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3 years ago

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A ball is dropped from 215m. How long will it take to reach the ground? Use 9.8

1 answer:

valkas [14]3 years ago

7 0

Answer:

The time is $t = 6.62 \ s$

Explanation:

From the question we are told that

The height of the platform where the ball was dropped from is s = 215 m

The acceleration due to gravity is $g = 9.8 \ m/s^2$

Generally from kinematic equations

$s = ut + \frac{1}{2} gt^2$

Here u is the initial velocity of the ball and the value is u = 0 m/s given that the ball was at rest before it was dropped

So

$21 5 = 0 * t + \frac{1}{2} * 9.8t^2$

=> $21 5 = 4.9t^2$

=> $t^2 = \frac{215}{4.9}$

=> $t = 6.62 \ s$

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Answer:

d

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A space vehicle accelerates uniformly from 85 m/s at t = 0 to 164 m/s at t = 10.0 s .How far did it move between t = 2.0 s and t

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$f=\frac{v}{\lambda}$

where

v is the wave's speed

$\lambda$ is the wavelength

For the red laser light in this problem, we have

$v=c=3\cdot 10^8 m/s$ (speed of light)

$\lambda=632.8 nm=632.8\cdot 10^{-9} m$

Substituting,

$f=\frac{3\cdot 10^8 m/s}{632.8 \cdot 10^{-9} m}=4.74 \cdot 10^{14}Hz$

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The wavelength of the wave in the glass is given by

$\lambda=\frac{\lambda_0}{n}$

where

$\lambda_0 = 632.8\cdot 10^{-9} m$ is the original wavelength of the wave in air

n = 1.48 is the refractive index of glass

Substituting into the formula,

$\lambda=\frac{632.8\cdot 10^{-9}m}{1.48}=427.6\cdot 10^{-9}m=427.6 nm$

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The speed of the wave in the glass is given by

$v=\frac{c}{n}$

where

$c = 3\cdot 10^8 m/s$ is the original speed of the wave in air

n = 1.48 is the refractive index of glass

Substituting into the formula,

$v=\frac{3\cdot 10^8 m/s}{1.48}=2.02\cdot 10^8 m/s$

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kiruha [24]

Answer:

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Explanation:

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