1 = V2 * P2

Solution

20 * 400 = 80 * P2 Divide by 80

20 * 400/80 = P2

P2 = 8000 / 80

P2 = 100 mmHg

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3 years ago

A small sandbag is dropped from rest from a hovering hot-air balloon. (assume the positive direction is upward.) (a) after 1.5 s

marishachu [46]

solution:

We know v0 = 0, a = 9.8, t = 4.0. We need to solve for v

so,

we use the equation:

v = v0 + at

v = 0 + 9.8*4.0

v = 39.2 m/s

Now we just need to solve for d, so we use the equation:

d = v0t + 1/2*a*t^2

d = 0*4.0 + 1/2*9.8*4.0^2

d = 78.4 m

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3 years ago

once you decide that it is safe to move out of traffic, what steps should you follow to ensure safe passage?

Solnce55 [7]

First put your turn signal on, next check for any ongoing traffic and wait until it is clear lastly start to drift into the lane you need to clear away from traffic

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3 years ago

The low-frequency speaker of a stereo set has a surface area of 0.06 m2 and produces 2.03 W of acoustical power. What is the int

zimovet [89]

Answer:

33.83W/m²

Explanation:

The intensity of the speake at the surface is

I = P/A

I = 2.03W / 0.06m²

I = 33.83W/m²

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3 years ago

A soft tennis ball is dropped onto a hard floor from a height of 1.50 m and rebounds to a height of 1.10 m. (a) Calculate its ve

Gemiola [76]

Answer:

(a) v = 5.42m/s

(b) vo = 4.64m/s

(d) Δy = 5.11*10^-3m

Explanation:

(a) The velocity of the ball before it hits the floor is given by:

$v=\sqrt{2gh}$ (1)

g: gravitational acceleration = 9.8m/s^2

h: height where the ball falls down = 1.50m

$v=\sqrt{2(9.8m/s^2)(1.50m)}=5.42\frac{m}{s}$

The speed of the ball is 5.42m/s

(b) To calculate the velocity of the ball, after it leaves the floor, you use the information of the maximum height reached by the ball after it leaves the floor.

You use the following formula:

$h_{max}=\frac{v_o^2}{2g}$ (2)