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marissa [1.9K]
3 years ago
15

a particle of mass 1.3kg is sliding down a frictionless slope inclined at 30 to the horizontal. the acceleration of the particle

down the slope is
Physics
1 answer:
OverLord2011 [107]3 years ago
4 0

Answer:

4.9\ m/sec^2

Explanation:

The computation of acceleration of the particle down the slope is shown below:-

data provided in the question

Particle of mass = 1.3 kg i,e sliding down

Inclined = 30 to the horizontal

based on the above information

Force is given by

N = mg\ cos \theta ............ 1

and sliding force is given by

F = mg\ sin\alpha

a = g(sin\ 30^{\circ})

= 9.8\times \frac{1}{2} m/sec^2

= = 4.9\ m/sec^2

Hence, the acceleration of the particle down the slope is 4.9 m/sec^2

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A 975-kg sports car (including driver) crosses the rounded top of a hill at determine (a) the normal force exerted by the road o
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There are missing data in the text of the problem (found them on internet):
- speed of the car at the top of the hill: v=15 m/s
- radius of the hill: r=100 m

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(a) The car is moving by circular motion. There are two forces acting on the car: the weight of the car W=mg (downwards) and the normal force N exerted by the road (upwards). The resultant of these two forces is equal to the centripetal force, m \frac{v^2}{r}, so we can write:
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(b) The problem is exactly identical to step (a), but this time we have to use the mass of the driver instead of the mass of the car. Therefore, we find:
N=mg-m \frac{v^2}{r}=(62 kg)(9.81 m/s^2)-(62 kg) \frac{(15 m/s)^2}{100 m}=469 N

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mg=m \frac{v^2}{r}
from which we find
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