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tatyana61 [14]
3 years ago
8

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:

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Then acceleration = change in velocity/Time.

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

Acceleration = (9-0)/3=9/3=3 m/s².

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Taiga is the answer.

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