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liberstina [14]
3 years ago
12

Two cars (with masses 2000 kg and 1000 kg) collide head on an icy road. Before the collision, the more massive vehicle was movin

g at 10 m/s to the right and the less massive vehicle was moving at 15 m/s to the left. Immediately after the collision, the more massive vehicle was moving at 7.0 m/s to the left. How fast was the less massive Vehicle immediately after the collision?
Physics
1 answer:
mezya [45]3 years ago
5 0

Answer:

v₂' = 19 m/s

Explanation:

given,                    

mass of car one = m₁ = 2000 Kg

mass of car two = m₂ = 1000 Kg

velocity of car one in right (v₁) = 10 m/s

velocity of car two in left (v₂)= 15 m/s

after collision                        

velocity(v₁') = 7 m/s

v₂' = ?                            

using conservation of momentum

m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'

m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'

2000 x 10 - 1000 x 15 = - 2000 x 7 + 1000 v₂'

1000 v₂' = 19000

v₂' = 19 m/s

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Answer and Explanation: Kinetic energy is related to movement: it is the energy an object possesses during the movement. it is calculated as:

K=\frac{1}{2}mv^{2}

For the object thrown in the air:

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K=(-9.8t+24)^{2}

K=96.04t^{2}-470.4t+576

Kinetic energy of the object as a function of time: K=96.04t^{2}-470.4t+576

Potential energy is the energy an object possesses due to its position in relation to other objects. It is calculated as:

U=mgh

For the object thrown in the air:

U=9.8.2.h(t)

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U=-96.04t^{2}+470.4t+1176

Potential energy as function of time: U=-96.04t^{2}+470.4t+1176

Total kinetic and potential energy, also known as mechanical energy is

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A spherical capacitor contains a charge of 3.50 nC when connected to a potential difference of 210.0 V. Its plates are separated
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Incomplete question as we have not told to find what quantity.The complete question is here

A spherical capacitor contains a charge of 3.50 nC when connected to a potential difference of 210.0 V. Its plates are separated by vacuum and the inner radius of the outer shell is 5.00 cm.calculate: (a) the capacitance; (b) the radius of the inner sphere; (c) the electric field just outside the surface of the inner sphere.

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Explanation:

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Q=3.50nC\\V=210V\\r_{b}=5.0cm

For part (a)

The Capacitance given by:

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For part (b)

The Capacitance of coordinates is given as

C=\frac{4\pi e}{\frac{1}{r_{a} }-\frac{1}{r_{b} } }\\ So\\{\frac{1}{r_{a} }-\frac{1}{r_{b} } }=\frac{4\pi *8.85*10^{-12} }{1.666*10^{-11}}=6.672m^{-1} \\ \frac{1}{r_{a} }=6.672+(1 /0.05)\\\frac{1}{r_{a} }=26.672\\r_{a} =1/26.672\\r_{a} =0.0375m\\r_{a} =3.749cm

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