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Alinara [238K]
3 years ago
7

When hitting the golf ball the golfer swings the golf club to keep it in contact with the golf ball for as long as possible. The

force acting on the golf ball is constant during this time. Explain the effect that the time of contact between the golf club and the golf ball has on the distance the golf ball travels
Physics
1 answer:
Gnesinka [82]3 years ago
8 0

Answer:

Explanation:

We shall apply the concept of impulse which is given as follows .

Impulse = force x time

Impulse = change in momentum

If u be the initial velocity of golf ball and v be the final velocity , m be the mass

change in momentum

= mu - ( - mv )

= mu+ mv

If F be the force applied and t be the duration of touch with the ball

Impulse = F x t

F x t = mu + mv

mv = Ft - mu

For given mu , greater the value of t , greater will be the value of v

so v is increased when t is increased .

Increased value of v will help in achieving greater distance attained by

golf ball

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Explanation:

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An investigation has been completed similar to the one on latent heat of fusion, where steam is bubbled through a container of w
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Answer:

Q_a=330 cal

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Explanation:

From the question we are told that

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Mass of the container and water 250 g

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Final temperature of the container, water, and condensed steam 50°C

Mass of the container, water, and condensed steam 261 g

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a) Heat energy on container

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Therefore imputing variables we have

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b) Heat energy on water

Generally the formula for mathematically solving heat gain

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Therefore imputing variables we have

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A scientist performed an investigation involving a reaction that produce AI 2 (SO 4) 3 how many sulfur atoms are represented in
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The electron's velocity at that instant is purely horizontal with a magnitude of 2 \times 10^5 ~\text{m/s}2×10 ​5 ​​ m/s then ho
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Complete question:

At a particular instant, an electron is located at point (P) in a region of space with a uniform magnetic field that is directed vertically and has a magnitude of 3.47 mT. The electron's velocity at that instant is purely horizontal with a magnitude of 2×10​⁵​​ m/s then how long will it take for the particle to pass through point (P) again? Give your answer in nanoseconds.

[<em>Assume that this experiment takes place in deep space so that the effect of gravity is negligible.</em>]

Answer:

The time it will take the particle to pass through point (P) again is 1.639 ns.

Explanation:

F = qvB

Also;

F = \frac{MV}{t}

solving this two equations together;

\frac{MV}{t} = qVB\\\\t = \frac{MV}{qVB} = \frac{M}{qB}

where;

m is the mass of electron = 9.11 x 10⁻³¹ kg

q is the charge of electron = 1.602 x 10⁻¹⁹ C

B is the strength of the magnetic field = 3.47 x 10⁻³ T

substitute these values and solve for t

t = \frac{M}{qB} = \frac{9.11 *10^{-31}}{1.602*10^{-19}*3.47*10^{-3}} = 1.639 *10^{-9}  \ s \ = 1.639 \ ns

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Answer:

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Explanation:

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