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PolarNik [594]
3 years ago
15

The acceleration due to the earth's gravity, in si units, is 9.8 m/s2. in the absence of air friction, a ball is dropped from re

st. its speed on striking the ground is exactly 60 m/s. for what time interval was the ball falling?
Physics
1 answer:
gogolik [260]3 years ago
7 0
We don't know anything about the amount of distance it travels, but that's okay. The only equation we need here is 

velocity(final) = velocity(initial) + acceleration * time
vf = vi + (a * t)

The ball is dropped from rest, so vi = 0 m/s.
We want it so that the ball hits the ground with a final velocity of 60 m/s, so vf = 60 m/s. 
We are given the acceleration due to gravity, a = 9.8 m/s^2.
We are solving for the time, t = ?.

Now we just plug in the values.
vf = vi + (a * t)
60 m/s = 0 m/s + (9.8 m/s^2)*(t)

60 = 9.8t

60 / 9.8 = t

t = 6.122 s

Hopefully this is the right answer.

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A stone is thrown, run a velocity of 15m/s is projected of an elevation of 30° to the horizontal calculate the time rate of flig
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Answer:

1.53 seconds

Explanation:

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T = 2usin∅/g................ Equation 1

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From the question,

Given: u = 15 m/s, ∅ = 30°

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A wheel initially has an angular velocity of 18 rad/s, but it is slowing at a constant rate of 2 rad/s 2 . By the time it stops,
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Answer:

5) 13 revolutions (approximately)

Explanation:

We apply the equations of circular motion uniformly accelerated :

ωf²= ω₀² + 2α*θ Formula (1)

Where:

θ : angle that the body has rotated in a given time interval (rad)

α : angular acceleration (rad/s²)

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ωf : final angular speed  ( rad/s)

Data:

ω₀ = 18 rad/s

ωf = 0

α = -2 rad/s²  ; (-) indicates that the wheel is slowing

Revolutions calculation that turns the wheel until it stops

We apply the formula (1)

ωf²= ω₀² + 2α*θ

0 = (18)² + 2( -2)*θ

4*θ =  (18)²

θ =  (18)²/4 = 81 rad

1 revolution = 2π rad

θ = 81 rad * 1 revolution / 2πrad

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A decreases while B increases because the equilebrium only reacts to different sphers of this substance.

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