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2 years ago

What is the value of acceleration due to gravity at the pole, equator and the centre of the earth

Physics

1 answer:

fgiga [73]2 years ago

7 0

Answer:

In combination, the equatorial bulge and the effects of the surface centrifugal force due to rotation mean that sea-level gravity increases from about 9.780 m/s2 at the Equator to about 9.832 m/s2 at the poles, so an object will weigh approximately 0.5% more at the poles than at the Equator.

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In the "before" part of Fig. 9-60, car A (mass 1100 kg) is stopped at a traffic light when it is rear-ended by car B (mass 1400

liq [111]

Complete Question:

In the "before" part of Fig. 9-60, car A (mass 1100 kg) is stopped at a traffic light when it is rear-ended by car B (mass 1400 kg). Both cars then slide with locked wheels until the frictional force from the slick road (with a low ?k of 0.15) stops them, at distances dA = 6.1 m and dB = 4.4 m. What are the speeds of (a) car A and (b) car B at the start of the sliding, just after the collision? (c) Assuming that linear momentum is conserved during the collision, find the speed of car B just before the collision.

Answer:

a) Speed of car A at the start of sliding = 4.23 m/s

b) speed of car B at the start of sliding = 3.957 m/s

c) Speed of car B before the collision = 7.28 m/s

Explanation:

NB: The figure is not provided but all the parameters needed to solve the question have been given.

Let the frictional force acting on car A, $f_{ra} = \mu mg\\$ ............(1)

Since frictional force is a type of force, we are safe to say $f_{ra} = ma$ .......(2)

Equating (1) and (2)

$ma = \mu mg\\a = \mu g\\\mu = 0.15\\a = 0.15 * 9.8 = 1.47 m/s^{2}$

a) Speed of A at the start of the sliding

$d_{A} = 6.1 m\\Speed of A at the start of sliding, v_{A} = \sqrt{2ad_{A} }\\ v_{A} = \sqrt{2*1.47*6.1 } \\v_{A} = \sqrt{17.934 } \\v_{A} = 4.23 m/s$

b) Speed of B at the start of the sliding

$d_{A} = 4.4 m\\Speed of A at the start of sliding, v_{B} = \sqrt{2ad_{B} }\\ v_{B} = \sqrt{2*1.47*4.4 } \\v_{B} = \sqrt{12.936 } \\v_{B} = 3.957 m/s$

Let the speed of car B before collision = $v_{B1}$

Momentum of car B before collision = $m_{B} v_{B1}$

Momentum after collision = $m_{A} v_{A} + m_{B} v_{B2}$

Applying the law of conservation of momentum:

$m_{B} v_{B1} = m_{A} v_{A} +m_{B} v_{B2}$

$m_{A} = 1100 kg\\m_{B} = 1400 kg$

$(1400*v_{B1} ) = (1100 * 4.23) + ( 1400 * 3.957)\\(1400*v_{B1} ) = 10192.8\\v_{B1} = 10192.8/1400\\v_{B1 = 7.28 m/s$

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