Answer:
it comes from your knowledge and the information you have to get the reason why that is the answer so you are putting together things that you already know what the new information you have
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: C
high; large
Explanation:
The wave energy is related to its amplitude and frequency.
The wave energy is proportional to the amplitude of the wave. So, wave with the most energy will have high amplitude.
Also, frequency is related to wave energy. The larger the frequency, the more the energy of the wave.
Therefore, The waves with the MOST energy have high amplitudes and large
frequencies.
Answer:
The answer to your question is:
a) t = 3.81 s
b) vf = 37.4 m/s
Explanation:
Data
height = 71.3 m = 234 feet
t = 0 m/s
vf = ?
vo = 0 m/s
Formula
h = vot + 1/2gt²
vf = vo + gt
Process
a)
h = vot + 1/2gt²
71.3 = 0t + 1/2(9.81)t²
2(71.3) = 9,81t²
t² = 2(71.3)/9.81
t² = 14.53
t = 3.81 s
b)
vf = 0 + (9.81)(3.81)
vf = 37.4 m/s