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Varvara68 [4.7K]
3 years ago
8

A car, X, with a mass of 1500 kg is moving west with an acceleration of 0.50 m/s2. A second car, Y, of equal mass is heading eas

t with an acceleration of 1.0 m/s2. Which of the following statements best describes the net force acting on each car?
Physics
1 answer:
ehidna [41]3 years ago
8 0

Answer:

A

Explanation:

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The force generated by a single muscle fiber can be increased by increasing is called
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The force generated by a single muscle fiber can be increased by increasing the frequency of action potentials

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What is the name of the device that regulates the amount of fuel going into the engine of a personal watercraft (pwc)?
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2 years ago
A 500 kg dragster accelerates from rest to a final speed of 100 m/s in 400 m (about a quarter of a mile) and encounters an avera
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In order to accelerate the dragster at a speed v_f = 100 m/s, its engine must do a work equal to the increase in kinetic energy of the dragster. Since it starts from rest, the initial kinetic energy is zero, so the work done by the engine to accelerate the dragster to 100 m/s is
W= K_f - K_i = K_f =  \frac{1}{2}mv_f^2=2.5 \cdot 10^6 J

however, we must take into account also the fact that there is a frictional force doing work against the dragster, and the work done by the frictional force is:
W_f = F_f d = -(1200 N)(400 m)= -4.8 \cdot 10^5 J
and the sign is negative because the frictional force acts against the direction of motion of the dragster.

This means that the total work done by the dragster engine is equal to the work done to accelerate the dragster plus the energy lost because of the frictional force, which is -W_f:
W_t = W + (-W_f)=2.5 \cdot 10^6 J+4.8 \cdot 10^5 J=2.98 \cdot 10^6 J

So, the power delivered by the engine is the total work divided by the time, t=7.30 s:
P= \frac{W}{t}= \frac{2.98 \cdot 10^6 J}{7.30 s}=4.08 \cdot 10^6 W

And since 1 horsepower is equal to 746 W, we can rewrite the power as
P=4.08 \cdot 10^6 W \cdot  \frac{1 hp}{746 W} =547 hp



3 0
3 years ago
Read 2 more answers
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