Answer:
0.913
Explanation:
k.e=1/2mv square
k.e=1/2×0.078g×23.4256m/s square
k.e=0.913J
Answer: Gravity
Explanation:A projectile is an object upon which the only force is gravity. Gravity acts to influence the vertical motion of the projectile, thus causing a vertical acceleration. The horizontal motion of the projectile is the result of the tendency of any object in motion to remain in motion at constant velocity.
a. The direction of the stone's velocity changes as it moves around the circle.
b. The magnitude of the stone's velocity does not change.
d. The change in direction of the stone's motion is due to the centripetal force acting on the stone.
Above given are true for the given situation.
<u>Answer:</u> Option A, B and D
<u>Explanation:</u>
Circular motion may be characterized as the moving of an objects along the diameter of the circle or any circular direction. It may be standardized and non-uniform based on whether or not the rate of rotation is unchanged.
The velocity, a vector quantity is constant in a uniform circle motion speed is constant as its direction continues to change. Centripetal force works inward toward the core to counterbalance the centrifugal force from the center moving outward.
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
Answer:
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Explanation:
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