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
To calculate the force of impact F, first lets calculate the acceleration a of the ball:
a=v/t where v is the velocity of the ball and t is time
a=32/0.8=40 m/s²
To get the force F we need the Newtons second law:
F=m*a where m is the mass of the ball and a is the acceleration.
F=m*a= 0.2*40 = 8 N
So the impact force is F= 8 N.
No because there must be an even # if their is an even amount one of the forces isn’t being cancelled
Answer:
I think the answer is B. amount of energy present but I'm not 100% sure
Explanation:
