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 answer is either
b A system in which Newton's Laws are valid
or
c A system in which there are no external forces.
Explanation:
not a, and not d
There are energy changes in a closed system.
A closed system obeys the conservation laws in its physical description.
Answer:
400ft. 32ft/s -32ft/s
Explanation:
In reality the gravitational acceleration is 9.81 so the quadratic coefficient of the function should be 9.81/2
Anyway for the sake of assumtion let us takes=160t-16t^2
ds/dt=160-32t=0
t=160/32= 5 seconds.
s=160*160/32-16*(160/32)^2= 400 mts
s=384 mts
160t-16t^2=384
i.e
16t^2-160t+384=0
t^2-10t+24=0
(t-6)(t-4)=0
t=[4,6]
we have to take t=4 because it is all the up i.e <5
velocity =v=ds/dt=160-32t
v=160-32*4=32 ft/sec still going up
for all the way down take t=6 whuch is >5
v=160-6*32=-32 ft/sec (falling down!!!)
