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STALIN [3.7K]
3 years ago
11

Ellus

Physics
1 answer:
Lesechka [4]3 years ago
6 0

Answer:

W'=125.44 N

Explanation:

The weight of a person on the surface of Earth is 784 N

Weight is given by :

W = mg

m is mass of the person and g is acceleration due to gravity on the surface of Earth (10 m/s²)

m=\dfrac{W}{g}\\\\m=\dfrac{784}{10}\\\\m=78.4\ kg

The acceleration due to gravity on the surface of Moon, g' = 1.6 m/s²

Weight of the person on the moon is :

W'=mg'

W'=78.4\ kg\times 1.6\ m/s^2\\\\W'=125.44\ N

Hence, the person would weigh 125.44 N on the Moon.

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Refer to the diagram shown below.

m =  the mass of the object
x = the distance of the object from the equilibrium position at time t.
v = the velocity of the object at time t
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A =  the amplitude ( the maximum distance) of the mass from the equilibrium
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The oscillatory motion of the object (without damping) is given by
x(t) = A sin(ωt)
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*A car is going through a dip in the road whose curvature approximates a circle of radius 150m. At what velocity will the occupa
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Explanation:

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  • In this case, is just the difference between the normal force (always perpendicular to the surface, pointing upward) and the force that gravity exerts on the car (which is known as the weight), pointing downward.
  • So, we can write the following expression:

       F_{cent} = F_{n} - F_{g}  (1)

  • It can be showed that the centripetal force is related to the speed by the following expression:
  • F_{cent} = m*\frac{v^{2}}{r} (2)
  • The normal force, it is called the apparent weight, because it would be the weight as measured by a scale.
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       F_{n} = m*\frac{v^{2} }{r} + m*g (3)

  • Now, we need to find the value of v that makes Fn, exactly 15% more than the weight m*g, so we can write the following equation:

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  • Replacing Fg by its value, simplifying, and solving for v, we get:

       v = \sqrt{0.15*g*r} = \sqrt{9.8 m/s2*0.15*150m} = 14.85 m/s (5)

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3 years ago
If velocity is in m/s and time is in sec, what is the unit of acceleration?​
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