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Sergio039 [100]
4 years ago
6

I WIL A 2.00-kg object traveling east at 20.0 m/s collides with a 3.00-kg object traveling west at 10.0 m/s. After the collision

, the 2.00-kg object has a velocity 5.00 m/s to the west. How much kinetic energy was lost during the collision? . .
Physics
1 answer:
pishuonlain [190]4 years ago
4 0
Using the following given values:

Object 1:
Mass = M1 = 2 kg
Velocity before collision = Vb1 = 20 m/s
Velocity after collision = Va1 = -5 m/s 

Object 2:
Mass = M2 = 3 kg
Velocity before collision = Vb2 = -10 m/s
Velocity after collision = Va2 = ? m/s<span> 
</span>
Obtaining Va2 via law of conservation of momentum:

total momentum after collision = total momentum before collision
M1 * Va1 + M2 * Va2 = M1 * Vb1 + M2 * Vb2
2*-5 + 3Va2 = 2*20 + 3*-10
Va2 = 6.67

Total kinetic energy before collision:

KE1 = (1/2)*M1*Vb1^2 + (1/2)*M2*Vb2^2
<span>KE1 = (1/2)*2*(20)^2 + (1/2)*3*(-10)^2
KE1 = 550 J

</span>Total kinetic energy after collision:

KE2 = (1/2)*M1*Va1^2 + (1/2)*M2*Va2^2
<span>KE2 = (1/2)*2*(-5)^2 + (1/2)*3*(6.67)^2
KE2 = 91.73 J
</span>
Total kinetic energy lost:

Energy lost = KE1 - KE2 = 550 - 91.73 = 458.27 J
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Explanation:

Given that

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That is why

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Answer:

The charge density in the system is 4.25*10^4C/m

Explanation:

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Our data given correspond to:

r=1*10^{-3}m\\v = 5.2*10^{-4}m/s\\e= 1.6*10^{-19}C

We need to asume here the number of free electrons in a copper conductor, at which is generally of 8.5 *10^{28}m^{-3}

The equation to find the current is

I = VenA

Where

I =Current

V=Velocity

A = Cross-Section Area

e= Charge for a electron

n= Number of free electrons

Then replacing,

I = (5.2*10^{-4})(1.6*10^{-19})(88.5 *10^{28})(\pi(1*10^{-3})^2)

I= 22.11a

Now to find the linear charge density, we know that

I = \frac{Q}{t} \rightarrow Q = It

Where:

I: current intensity

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t: time in which electrical charges circulate through the conductor

And also that the velocity is given in proportion with length and time,

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The charge density is defined as

\lambda = \frac{Q}{l}\\\lambda = \frac{It}{V_d t}\\\lambda = \frac{I}{V_d}

Replacing our values

\lambda = \frac{22.11}{5.20*10{-4}}

\lambda= 4.25*10^4C/m

Therefore the charge density in the system is 4.25*10^4C/m

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