Arace car accelerates uniformly at 11.3 m/s2. If the race car starts from rest how fast will it
1 answer:
Answer:
75.71 m/s
Explanation:
From equation of motion, acceleration is given by
where v is the final velocity, u is the initial velocity and t is time taken.
Making v the subject of the above formula
v=at+u
Substituting 6.7 s for time, t and 11.3 for a and taking u as zero since it starts from rest
v=11.3*6.7=75.71 m/s
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Answer:
(A) -2940 J
(B) 392 J
(C) 212.33 N
Explanation:
mass of bear (m) = 25 kg
height of the pole (h) = 12 m
speed (v) = 5.6 m/s
acceleration due to gravity (g) = 9.8 m/s
(A) change in gravitational potential energy (ΔU) = mg(height at the bottom- height at the top)
height at the bottom = 0
= 25 x 9.8 x (0-12) = -2940 J
(B) kinetic energy of the Bear (KE) =
= = 392 J
(C) average frictional force =
- change in KE (ΔKE) = initial KE - final KE
- ΔKE =
-
- when the Bear reaches the bottom of the pole, the final velocity (Vf) is 0, therefore the change in kinetic energy becomes ΔKE =
- 0 = 392 J
\frac{-(ΔKE+ΔU)}{h}[/tex] =
= = 212.33 N