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Aleksandr-060686 [28]
3 years ago
6

A 430 g soccer ball lying on the ground is kicked and flies off at 25 m/s. If the duration of the impact was .01 s, what was the

average force acting on the ball?
Physics
1 answer:
maw [93]3 years ago
4 0
<h2>The average force acting on the ball is 1075 N in the direction of travel of ball.</h2>

Explanation:

Force is given by rate of change of momentum.

Mass of soccer ball = 430 g = 0.43 kg

Initial velocity = 0 m/s

Final velocity = 25 m/s

Change in momentum = 0.43 x 25 - 0.43 x 0

Change in momentum = 10.75 kg m/s

Time taken = 0.01 s

Rate of change of momentum = Change in momentum ÷ Time

Rate of change of momentum = 10.75 ÷ 0.01

Rate of change of momentum = 1075 N

Force = 1075 N

The average force acting on the ball is 1075 N in the direction of travel of ball.

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Coulomb's law for the magnitude of the force FFF between two particles with charges QQQ and Q′Q′Q^\prime separated by a distance
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Answer:

Explanation:

Force between two charges of q₁ and q₂ at distance d is given by the expression

F = k q₁ q₂ / d₂

Here force between charge q₁ = - 15 x 10⁻⁹ C and q₃ = 47 x 10⁻⁹ C when distance between them d = (1.66 - 1.24 ) = .42 mm

k = 1/ 4π x 8.85 x 10⁻¹²

putting the values in the expression

F = 1/ 4π x 8.85 x 10⁻¹²  x - 15 x 10⁻⁹ x 47 x 10⁻⁹ /( .42 x 10⁻³)²

= 9 x 10⁹ x  - 15 x 10⁻⁹ x 47 x 10⁻⁹ /( .42 x 10⁻³)²

= 35969.4 x 10⁻³ N .

force between charge q₂ =  34.5 x 10⁻⁹ C and q₃ = 47 x 10⁻⁹ C when distance between them d = ( 1.24 - 0 ) = 1.24 mm .

putting the values in the expression

F = 1/ 4π x 8.85 x 10⁻¹²  x  34.5 x 10⁻⁹ x 47 x 10⁻⁹ /( .42 x 10⁻³)²

= 9 x 10⁹ x  - 34.5 x 10⁻⁹ x 47 x 10⁻⁹ /( .42 x 10⁻³)²

= 82729.6  x 10⁻³ N

Both these forces will act in the same direction towards the left (away from the origin towards - ve x axis)

Total force = 118699 x 10⁻³

= 118.7 N.

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