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3 years ago

A 55 kilogram person jumps off a cliff and hits the water 5.8 seconds later, how high is the cliff above the water?

1 answer:

8 0

R=ut+gt^2/2
r- displacement (height to find)
u - initial speed (zero)
t - time taken

r=0*5.8 + 10*5.8^2 /2 = 168.2 meters

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Look at the map of the United States. Which of the following would be a reasonable estimate for the amount of the U.S. coastline

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4 years ago

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Two small objects each with a net charge of Q (positive) exert a force of magnitude F on each other. We replace one of the objec

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Answer:

option B

Explanation:

we know,

Force between two charge is calculated by

$F = \dfrac{kq_1q_2}{r^2}$

r is the distance between the two charges

k is Coulomb constant.

give two small object with charge Q

now, Force

$F = \dfrac{kQQ}{r^2}$ ......(1)

now,

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3 years ago

An unwary football player collides with a padded goalpost while running at a velocity of 9.50 m/s and comes to a full stop after

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Answer:

(a) The collision lasts for 0.053 s.

(b) The deceleration is 180.5 m/s².

Explanation:

Given:

Initial velocity of the player (u) = 9.50 m/s

Final velocity of the player (v) = 0 m/s (Comes to a stop)

Displacement of the player (S) = 0.250 m

We know that, using equation of motion relating displacement (S), acceleration (a), initial velocity (u) and final velocity (v), we have:

$v^2=u^2+2aS$

Expressing in terms of 'a', we get:

$a=\frac{v^2-u^2}{2S}$

Plug in the given values and solve for 'a'. This gives,

$a=\frac{0-9.50^2}{2\times 0.250}\\\\a=\frac{-90.25}{0.5}=-180.5\ m/s^2$

Therefore, the acceleration of the player is -180.5 m/s². So, the deceleration is 180.5 m/s².

Now, using the first equation of motion, we have:

$v=u+at\\\\t=\frac{v-u}{a}$

Plug in the given values and solve for 't'. This gives,

$t=\frac{0-9.5}{-180.5}\\\\t=0.053\ s$

Therefore, the the collision will last for 0.053 s.

(a) The collision lasts for 0.053 s.

(b) The deceleration is 180.5 m/s².

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3 years ago

Five lamp, each labbled "6V,3W" are operated at normal brightness. What is the total energy supplied to the lamps in five second

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Answer:

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