A space probe is launched from Earth, headed for deep space. At a distance of 10,000 miles from the Earth's center, the gravitat
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Answer:
The centripetal force acting on the child is 39400.56 N.
Explanation:
Given:
Mass of the child is,
Radius of the barrel is,
Number of revolutions are,
Time taken for 10 revolutions is,
Therefore, the time period of the child is given as:
Now, angular velocity is related to time period as:
Now, centripetal force acting on the child is given as:
Therefore, the centripetal force acting on the child is 39400.56 N.
Answer:
the minimum value of the coefficient of static friction between the ground and the cheetah's feet is 1.94
Explanation:
Given that ;
the top speed of Cheetahs is almost 60 mph
In cornering abilities ; the maximum centripetal acceleration of a cheetah was measured to be = 19 m/s^2
The objective of this question is to determine the what minimum value of the coefficient of static friction between the ground and the cheetah's feet is necessary to provide this acceleration?
From the knowledge of Newton's Law;
we knew that ;
Force F = mass m × acceleration a
Also;
The net force = frictional force
so we can say that;
m×a =
where;
the coefficient of static friction is:
= 1.94
Hence; the minimum value of the coefficient of static friction between the ground and the cheetah's feet is 1.94