4 years ago

A bicyclist is in a 50-km race. She says she had an average velocity of 35.

Physics

1 answer:

zlopas [31]4 years ago

8 0

The correct answer to this is (A. Units Only).

It shows that there is a velocity of 35, but the units are missing.

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The equation for free fall at the surface of a celestial body in outer space (s in meters, t in seconds) is sequals10.04tsqua

Sindrei [870]

Answer:

1.42 s

Explanation:

The equation for free fall of an object starting from rest is generally written as

$s=\frac{1}{2}at^2$

where

s is the vertical distance covered

a is the acceleration due to gravity

t is the time

On this celestial body, the equation is

$s=10.04 t^2$

this means that

$\frac{1}{2}g = 10.04$

so the acceleration of gravity on the body is

$g=2\cdot 10.04 = 20.08 m/s^2$

The velocity of an object in free fall starting from rest is given by

$v=gt$

In this case,

g = 20.08 m/s^2

So the time taken to reach a velocity of

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$t=\frac{v}{g}=\frac{28.6 m/s}{20.08 m/s^2}=1.42 s$

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