Consider two identical objects released from rest high above the surface of Earth. (Neglect air resistance for this question.)
1 answer:
The question is missing its alternatives. Here is the complete question.
Consider two identical objects released from rest high above the surface of Earth. In Case 1, the object is released from a height above the surface of Earth equal to 1 Earth radius, and we measure its kinetic energy just before it hits the Earth to be K1. In Case 2, the obejct is released from a height above the surface of Earth equal to 2 Earth radii and its kinetic energy just before it hits is K2.
1. Compare the kinetic energy of the two objects just before they hit the surface of the earth.
a) K2 = 2K1; b) K2 = 4K; c) K2 = (4/3)K1; d) K2 = (3/2)K1;
Answer: C) K2 = (4/3)K1
Explanation: As it is related to the gravity of the Earth, the potencial energy is: U(r)= - + U₀
In this case, U₀=0, G is the universal gravitational constant, Me is the mass of Earth, m is the mass of the object and r is the distance between the center of the Earth and the object.
The potencial energy of an object of mass m on the surface of the Earth is:
Usurface = -
The potencial energy of the object in Case 1 is
U1 = -
For the Case 2:
U2 = -
The potencial change in Case 1:
ΔU1 = - G.Me.m.() = -
For Case 2:
ΔU2 = - G.Me.m() = -
Comparing ΔK1 and ΔK2 equals comparing ΔU1 and ΔU2:
Δ = (-2/3)(-1/2) = 4/3
So, comparing kinetic energies, K2 is 4/3 of K1.
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Answer:
Force, |F| = 2100 N
Explanation:
It is given that,
Water from a fire hose is directed horizontally against at a rate of 50.0 kg/s,
Initial speed, v = 42 m/s
The momentum is reduced to zero, final speed, v = 0
The relation between the force and the momentum is given by :
|F| = 2100 N
So, the magnitude of the force exerted on the wall is 2100 N. Hence, this is the required solution.
at the lowest point in the trajector, the kinetic energy of the bob is 300 J.
Explanation:
The total mechanical energy of the bob at any point of its motion is given by
Where
is the kinetic energy, where
m is the mass of the bob
v is its speed
is the gravitational potential energy, where
g is the acceleration of gravity
h is the height of the bob, measured with respect to the lowest point of the trajector
In absence of friction, the total mechanical energy E remains constant. So we have:
- When the bob swings upward, the PE increases (because h increases) and the KE decreases (so the speed decreases). At the highest point in the trajector, the speed of the bob is zero (v=0), so its KE is also zero and all the mechanical energy is potential energy: U = 300 J
- When the bob swings downward, the PE decreases (because h decreases) and the KE increases (so the speed increases). At the lowest point in the trajectory, the height has become zero (h=0), so the PE is zero and all the mechanical energy is kinetic energy: KE = 300 J
Therefore, at the lowest point in the trajector, the kinetic energy of the bob is 300 J.
Learn more about kinetic energy:
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