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timofeeve [1]
3 years ago
9

i throw a tennis ball straight down from the roof of a building 400 meters tall.id it leaves my hand at 10 m/s, how fast will it

be moving when it hits the ground
Physics
1 answer:
Sidana [21]3 years ago
6 0

consider the motion of the tennis ball in downward direction

Y = vertical displacement = 400 m

a = acceleration = acceleration due to gravity = 9.8 m/s²

v₀ = initial velocity of the ball at the top of building = 10 m/s

v = final velocity of the ball when it hits the ground = ?

using the kinematics equation

v² = v²₀ + 2 a Y

inserting the values

v² = 10² + 2 (9.8) (400)

v = 89.11 m/s

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Answer: Force F will be one-sixteenth of the new force when the charges are doubled and distance halved

Explanation:

Let the charges be q1 and q2 and the distance between the charges be 'd'

Mathematical representation of coulombs law will be;

F1=kq1q2/d²...(1)

Where k is the electrostatic constant.

If q1 and q2 is doubled and the distance halved, we will have;

F2 = k(2q1)(2q2)/(d/2)²

F2 = 4kq1q2/(d²/4)

F2 = 16kq1q2/d²...(2)

Dividing equation 1 by 2

F1/F2 = kq1q2/d² ÷ 16kq1q2/d²

F1/F2 = kq1q2/d² × d²/16kq1q2

F1/F2 = 1/16

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Andrea's near point is 20.0 cm and her far point is 2.0 m. Her contact lenses are designed so that she can see objects that are
Shalnov [3]

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the closest distance that she can see an object clearly when she wears her contacts is 22.2 cm

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here, far point i = 2 m = 200 cm  and v is ∞

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1/f = 1/(-200 cm)  +  1/∞

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o = ( i × f ) / ( i - f )

given that near point i = 20 cm

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o = ( -20 × -200 ) / ( -20 - (-200) )

o = 4000 / 180

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Therefore, the closest distance that she can see an object clearly when she wears her contacts is 22.2 cm

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