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3 years ago

A 50Kg girl jumps off a 5-meter-high diving board. What is her kinetic energy right before she hits the water?

Physics

1 answer:

Kryger [21]3 years ago

6 0

Answer:

2450 J

Explanation:

Given that,

A 50Kg girl jumps off a 5-meter-high diving board.

We need to find the kinetic energy of the girl before she hits the water. At this point the kinetic energy becomes equal to the potential energy such that,

$K=mgh\\\\K=50\times 9.8\times 5\\\\K=2450\ J$

So, her kinetic energy right before she hits the water is equal to 2450 J.

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A meter stick A hurtles through space at a speed v = 0.25c relative to you, with its length aligned with the direction of motion

yaroslaw [1]

Answer:

$L_0\approx1.0328\ m$

Explanation:

Given:

relativistic length of stick A, $L=1\ m$

relativistic velocity of stick A with respect to observer, $v=0.25c=7.5\times 10^{7}\ m.s^{-1}$

Since the object is moving with a velocity comparable to the velocity of light with respect to the observer therefore the length will appear shorter according to the theory of relativity.

Mathematical expression of the theory of relativity for length contraction:

$L=\frac{L_0}{\gamma}$

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L = relativistic length

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$\gamma =$ Lorentz factor $=\frac{1}{\sqrt{1-\frac{v^2}{c^2} } }$

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A 5.0-kg rock and a 3.0 × 10−4-kg pebble are held near the surface of the earth.(a)Determine the magnitude of the gravitational

a_sh-v [17]

Answer:

a). Determine the magnitude of the gravitational force exerted on each by the earth.

Rock: $F = 49.06N$

Pebble: $F = 29.44N$

(b)Calculate the magnitude of the acceleration of each object when released.

Rock: $a =9.8m/s^{2}$

Pebble: $a =9.8m/s^{2}$

Explanation:

The universal law of gravitation is defined as:

$F = G\frac{m1m2}{r^{2}}$ (1)

Where G is the gravitational constant, m1 and m2 are the masses of the two objects and r is the distance between them.

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m1 will be equal to the mass of the Earth $m1 = 5.972×10^{24} Kg$ and since the rock and the pebble are held near the surface of the Earth, then, r will be equal to the radius of the Earth $r = 6371000m$ .