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Minchanka [31]
3 years ago
15

If the force of a golf club on a golf ball is 200 N forward, what will the force of the ball on the club be?

Physics
2 answers:
AVprozaik [17]3 years ago
8 0

Answer:

it's b 200 n backward

Explanation:

Afina-wow [57]3 years ago
5 0

Newton's Third Law is every action has an equal and opposite reaction.

B. 200 N Backwards is the correct answer

The ball should put 200 N of force towards the golfer.

The ball is exerting 200 N of force towards the club as well, but the opposite reaction is that it flies away.

For more details: For each reaction there must be a opposite reaction(Which is the ball going 200 N backwards) and an equal reaction(The ball putting 200 N of force towards the golfer and the golfer)

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Answer:

∆U = 2.296×10^10Joules

Explanation:

Gravitational potential energy is defined as the energy possessed by an object under the influence of gravity due to its virtue of position.

Potential energy U = Fr where;

F is the force of attraction between the masses of the moon and the rocket.

r is the radius or height of the object.

From Newton's law of universal gravitation, F = GMm/r²

Potential energy U = (-GMm/r²)×r

Potential energy U = -GMm/r

The force is negative because the objects act upward.

M is the mass of the rocket

m is the mass of the moon

Gravitational potential energy possessed by the rocket

U1 = -GMm/r1

r1 is the altitude covered by the rocket

Gravitational potential energy possessed by the Moon

U2 = -GMm/(r2+r1)

r2 is the radius of the moon

Change in gravitational potential energy ∆U = U2-U1

∆U = -GMm/(r2+r1)-(-GMm/r1)

∆U = -GMm/(r2+r1) + GMm/r1

∆U = -GMm{1/(r2+r1)-1/r1}

Given

G = 6.67×10^-11m³/kgs²

M = 1130kg

m = 7.36×10²²kg

r1 = 215km = 215,000m

r2 = 1740km = 1,740,000m

∆U = -6.67×10^-11× 7.36×10²² × 1130{1/(215,000+1,740,000)-1/215000}

∆U= -55.47×10¹⁴{1/1955000-1/215000}

∆U = -55.47×10¹⁴{5.12×10^-7 - 4.65×10^-6}

∆U = -284×10^7 + 257.94×10^8

∆U = 22,954,000,000Joules

∆U = 2.296×10^10Joules

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3 years ago
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The barometric pressure at sea level is 30 in of mercury when that on a mountain top is 29 in. If the specific weight of air is
stealth61 [152]

To solve this problem we will apply the concepts related to pressure, depending on the product between the density of the fluid, the gravity and the depth / height at which it is located.

For mercury, density, gravity and height are defined as

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For the air the defined properties would be

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Replacing,

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Rearranging to find h_2

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