Answer:
a. the magnitude of the force experienced by the muon is 2.55 × 10⁻¹⁹N
b. this force compare to the weight of the muon; the force is 1.38 × 10⁸ greater than muon
Explanation:
F= ma
v²=u² -2aS
(1.56 ✕ 10⁶)²=(2.40 ✕ 10⁶)²-2a(1220)
a=1.36×10⁹m/s²
recall
F=ma
F = 1.88 ✕ 10⁻²⁸ kg × 1.36×10⁹m/s²
F= 2.55 × 10⁻¹⁹N
the magnitude of the force experienced by the muon is 2.55 × 10⁻¹⁹N
b. this force compare to the weight of the muon
F/mg= 2.55 × 10⁻¹⁹/ (1.88 ✕ 10⁻²⁸ × 9.8)
= 1.38 × 10⁸
The average acceleration of the cheetah is 7.75 m/s²
The given parameters;
velocity of the cheetah, v = 31 m/s
time of motion, t = 4 seconds
The average acceleration of the cheetah is calculated as follows;

where;
v is the final velocity
u is the initial velocity
t is the time of motion
Assuming the cheetah started from rest, the initial velocity, u = 0
The average acceleration is calculated as follows;

Thus, the average acceleration of the cheetah is 7.75 m/s²
Learn more here: brainly.com/question/17280180
Answer:
Part a)
When there is no friction then acceleration is

Part b)
if there is friction force along the inclined then acceleration is

Explanation:
Part a)
As we know that the skier is on inclined plane
So here if there is no friction then net force along the inclined plane is given as

now acceleration of the skier is given as




Part b)
if there is friction force along the inclined then net force along the inclined plane is given as

now acceleration of the skier is given as



