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yanalaym [24]
4 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]4 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]4 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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A 1200 kg car is being driven down a road. If it has 101 kJ of kinetic energy, what is its speed?
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The weight of the body varies from place to place but the inertia of the body remains the same. Why?
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The weight changes from place to place as the weight is a vector quantity as well as a dependent quantity. Therefore, weight seems to be varying but the inertia is not.

<u>Explanation: </u>

The weight as already known well, is not a universal constant, whereas the mass is. As the weight is a product of mass of the body and the gravitational acceleration working on it, it varies as the factor of acceleration involved makes it a vector quantity.

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3 years ago
"A car is travelling at 70 m/s. The driver sees a cow in the road and slows down with a constant acceleration for 10 seconds bef
Veronika [31]

<u>The question does not ask for anything in particular. I will calculate the acceleration and distance.</u>

Answer:

<em>The acceleration of the car is</em> -7\ m/s^2

<em>The distance traveled before stopping is 350 m</em>

Explanation:

<u>Constant Acceleration Motion</u>

It's a type of motion in which the velocity of an object changes by an equal amount in every equal period of time.

Being a the constant acceleration, vo the initial speed, vf the final speed, and t the time, the following relation applies:

v_f=v_o+at\qquad [1]

The distance traveled by the object is given by:

\displaystyle x=v_o.t+\frac{a.t^2}{2}\qquad [2]

The car has an initial speed of v0=70 m/s when the driver sees a cow in the road and steps on the brakes with an (assumed) constant acceleration during t=10 seconds before stopping (vf=0).

With the information provided, we can calculate the value of the acceleration and the distance traveled.

Using the equation [1] we can solve for a:

\displaystyle a=\frac{v_f-v_o}{t}

Substituting the numerical values:

\displaystyle a=\frac{0-70}{10}=-7\ m/s^2

The acceleration of the car is -7\ m/s^2

The distance is now calculated by using [2]

\displaystyle x=70\cdot 10+\frac{(-7)\cdot 10^2}{2}

\displaystyle x=700-\frac{700}{2}=350\ m

The distance traveled before stopping is 350 m

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4 years ago
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