An automobile having a mass of 1,000 kg is driven into a brick wall in a safety test. The bumper behaves like a spring with cons
1 answer:
Answer:
Explanation:
Assuming no energy lost, according to the law of conservation of energy, the kinetic energy of the automobile becomes potential energy after the crash:
Here m is the automobile's mass, v is the speed of the car before impact, k is the "bumper" constant and x is the compression of the bumper due to the collision. Solving for v:
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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
Answer:
T = 693.147 minutes
Explanation:
The tank is being continuously stirred. So let the salt concentration of the tank at some time t be x in units of kg/L.
Therefore, the total salt in the tank at time t = 1000x kg
Brine water flows into the tank at a rate of 6 L/min which has a concentration of 0.1 kg/L
Hence, the amount of salt that is added to the tank per minute =
Also, there is a continuous outflow from the tank at a rate of 6 L/min.
Hence, amount of salt subtracted from the tank per minute = 6x kg/min
Now, the rate of change of salt concentration in the tank =
So, the rate of change of salt in the tank can be given by the following equation,
or,
or, T = 693.147 min (time taken for the tank to reach a salt concentration
of 0.05 kg/L)