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yanalaym [24]
3 years ago
8

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? A. 200 N forward B.

200 N backward C. 400 N forward D. 400 N backward
Physics
2 answers:
Korolek [52]3 years ago
8 0

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

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

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

iren [92.7K]3 years ago
3 0

Explanation :

According to Newton's third law of motion, every action has an equal and opposite reaction.

If the force of a golf club on a golf ball is 200 N forward, then the force of the ball on the club will be same as 200 N. But this force will act in opposite direction i.e. backward.

So, the correct option is (B) i.e. 200 N backward.

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3 years ago
Which of the following statements explains how total time spent in the air is affected as a projectile's angle of launch is incr
charle [14.2K]

Answer:

Therefore letter <u>C is the correct answer.</u>

Explanation:

In a projectile motion the total time in the air can be calculated using the following equation:

We analyze the y-component motion.

v_{fy}=v_{iy}-gt

When the final velocity (v(f)) is equal to zero we calculate the upward time and multiplying it by 2 we find the total time in the air. So we will have:

t_{tot}=2\frac{v_{iy}}{g}

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We can see that the <u>total time is directly proportional to the angle</u>, then when <u>θ increase t increase.</u>

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3 0
3 years ago
How much kinetic energy does a system containing 3 moles of an ideal gas at 300 K possess? What is the heat capacity at constant
ElenaW [278]

Answer:

1. K.E = 11.2239 kJ ≈ 11.224 kJ

2. C_{V} = 37.413 JK^{- 1}

3. Q = 10.7749 kJ

Solution:

Now, the kinetic energy of an ideal gas per mole is given by:

K.E = \frac{3}{2}mRT

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m = no. of moles = 3

R = Rydberg's constant = 8.314 J/mol.K

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Therefore,

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K.E = 11223.9 J = 11.2239 kJ ≈ 11.224 kJ

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C_{V} = \frac{3}{2}mR

C_{V} = \frac{3}{2}\times 3\times 8.314 = 37.413 JK^{- 1}

Now,

Required heat transfer to raise the temperature by 15^{\circ} is:

Q = C_{V}\Delta T

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Q = 37.413\times 288 = 10774.9 J = 10.7749 kJ

8 0
3 years ago
The jet fuel in an airplane has a mass of 97.5 kg and a density of 0.804 g/cm3. What is the volume of this jet fuel
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The volume of the jet fuel is 1.213 x 10⁵ cm³.

<h3>What is density?</h3>

Density is a physical property that measures the ratio of mass to volume occupied by an object.

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The volume of the jet fuel is calculated as follows;

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