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statuscvo [17]
3 years ago
8

Using energy considerations, calculate the average force (in N) a 67.0 kg sprinter exerts backward on the track to accelerate fr

om 2.00 to 8.00 m/s in a distance of 25.0 m, if he encounters a headwind that exerts an average force of 30.0 N against him.

Physics
2 answers:
Illusion [34]3 years ago
7 0

Answer:

F_{sprinter}=110.4N

Explanation:

Given data

Mass m=67.0 kg

Final Speed vf=8.00 m/s

Initial Speed vi=2.00 m/s

Distance d=25.0 m

Force F=30.0 N

From work-energy theorem we know that the work done equals the change in kinetic energy

W=ΔK=Kf-Ki=1/2mvf²-1/2mvi²

And

W=F_{total}.d

So

W=1/2mv_{f}^2-1/2mv_{i}^2\\F_{total}=\frac{1/2mv_{f}^2-1/2mv_{i}^2}{d} \\F_{total}=\frac{1/2(67.0kg)(8.00m/s)^2-1/2(67.0kg)(2.00m/s)^2}{25.0m} \\F_{total}=80.4N

and we know that the force the sprinter exerted Fsprinter the force of the headwind Fwind=30.0N

So

F_{sprinter}=F_{total}+F_{wind}\\F_{sprinter}=80.4N+30N\\F_{sprinter}=110.4N  

dsp733 years ago
5 0

Explanation:

Below is an attachment containing the solution.

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3 years ago
An electron moves in a circular path perpendicular to a constant magnetic field with a magnitude of 2.50 mT. The angular momentu
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Answer:

Explanation:

The angular momentum of electron mvR = 6 x 10⁻²⁵  Js

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R = 3.87 x 10⁻² m

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Answer:

\Delta S=1.69J/K

Explanation:

We know,

\eta=1-\frac{T_2}{T_1}=1-\frac{Q_2}{Q_1}      ..............(1)

where,

η = Efficiency of the engine

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substituting the values in the above equation we get

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