Acceleration is the force of something moving, or acceleration can be how fast an object is going!
Answer:
The speed of the two cars after coupling is 0.46 m/s.
Explanation:
It is given that,
Mass of car 1, m₁ = 15,000 kg
Mass of car 2, m₂ = 50,000 kg
Speed of car 1, u₁ = 2 m/s
Initial speed of car 2, u₂ = 0
Let V is the speed of the two cars after coupling. It is the case of inelastic collision. Applying the conservation of momentum as :
V = 0.46 m/s
So, the speed of the two cars after coupling is 0.46 m/s. Hence, this is the required solution.
Answer:
vf = √(vi²+2*(F/m)*D)
Explanation:
Given
Mass of the particle: M
Initial speed of the particle: vi
Force: F
Distance: D
We can apply the formula
F = M*a ⇒ a = F/m
then we use the equation
vf = √(vi²+2*a*D)
⇒ vf = √(vi²+2*(F/m)*D)
from one energy form to one
Answer:
the earthquake's epicenter is L = 1091.2 km away
Explanation:
if the P and S waves start from the same place (x=0) :
P would travel the distance L in , L = vP * tP
P would travel the distance L in , L = vS * tS
therefore
vP * tP= vS * tS
if the difference in time d= tS-tP → tS=d+tP , then
vS * tS= vP * tP
vS *(d+tP) = vP * tP
vS*d + vS*tP = vP * tP
vS*d = vP * tP - vS*tP = (vP - vS) * tP
tP = vS*d / (vP - vS)
replacing values
tP = vS*d / (vS - vP) = 5.1 km/s * 1.5 min*60 s/min / ( 8.8 km/s - 5.1 Km/s) = 124 s
therefore
L = vP*tP = 8.8 km/s * 124 s = 1091.2 km
