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azamat
2 years ago
10

A cart of mass m = 0.12 kg moves with a speed v = 0.45 m/s on a frictionless air track and collides with an identical cart that

is stationary. The carts stick together after the collision. What is the final kinetic energy of the system?
Physics
1 answer:
lina2011 [118]2 years ago
5 0

Answer:

0.006075Joules

Explanation:

The final kinetic energy of the system is expressed as;

KE = 1/2(m1+m2)v²

m1 and m2 are the masses of the two bodies

v is the final velocity of the bodies after collision

get the final velocity using the law of conservation of momentum

m1u1 + m2u2 = (m1+m2)v

0.12(0.45) + 0/12(0) = (0.12+0.12)v

0.054 = 0.24v

v = 0.054/0.24

v = 0.225m/s

Get the final kinetic energy;

KE = 1/2(m1+m2)v

KE = 1/2(0.12+0.12)(0.225)²

KE = 1/2(0.24)(0.050625)

KE = 0.12*0.050625

KE = 0.006075Joules

Hence the final kinetic energy of the system is 0.006075Joules

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Answer:

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As we know by Newton's II law that net force on the system of mass is given as product of mass and acceleration

Here we know that

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What is the definition of cell ?
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2 years ago
The suspension cable of a 1,000 kg elevator snaps, sending the elevator moving downward through its shaft. The emergency brakes
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Answer:

option (E) 1,000,000 J

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3 years ago
a certain projetor uses a concave mirror for projecting an object's image on a screen .it produces on image that is 4 times bigg
IrinaK [193]

Answer:

f = 1 m

Explanation:

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M = Magnification = 4

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Therefore,

4 = \frac{5\ m}{p}\\\\p = \frac{5\ m}{4}\\\\p = 1.25\ m

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2 years ago
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