What is the absolute value of the horizontal force that each athlete exerts against the ground?
1 answer:
Refer to the diagram shown below.
When an athlete is in motion, he/she exerts a vertical force (the person's weight, W) on the ground. The ground exerts an equal and opposite force, N, the normal reaction on the athlete, so that W = N.
At the same time, the ground exerts a horizontal force, F, o n the athlete so that he/she does not slip.
The magnitude of the horizontal force is
F = μN = μW
where μ = the dynamic coefficient of friction.
Answer:
The horizontal force is μW,
where
W = the weight of the athlete and,
μ = the dynamic coefficient of friction.
When an athlete is in motion, he/she exerts a vertical force (the person's weight, W) on the ground. The ground exerts an equal and opposite force, N, the normal reaction on the athlete, so that W = N.
At the same time, the ground exerts a horizontal force, F, o n the athlete so that he/she does not slip.
The magnitude of the horizontal force is
F = μN = μW
where μ = the dynamic coefficient of friction.
Answer:
The horizontal force is μW,
where
W = the weight of the athlete and,
μ = the dynamic coefficient of friction.
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Answer:
32.76 Volt
Explanation:
frequency, f = 400 Hz
Area of crossection, A = 13 cm²
Maximum flux density, B = 0.9 tesla
Number of turns in secondary coil, N = 70
Let the maximum induced voltage is e.
According to the Faraday's law of electromagnetic induction, the induced emf is equal to the rate of change of magnetic flux.
e = dФ/dt
Time is defined as the reciprocal of frequency.
So, e = N B A f
e = 70 x 0.9 x 13 x 10^-4 x 400
e = 32.76 volt