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Paha777 [63]
3 years ago
12

An object is dropped from a bridge. A second object is thrown downwards 1.0 s later. They both reach the water 20 m below at the

same instant. What was the initial speed of the second object
Physics
1 answer:
olga2289 [7]3 years ago
6 0

Answer:

vi = 14.610

Explanation:

initial velocity of the first object (vi) = 0 m/s, because it was dropped

distance (y) = -20 m

initial position (y0) = 0 m

acceleration due to gravity (g) = -9.8 m/s^2

y = 1/2 gt^2 + vi*t + y0

-20 = 1/2(-9.8)t^2 + 0 + 0

-20 = -4.9t^2

4.081 = t^2

+√4.081 = t

t = 2.020

time of second object = 2.020 - 1 = 1.020

Now we can plug in the new time to solve for vi of the second object.

y = 1/2 gt^2 + vi*t + y0

-20 = 1/2(-9.8)1.020^2 + vi*1.020 + 0

-20 = -5.098 + 1.020vi

14.902 = 1.020vi

vi = 14.610 m/s

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Answer:

126.56 m

Explanation:

Applying,

-F = ma............. Equation 1

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F = mgμ.......... Equation 2

Where g = acceleration due to gravity, μ = coefficient of friction

Substitute equation 2 in equation 1

-mgμ = ma

a = -gμ.............. Equation 3

From the question,

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a = -9.8×0.735

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Finally,

Applying

v² = u²+2as.............. Equation 4

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From the question,

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What happens to an electromagnetic wave as it passes from space to matter?
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