<span>An automobile with a mass of 1450 kg is parked on a moving flatbed railcar; the flatbed is 1.5 m above the ground. The railcar has a mass of 38,500 kg and is moving to the right at a constant speed of 8.7 m/s on a frictionless rail...
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F=ma
F=125 N
m= 50 kg
125=50a
a=2.5 m/s^2
Answer:
v = 2 v₁ v₂ / (v₁ + v₂)
Explanation:
The body travels the first half of the distance with velocity v₁. The time it takes is:
t₁ = (d/2) / v₁
t₁ = d / (2v₁)
Similarly, the body travels the second half with velocity v₂, so the time is:
t₂ = (d/2) / v₂
t₂ = d / (2v₂)
The average velocity is the total displacement over total time:
v = d / t
v = d / (t₁ + t₂)
v = d / (d / (2v₁) + d / (2v₂))
v = d / (d/2 (1/v₁ + 1/v₂))
v = 2 / (1/v₁ + 1/v₂)
v = 2 / ((v₁ + v₂) / (v₁ v₂))
v = 2 v₁ v₂ / (v₁ + v₂)
Answer:
8kg
Explanation:
For the box to be in equilibrium. The clockwise moment ensued by the box on the right should be same as that ensued by the one on the right. Hence :
M ×3 = 12 ×2
M = 24/3 = 8kg
Note mass is used because trying to compute the weight by multiplying by the acceleration of free fall due to gravity on both sides will cancel out.
Answer:
1.137278672 m/s
+5.9 cm or -5.9 cm
Explanation:
A = Amplitude = 6.25 cm
m = Mass of object = 225 g
k = Spring constant = 74.5 N/m
Maximum speed is given by
The maximum speed of the object is 1.137278672 m/s
Velocity is at any instant is given by
The locations are +5.9 cm or -5.9 cm
