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3 years ago

Describe another real-world example of a perfectly inelastic collision

Physics

2 answers:

Sidana [21]3 years ago

8 0

1. Newtons Cradle
2. When you play pool and one ball transfers all its momentum, making the first ball come to a halt.

Ivahew [28]3 years ago

5 0

<span>An example of a perfectly elastic collision is that it doesn't exist. But for the sake of saying it exist you would assume that a collision takes place where energy is completely conserved and is not lost as (non-usable energy) heat. Maybe you say that a gas molecule collides with a completely (perfectly) smooth surface and the gas molecules collides perpendicular to the surface (does not collide at an angle). The gas molecule then bounces off the wall with the same speed as when it colliding against the wall. This would assume a perfectly elastic collision as KE=1/2mv^2. When your gas molecule collides its mass doesn't change but when you say that the gas molecule had the same velocity when it bounces off the wall you ahve assumed that kinetic energy is maintained (which assumes perfectly elastic).</span>

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A 22 µF capacitor charged to 0.7 kV and a second 115 µF capacitor charged to 5.5 kV are connected to each other, with the positi

vesna_86 [32]

Answer:

0.099C

Explanation:

First, we need to get the common potential voltage using the formula

$V=\frac {C_2V_2-C_1V_1}{C_1+C_2}$

Where V is the common voltage, C and V represent capacitance and charge respectively. Subscripts 1 and 2 to represent the the first and second respectively. Substituting the above with the following given values then

$C_1=22\times 10^{-6} F\\ C_2=115\times 10^{-6} F\\ V_1= 0.7\times 10^{3}\\V_2=5.5\times 10^{3}$

Therefore

$V=\frac {115\times 10^{-6}\times 5.5\times 10^{3}-22\times 10^{6}\times 0.7\times 10^{3}}{22\times 10^{-6}+115\times 10^{-6}}=4504.3795620437$

Charge, Q is given by CV hence for the first capacitor charge will be $Q_1=C_1V$

Here, $Q_1=22\times 10^{-6}\times 4504.3795620437=0.0990963503649C\approx 0.099C$

8 0

3 years ago

1. A cyclist accelerates from 0 m/s to 9 m/s in 3 seconds. What is his<br>acceleration?

Lapatulllka [165]

The cyclist accelerates from 0 m/s to 9 m/s in 3 seconds with an acceleration of 3 m/s².

Answer:

Explanation:

Acceleration exerted by an object is the measure of change in speed or velocity of that object with respect to time. So the initial and final velocities play a major role in determining the acceleration of the cyclist. As here the initial velocity of the cyclist is the speed at rest and that is given as 0 m/s. Then after 3 seconds, the velocity of the cyclist changes to 9 m/s.

Then acceleration = change in velocity/Time.

$Acceleration = \frac{Change in velocity}{Time taken}$

Acceleration = (9-0)/3=9/3=3 m/s².

So the cyclist accelerates from 0 m/s to 9 m/s in 3 seconds with an acceleration of 3 m/s².

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sveticcg [70]

Taiga is the answer.

Hope it helps!

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3 years ago

What are the differences and the similarities between psychology and the social science

tiny-mole [99]

Answer:

psychology has to do with the mind and brain while social science has to do with the scientific study if human relationships

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3 years ago

What is night vision pls explain in long answers use<br> can use google

bulgar [2K]

Answer:

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3 years ago