A 910 kg car is approaching a loop-the-loop. The loop has a diameter of 50 m. Determine the minimum speed the car must have at the top of the loop to not fall. g

The minimum speed the car must have at the top of the loop to not fall = 35 m/s

Explanation:

Anywhere else on the loop, the speed needed to keep the car in the loop is obtained from the force that keeps the body in circular motion around the loop which has to just match the force of gravity on the car. (Given that frictional force = 0)

mv²/r = mg

v² = gr = 9.8 × 25 = 245

v = 15.65 m/s

But at the top, the change in kinetic energy of the car must match the potential energy at the very top of the loop-the-loop

Change in kinetic energy = potential energy at the top

Change in kinetic energy = (mv₂² - mv₁²)/2

v₁ = velocity required to stay in the loop anywhere else = 15.65 m/s

v₂ = minimum velocity the car must have at the top of the loop to not fall

And potential energy at the top of the loop = mgh (where h = the diameter of the loop)

(mv₂² - mv₁²)/2 = mgh

(v₂² - v₁²) = 2gh

(v₂² - (15.65)²) = 2×9.8×50

v₂² - 245 = 980

v₂² = 1225

v₂ = 35 m/s

Hence, the minimum speed the car must have at the top of the loop to not fall = 35 m/s

answer: 802.3 kilo joules of energy represents 191.75 kilo calorie of energy.

explanation:

energy released during combustion of one mole methane = 802.3 kilo joules

if 1 kilo joule = 0.239006   kilo calorie

then ,802.3 kilo joule will be = 802.30.239006 kilo calorie

802.3 kilo joule = 191.75 kilo calorie

hence, the energy in kilo calorie is 191.75 kilo calorie.

the answer is a)

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