Answer:
≈147.86 (m.)
Explanation:
1) if the acceleration is a=(V-V₀)/t, where V₀ - initial velocity (25), V - final velocity (2), t - elapsed time, then
t=(V-V₀)/a=23/2.1 (sec.);
2) the required distance can be calculated according to the formula:
S=V₀t+0.5(at²), where V₀=25; t=23/2.1; a= -2.1;
3)
Question: Initially, the car travels along a straight road with a speed of 35 m/s. If the brakes are applied and the speed of the car is reduced to 13 m/s in 17 s, determine the constant deceleration of the car.
Answer:
1.29 m/s²
Explanation:
From the question,
a = (v-u)/t............................ Equation 1
Where a = deceleration of the car, v = final velocity of the car, u = initial velocity of the car, t = time.
Given: v = 13 m/s, u = 35 m/s, t = 17 s.
a = (13-35)/17
a = -22/17
a = -1.29 m/s²
Hence the deceleration of the car is 1.29 m/s²
Answer:
B
Explanation:
This looks like a momentum question. It also looks like there is no horizontal acceleration.
Momentum before Momentum after
m1vo (m1 + m2)*vo/3 multiply through by 3
3 m1*vo m1* vo + m2*vo subtract m1*vo from both sides
3m1*vo - m1*vo m2*vo
2m1*vo m2*vo Divide by vo
2m1 = m2
Conclusion: It takes 2 m1's to equal 1 m2.
B is the answer.
Answer:
8400 J
Explanation:
That is the picture of the explanation
Answer:
W = 1493.9 J = 1.49 KJ
Explanation:
The work done by the elevator on the object will be equal to the gain in is potential energy:
W = ΔP.E
W = mgΔh
where,
W = Work = ?
m = mass of object = 7.4 kg
g = 9.8 m/s²
Δh = gain in height = 20.6 m
Therefore,
W = (7.4 kg)(9.8 m/s²)(20.6 m)
<u>W = 1493.9 J = 1.49 KJ</u>