By which method does the following move into the cell: movement of oxygen, carbon dioxide, and other small uncharged molecules a
1 answer:
You might be interested in
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:
Static friction, F=2.9988 N
Explanation:
Static Friction: It is the friction between two or more solid body that are not moving relative to each other.
When an external force is applied to a body that is in contact with another body, static friction opposes the applied force upto a certain limit, keeping the body at rest. When the applied force exceeds the certain limit the body begins to move. At that moment applied force is equal to limiting value of static friction F.
It varies linearly with normal force(N)
F∝N
F=μN (where μ is constant of proportionality and known as coefficient of static friction)
Now,
Mass, m=1.53 kg
Coefficient of static friction, μ=0.2
Normal force, N=m×g
= (g=9.8 m/s²)
=14.994 N
Static friction, F= μN
=
= 2.9988 N