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Andrej [43]
3 years ago
8

a man hits golf ball (2kg) which accelerates at a rate of 20 m/s/s what amount of force acted on the ball

Physics
1 answer:
sladkih [1.3K]3 years ago
5 0

Force = (mass) x (acceleration)

         = (2 kg) x (20 m/s²)  =  40 newtons .

                                     
about 9 pounds .

Note:  The ball can only accelerate during the short time
that the clubface is in contact with it.  Once it leaves the
clubface, it can't accelerate any more.

Also ... that's one heckuva golf ball.  It weighs about 4.4 pounds !

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linear charge density of system of two line charges is given as

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now as we know that electric field due to a line charge at some distance from it is given by

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so here we will first find the electric field of first line charge at the position of other line charge

E = \frac{5.20 * 10^{-6}}{2 \pi * 8.85 * 10^{-12}* 0.300}

E = 312000 N/C

now as we know that

F = qE

here q = charge on the line charge system at which force is required

E = electric field on that system of charge where force is required

now we can find the charge by

q = \lambda * L

q = 5.20 * 10^{-6}* 0.05 = 0.26 * 10^{-6} C

Now using the above formula

F = qE

F = 0.26 * 10^{-6} * 312000

F = 0.0811 N

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<h3><u>Answer;</u></h3>

C. Supersaturated

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Answer:

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Object is initially at rest, so its initial speed u = 0

Object falls at half the distance, so h = h/2 where t = t1

Hence, we have

h/2 = at1^2/2 - equation 1

<u>For second half distance: </u>

Similarly,

h = a(t1 + t2)^2/2 - equation 2 where t = t1 + t2 and u= 0

Using equation 2 by equation 1

we obtain 2 = (t1 + t2)^2/t1^2

Hence t2/t1 + 1 = \sqrt{2}

Hence t2/t1 = \sqrt{2} - 1

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