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dsp73
3 years ago
7

A car with a mass of 880 kg is moving on the road with constant velocity. The force exerted by engine is 1600N. What is the fric

tional force between the road an the car?​
Physics
1 answer:
vovangra [49]3 years ago
4 0

Answer:

1600 N

Explanation:

F-F' = ma.................. Equation 1

Where F = Force exerted by the engine, F' = Frictional Force, m = mass of the car, a = acceleration of the car.

make F' the subject of the equation

F' = F-ma............... Equation 2

Given: F = 1600 N, m = 880 kg, a = 0 m/s (moving with a constant velocity).

Substitite these values into equation 2

F' = 1600+880(0)

F' = 1600 N.

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3 0
2 years ago
awhite billiard ball with mass mw = 1.47 kg is moving directly to the right with a speed of v = 3.01 m/s and collides elasticall
pochemuha

Answer:

speed of white ball is 1.13 m/s and speed of black ball is 2.78 m/s

initial kinetic energy = final kinetic energy

KE = 6.66 J

Explanation:

Since there is no external force on the system of two balls so here total momentum of two balls initially must be equal to the total momentum of two balls after collision

So we will have

momentum conservation along x direction

m_1v_{1i} + m_2v_{2i} = m_1v_{1x} + m_2v_{2x}

now plug in all values in it

1.47 \times 3.01 + 0 = 1.47 v_1cos68 + 1.47 v_2cos22

so we have

3.01 = 0.375v_1 + 0.927v_2

similarly in Y direction we have

m_1v_{1i} + m_2v_{2i} = m_1v_{1y} + m_2v_{2y}

now plug in all values in it

0 + 0 = 1.47 v_1sin68 - 1.47 v_2sin22

so we have

0 = 0.927v_1 - 0.375v_2

v_2 = 2.47 v_1

now from 1st equation we have

3.01 = 0.375 v_1 + 0.927(2.47 v_1)

v_1 = 1.13 m/s

v_2 = 2.78 m/s

so speed of white ball is 1.13 m/s and speed of black ball is 2.78 m/s

Also we know that since this is an elastic collision so here kinetic energy is always conserved to

initial kinetic energy = final kinetic energy

KE = \frac{1}{2}(1.47)(3.01^2)

KE = 6.66 J

5 0
4 years ago
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