Gtwo balls of equal size are dropped from the same height from the roof of a building. the mass of ball a is twice that of ball b, m subscript a equals 2 m subscript b. when the two balls reach the ground, how do their kinetic energies compare? group of answer choices when reaching the ground, latex: ke_a=\sqrt{2}ke_b k e a = 2 k e b when reaching the ground, latex: ke_a=\frac{1}{2}ke_b k e a = 1 2 k e b when reaching the ground, latex: ke_a=2ke_b k e a = 2 k e b when reaching the ground, latex: ke_a=ke_b k e a = k e b when reaching the ground, latex: ke_a=4ke_b

Explanation:

As we know that initial height of both the balls are same

also the mass of the two balls is given as

so here we can say by mechanical energy conservation

initial total mechanical energy = final total mechanical energy

since both balls are initially at rest so initial total kinetic energy of the balls will be zero

now final total kinetic energy = initial total potential energy

now we say

so we have

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2

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