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Karolina [17]
3 years ago
6

Justin strikes a 0.058-kg golf ball with a force of 260 N. If the ball moves with a velocity of 63 m/s, calculate the time the b

all is in contact with the club
Physics
1 answer:
Ymorist [56]3 years ago
3 0

Answer:

0.014s

Explanation:

Given parameters:

Mass of the ball  = 0.058kg

Force  = 260N

Velocity of the ball  = 63m/s

Unknown:

Time of contact  = ?

Solution:

To solve this problem, we apply Newton's second law of motion which is mathematically expressed as;

    Ft  = m(v - u)

F is the force

t is time

m is the mass

v is the final velocity

u is the initial velocity = 0m/s

   Insert the parameters and find t;

      260 x t  = 0.058(63 - 0)

       260t  = 3.65

            t = 0.014s

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A object with a mass of 1.5 kg is lifted from the ground to a height of 0.22 m what is the objects potential energy
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<h2>3.3 J</h2>

Explanation:

The potential energy of a body can be found by using the formula

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where

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From the question we have

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We have the final answer as

<h3>3.3 J</h3>

Hope this helps you

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Two point charges are separated by 10 cm, with an attractive force between them of 15 N. Find the force between them when they a
suter [353]

Answer:

(a) the force is 8.876 N

(b) the magnitude of each charge is 4.085 μC

Explanation:

Part (a)

Given;

coulomb's constant, K = 8.99 x 10⁹ N.m²/C²

distance between two charges, r = 10 cm = 0.1 m

force between the two charges, F = 15 N

when the distance between the charges changes to 13 cm (0.13 m)

force between the two charges, F = ?

Apply Coulomb's law;

F = \frac{Kq_1q_2}{r^2} \\\\let \ Kq_1q_2 = C\\\\F =\frac{C}{r^2} \\\\C = Fr^2\\\\F_1r_1^2 = F_2r_2^2\\\\F_2 =\frac{F_1r_1^2}{r_2^2} \\\\F_2 = \frac{15*0.1^2}{0.13^2} \\\\F_2 = 8.876 \ N

Part (b)

the magnitude of each charge, if they have equal magnitude

F = \frac{KQ^2}{r^2}

where;

F is the force between the charges

K is Coulomb's constant

Q is the charge

r is the distance between the charges

F = \frac{KQ^2}{r^2} \\\\Q = \sqrt{\frac{Fr^2}{K} } \\\\Q =  \sqrt{\frac{15*(0.1)^2}{8.99*10^9} } = 4.085 *10^{-6} \ C\\\\Q = 4.085 \ \mu C

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3 years ago
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