Physics, asked by salomesalosalome123, 7 months ago

A car moving at a steady 10 m/s on a level highway encounters a bump that has a circular cross-section with a radius of 30 m. The car maintains its speed over the bump. a) What is the centripetal acceleration of the car? (b) What is the normal force exerted by the seat of the car on a 60.0-kg passenger when the car is at the top of the bump?

Answers

Answered by kennys24
3
Answer:
(a) 388 newton
(b) 17.15 m/s
Explanation:
r = 30 m
v = 10 m/s
m = 60 kg
(a) Let N be the normal reaction.
At the bump
N = mg - mv^2 / r
N = 60 x 9.8 - 60 x 10 x 10 / 30 = 588 - 200 = 388 newton
(b) Then the contact loose, N = 0
So, mg = mv^2 / r
v^2 = r x g = 30 x 9.8 = 294
v = 17.15 m/s
Answered by syedtahir20
0

As we know that,

Given:

A car moving at a steady 10 m/s on a level highway.

with a radius of 30 m.

Find out:

a) What is the centripetal acceleration of the car?

(b) What is the normal force exerted by the seat of the car on a 60.0-kg passenger when the car is at the top of the bump?

Let us:

r = 30 m

v = 10 m/s

m = 60 kg

so,

a) N be the normal reaction.

At the bump

N = mg - mv^\frac{2}{r}

N = 60 x 9.8 - 60 x 10x\frac{10}{30}

= 588 - 200

= 388newton

The centripetal acceleration of the car = 388newton

then,

b) the contact loose, N = 0

mg = mv^\frac{2}{r}

v^2 =r x g = 30 x 9.8 = 294

v = 17.15 m/s

The normal force 17.15m/s

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