Answer:
The minimum stopping distance when the car is moving at
29.0 m/sec = 285.94 m
Explanation:
We know by equation of motion that,

Where, v= final velocity m/sec
u=initial velocity m/sec
a=Acceleration m/
s= Distance traveled before stop m
Case 1
u= 13 m/sec, v=0, s= 57.46 m, a=?

a = -1.47 m/
(a is negative since final velocity is less then initial velocity)
Case 2
u=29 m/sec, v=0, s= ?, a=-1.47 m/
(since same friction force is applied)

s = 285.94 m
Hence the minimum stopping distance when the car is moving at
29.0 m/sec = 285.94 m
Answer:
ok so
Explanation:
Im not sure rn but ill get back to you.
Answer:
11 m/s south
Explanation:
The velocity of the passenger relative to the river bank is equal to the velocity of the passenger relative to the ferry, plus the velocity of the ferry relative to the river, plus the velocity of the river relative to the river bank.
v_passenger,bank = v_passenger,ferry + v_ferry,river + v_river,bank
If we take north to be positive and south to be negative:
v = 1.0 m/s + (-10 m/s) + (-2 m/s)
v = -11 m/s
v = 11 m/s south
Change in temperature = final temperature - Initial temperature
Δt = t₂ - t₁
Δt = 17 - (-6)
Δt = 17 + 6 = 23 f
In short, Your Answer would be Option D
Hope this helps!