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1 year ago

A 10kg object is 15 meters up a hill. Find its potential energy

Physics

1 answer:

Tema [17]1 year ago

5 0

Answer:

Explanation:

Relative to an origin at the bottom of the hill,

PE = mgh = 10(9.8)(15) = 1470 J

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Blocks A (mass 2.00 kg) and B (mass 10.00 kg, to the right of A) move on a frictionless, horizontal surface. Initially, block B

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2 years ago

Which of the following is a true statement about mass and weight? Mass is a measure of how much matter an object has, while weig

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Answer:

Mass will not change based on location, while weight will change based on gravitational pull.

Explanation:

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2 years ago

Diamonds are a very dense material. Predict what would happen to the light ray if you projected it from air through a diamond

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It [the diamond] would act like a prism, and make a rainbow, or, the light would break up and disappear

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1 year ago

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A bowling ball encounters a 0.760 m vertical rise on the way back to the ball rack. Ignore frictional losses and assume the mass

Gre4nikov [31]

To solve this exercise we need the concept of Kinetic Energy and its respective change: Initial and final kinetic energy.

Let's start considering that the angular velocity is given by,

$\omega = \frac{v}{R}$

Where,

V = linear speed

R = the radius

In the case of the initial kinetic energy:

$KE_i=\frac{1}{2} mv^2 + \frac{1}{2}I \omega^2$

Where I is the moment of inertia previously defined.

$KE_i = \frac{1}{2}(m)3.5^2 + \frac{1}{2}* (\frac{2}{5} m R^2) (\frac{3.5}{R})^2$

In the case of the final kinetic energy, we have to,

$KE_f= mgh+ \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2$

$KE_f = m * 9.81 * 0.76 + \frac{1}{2} m v^2 + \frac{1}{2} (\frac{2}{5} m R^2) (\frac{v}{R})^2$

For conservation of Energy we have, that

$KE_f = KE_i$ , then (canceling the mass and the radius)

$\frac{1}{2} 3.5^2 + \frac{1}{2}(\frac{2}{5})(3.5)^2= 9.81 * 0.76 + \frac{1}{2} v^2 + \frac{1}{2} (\frac{2}{5}) (v)^2$

$8.575= 7.4556+ \frac{1}{2} v^2 + \frac{1}{2} (\frac{2}{5}) (v)^2$

$1.1194= \frac{1}{2}( v^2 + (\frac{2}{5}) (v)^2)$

$2.2388= (\frac{7}{5}) (v)^2$

$v=1.26m/s$

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3 years ago

Monochromatic light passes through two parallel narrow slits and forms an interference pattern on a screen. As the distance betw

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2 years ago

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