A 1.0 × 10^3kilogram 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 meter
Question
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
2 Answer

1. User Answers 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 
2. User Answers snehashish65
The required value of car's centripetal acceleration is [tex]8 \;\rm m/s^{2}[/tex]. Hence, option (3) is correct.
Given data:
The mass of car is, [tex]m = 1.0 \times 10^{3} \;\rm kg[/tex].
The speed of car is, v = 20 m/s.
The diameter of circular track is, [tex]d = 1.0 \times 10^{2} \;\rm m[/tex].
Let us calculate the radius of circular track first,
[tex]r = \dfrac{d}{2}\\\\r = \dfrac{1.0 \times 10^{2}}{2}\\\\r = 50 \;\rm m[/tex]
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,
[tex]a = \dfrac{v^{2}}{r}[/tex]
Solving as,
[tex]a = \dfrac{(20)^{2}}{50}\\\\a = 8 \;\rm m/s^{2}[/tex]
Thus, we can conclude that the required value of car's centripetal acceleration is [tex]8 \;\rm m/s^{2}[/tex].
Learn more about the centripetal acceleration here:
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