Answered

A 1.0 × 10^3-kilogram car travels at a constant

speed of 20. meters per second around a

horizontal circular track. The diameter of the

track is 1.0 × 10^2 meters. The magnitude of the

car’s centripetal acceleration is

(1) 0.20 m/s2 (3) 8.0 m/s2

(2) 2.0 m/s2 (4) 4.0 m/s2

Verified Answer :

TaylorBayley

Centripetal acceleration can be calculated with the formula F=v^2/r

Substituting the values in, we get:

a=20^2/50
=8

Therefore the magnitude of the centripetal acceleration of the car is 8 ms^-2

snehashish65

The required value of car's centripetal acceleration is \(8 \;\rm m/s^{2}\). Hence, option (3) is correct.

Given data:

The mass of car is, \(m = 1.0 \times 10^{3} \;\rm kg\).

The speed of car is, v = 20 m/s.

The diameter of circular track is, \(d = 1.0 \times 10^{2} \;\rm m\).

Let us calculate the radius of circular track first,

\(r = \dfrac{d}{2}\\\\r = \dfrac{1.0 \times 10^{2}}{2}\\\\r = 50 \;\rm m\)

Now, we know that the centripetal acceleration of any object corresponding to the center seeking force is known as centripetal acceleration. And the expression for the centripetal acceleration is given as,

\(a = \dfrac{v^{2}}{r}\)

Solving as,

\(a = \dfrac{(20)^{2}}{50}\\\\a = 8 \;\rm m/s^{2}\)

Thus, we can conclude that the required value of car's centripetal acceleration is \(8 \;\rm m/s^{2}\).

Learn more about the centripetal acceleration here:

https://brainly.com/question/14465119

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