For this trajectory, what would the vertical component of acceleration for the module be at time tm=t0−σ=325s? Recall that accel
1 answer:
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The correct answer is
B It increases.
In fact, the kinetic energy of a moving object is given by
where m is the mass of the object and v is its speed. We see that the kinetic energy is proportional to the mass and proportional to the square of the speed: in this problem, the speed of the object remains the same, while its mass increases, therefore the kinetic energy will increase as well.
B It increases.
In fact, the kinetic energy of a moving object is given by
where m is the mass of the object and v is its speed. We see that the kinetic energy is proportional to the mass and proportional to the square of the speed: in this problem, the speed of the object remains the same, while its mass increases, therefore the kinetic energy will increase as well.
Answer:
Power, P = 600 watts
Explanation:
It is given that,
Mass of sprinter, m = 54 kg
Speed, v = 10 m/s
Time taken, t = 3 s
We need to find the average power generated. The work done divided by time taken is called power generated by the sprinter i.e.
Work done is equal to the change in kinetic energy of the sprinter.
P = 900 watts
So, the average power generated by the sprinter is 900 watts. Hence, this is the required solution.