Answer:
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
The answer for this question, If I am correct, should be answer "D".
Answer:
Yes, the value of g affected by the radius.
Explanation:
The formula for the force of gravity of 2 objects is
, where m1 and m2 are the masses of the 2 objects, r is the radius, and G is the gravitational constant, which is approximately
.
Therefore, as the radius if bigger, the force of gravity is going to be smaller exponentially.
Explanation:
a) F = GmM / r²
F = (6.67×10⁻¹¹) (500) (5.98×10²⁴) / (6.4×10⁶ + 1.5×10⁶)²
F = 3200 N
b) F = ma
3200 = 500a
a = 6.4 m/s²
c) a = v² / r
640 = v² / (6.4×10⁶ + 1.5×10⁶)
v = 7100 m/s
a star that suddenly increases greatly in brightness because of a catastrophic explosion that ejects most of its mass.