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vovikov84 [41]
3 years ago
10

A 12 kg bowling ball would require what force to accelerate it down an alley at a rate of 2.5 m/s ^ 2

Physics
1 answer:
alisha [4.7K]3 years ago
8 0

Answer:

hi

Explanation:

hiijjjjjjjjjjjjjjj

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Biologists have studied the running ability of the northern quoll, a marsupial indigenous to Australia. In one set of experiment
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Answer:

u_K=0.862

Explanation:

The force of friction between the quails feet and the ground is:

F=m*a

F_K=m*a

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u_K*m*g=m*a_c

u_K*g=a

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a_c=\frac{v^2}{r}

So the coefficient of static is solve

u_K=\frac{\frac{v^2}{r}}{g}

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u_K=0.862

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3 years ago
State coulombs law in word​
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<h2><em>state coulombs law in word</em></h2>

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Write a note on unity of ant​
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3 0
3 years ago
Imagine that the earth and the-moon have positive charges of the same magnitud. How big isäºthe charge necesary to produce an el
lions [1.4K]

Answer:

5.7 x 10^12 C

Explanation:

Let the charge on earth and moon is q.

mass of earth, Me = 5.972 x 10^24 kg

mass of moon, Mm = 7.35 x 10^22 kg

Let d be the distance between earth and moon.

the gravitational force between them is

F_{g}=G\frac{M_{e} \times M_{m}}{d^{2}}

The electrostatic force between them is

F_{e}=\frac{Kq^{2}}{d^{2}}

According to the question

1 % of Fg = Fe

0.01 \times 6.67\times10^{-11}\frac{5.97 \times 10^{24}\times7.35 \times 10^{22}}{d^{2}}=9 \times 10^{9}\frac{q^{2}}{d^{2}}

2.927 \times 10^{35}=9 \times10^{9}q^{2}

3.25 \times 10^{25}=q^{2}

q = 5.7 x 10^12 C

Thus, the charge on earth and the moon is 5.7 x 10^12 C.

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3 years ago
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