Answer:
No
Explanation:
The vertical component of Jack's initial velocity is:
5.0
⋅
sin
30
∘
=
5.0
⋅
1
2
=
2.5
m/s
With gravitational acceleration
9.8
m/s
2
, he will reach the highest point of his trajectory after:
2.5
9.8
≈
0.255
s
The average vertical component of his velocity in that
0.255
s
will be:
1
2
⋅
2.5
=
1.25
m/s
So the highest point of his trajectory will be:
0.255
⋅
1.25
≈
0.32
m
So he will pass approximately
7
cm
above the top of the candle.
The horizontal component of his velocity will be a constant:
5.0
⋅
cos
30
∘
=
5.0
⋅
√
3
2
≈
4.33
m/s
So Jack's trajectory will be substantially longer than it is high and he will spend little time anywhere near above the candle.
22. a - (vf^2 - vi^2)/(2d)
a = (0 - 23^2)/(170)
a = -3.1 m/s^2
23. Find the time (t) to reach 33 m/s at 3 m/s^2
33-0/t = 3
33 = 3t
t = 11 sec to reach 33 m/s^2
Find the av velocuty: 33+0/2 = 16.5 m/s
Dist = 16.5 * 11 = 181.5 meters to each 33m/s speed. Runway has to be at least this long.
24. The sprinter starts from rest. The average acceleration is found from:
(Vf)^2 = (Vi)^2 = 2as ---> a = (Vf)^2 - (Vi)^2/2s = (11.5m/s)^2-0/2(15.0m) = 4.408m/s^2 estimated: 4.41m/s^2
The elapsed time is found by solving
Vf = Vi + at ----> t = vf-vi/a = 11.5m/s-0/4.408m/s^2 = 2.61s
25. Acceleration of car = v-u/t = 0ms^-1-21.0ms^-1/6.00s = -3.50ms^-2
S = v^2 - u^2/2a = (0ms^-1)^2-(21.0ms^-1)^2/2*-3.50ms^-2 = 63.0m
26. Assuming a constant deceleration of 7.00 m/s^2
final velocity, v = 0m/s
acceleration, a = -7.00m/s^2
displacement, s - 92m
Using v^2 = u^2 - 2as
0^2 - u^2 + 2 (-7.00) (92)
initial velocity, u = sqrt (1288) = 35.9 m/s
This is the speed pf the car just bore braking.
I hope this helps!!
The answer to your question is B
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Rolling friction is less than sliding friction