Boyle's Law mainly involves _______.
1 answer:
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Answer:
70.6 mph
Explanation:
Car A mass= 1515 lb
Car B mass=1125 lb
Speed of car B is 46 miles/h
Distance before locking, d=19.5 ft
Coefficient of kinetic friction is 0.75
Initial momentum of car B=mv where m is mass and v is velocity in ft/s
46 mph*1.46667=67.4666668 ft/s
Initial momentum of car A is given by
where
is velocity of A
Taking East as positive and west as negative then the sum of initial momentum is
The common velocity is represented as hence after collision, the final momentum is
From the law of conservation of linear momentum, sum of initial and final momentum equals each other hence
The acceleration of two cars
From kinematic equation
hence
Substituting the value of in equation
Answer:
6.86 m/s
Explanation:
This problem can be solved by doing the total energy balance, i.e:
initial (KE + PE) = final (KE + PE). { KE = Kinetic Energy and PE = Potential Energy}
Since the rod comes to a halt at the topmost position, the KE final is 0. Therefore, all the KE initial is changed to PE, i.e, ΔKE = ΔPE.
Now, at the initial position (the rod hanging vertically down), the bottom-most end is given a velocity of v0. The initial angular velocity(ω) of the rod is given by ω = v/r , where v is the velocity of a particle on the rod and r is the distance of this particle from the axis.
Now, taking v = v0 and r = length of the rod(L), we get ω = v0/ 0.8 rad/s
The rotational KE of the rod is given by KE = 0.5Iω², where I is the moment of inertia of the rod about the axis of rotation and this is given by I = 1/3mL², where L is the length of the rod. Therefore, KE = 1/2ω²1/3mL² = 1/6ω²mL². Also, ω = v0/L, hence KE = 1/6m(v0)²
This KE is equal to the change in PE of the rod. Since the rod is uniform, the center of mass of the rod is at its center and is therefore at a distane of L/2 from the axis of rotation in the downward direction and at the final position, it is at a distance of L/2 in the upward direction. Hence ΔPE = mgL/2 + mgL/2 = mgL. (g = 9.8 m/s²)
Now, 1/6m(v0)² = mgL ⇒ v0 =
Hence, v0 = 6.86 m/s