Answer: 0.2 hours
Explanation: In order to solve this question we have to considerer that a recargeable battery can supply 1800 mA in one hour then we have to determine how long could this battery drive current through a long, thin wire of resistance 34 Ω .
Besides, this battery has a voltage of 12 V
so by using the Ohm law we also know that V=R*I,
Fron this we can obtain:
I= V/R= 12 V/ 34 Ω=0.35 A= 350 mA
then considering that this battery can supply 1800 mA in one hour we have this battery can supply 350 mA in x time in the form:
1hour------- 1800 mA
x hour--------350 mA
time= 350/1800= 0.2 hour
The applicable equation:
P = F/A
P = pressure
F = Force or weight
A = surface area
Pressure on each cylinder = (W/n)/A
Where n = number of cylinders. Additionally, pressure in the reservoir is equivalent to the pressure in each cylinder.
Net pressure = 75 - 14.7 = 60.3 psi
Therefore,
60.3 = (W/n)/A = (450/n)/(πD^2/4) = (450/n)/(π*1.5^2/4) = (450/n)/(1.7671)
60.3*1.7671 = 450/n
106.03 = 450/n
n = 450/106.3 = 4.244 ≈ 5
The number of cylinders is 5.
The one at the base would be much older due to the law of super position, and the rock at the top would be much newer,again, due to the law of super position.
Answer:
K.E₂ = mg(h - 2R)
Explanation:
The diagram of the car at the top of the loop is given below. Considering the initial position of the car and the final position as the top of the loop. We apply law of conservation of energy:
K.E₁ + P.E₁ = K.E₂ + P.E₂
where,
K.E₁ = Initial Kinetic Energy = (1/2)mv² = (1/2)m(0 m/s)² = 0 (car initially at rest)
P.E₁ = Initial Potential Energy = mgh
K.E₂ = Final Kinetic Energy at the top of the loop = ?
P.E₂ = Final Potential Energy = mg(2R) (since, the height at top of loop is 2R)
Therefore,
0 + mgh = K.E₂ + mg(2R)
<u>K.E₂ = mg(h - 2R)</u>