A student hangs a balloon from a string then rubs it on their hair and releases it. They rub a second balloon on their hair then
1 answer:
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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
Answer:
Work done, W = 0 J
Explanation:
It is given that,
Mass of the bag, m = 7 kg
If a person carries a bag of groceries 1.2 m above the ground at constant speed across a 2.7 m room. We need to find the amount of work done on the bag he the process. It is given by :
Where
θ = angle between the force and the direction of motion. Here, θ = 90° (its weight is acting vertically downward and it is moving forward)
Since, cos(90) = 0
⇒ Work done, W = 0 J
So, the work done on the bag is zero. Hence, the correct option is (b) "0 J".