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A 805 kg test car travels around a 4.25 km circular track. If the magnitude of the force that maintains the car’s circular motion is 3140 N, what is the car’s tangential speed?

Respuesta :

Answer:

v = 51.36 m/s

Explanation:

given,

mass of the car = 805 Kg

circumference of the track = 4.25 Km

Force on the car in circular path, F = 3140 N

tangential speed of the car = ?

2 π r = 4250

r = 676.4 m

using formula

[tex]F = \dfrac{mv^2}{r}[/tex]

[tex]3140 = \dfrac{805\times v^2}{676.4}[/tex]

v² = 2638.38

v = 51.36 m/s

hence, tangential speed of the car is equal to 51.36 m/s

Answer:

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Explanation:

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