3 years ago

If the mass of a material is 99 grams and the volume of the material is 22 cm3, what would the density of the material be?

1 answer:

7 0

You are asked to give the answer in <span>g/cm3. So without knowing any single formulae you can just divide grams by cm3.

</span> $\frac{99 grams}{22 cm^{3} }$

= 4.5 g/cm3

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Identical charges A, B, C, and D, each with a charge of magnitude 1.5 micro Coulomb, are situated at the corners of a square who

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This the correct answer who will know this answer

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2 years ago

After you pick up a spare, your bowling ball rolls without slipping back toward the ball rack with a linear speed of 2.85 m/s. T

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Answer:

V' = 0.84 m/s

Explanation:

given,

Linear speed of the ball, v = 2.85 m/s

rise of the ball, h = 0.53 m

Linear speed of the ball, v' = ?

rotation kinetic energy of the ball

$KE_r = \dfrac{1}{2}I\omega^2$

I of the moment of inertia of the sphere

$I = \dfrac{2}{5}MR^2$

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using conservation of energy

$KE_r = \dfrac{1}{2}( \dfrac{2}{5}MR^2)(\dfrac{V}{R})2$

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Applying conservation of energy

Initial Linear KE + Initial roational KE = Final Linear KE + Final roational KE + Potential energy

$\dfrac{1}{2}MV^2 + \dfrac{1}{5}MV^2 = \dfrac{1}{2}MV'^2 + \dfrac{1}{5}MV'^2 + M g h$

$0.7 V^2 = 0.7 V'^2 + gh$

$0.7\times 2.85^2 = 0.7\times V'^2 +9.8\times 0.53$

V'² = 0.7025

V' = 0.84 m/s

the linear speed of the ball at the top of ramp is equal to 0.84 m/s

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3 years ago

If you know the distance of an earthquake epicenter from three seismic stations, how can you find the exact location of the epic

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You draw 3 circles around the stations with the size of the circle equal to the distance from the earthquake. Then you simply find where the edge circles all overlap.

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3 years ago

PLEASE HELP, PLEASE A CORRECT ANSWER!

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Answer: I like your profile picture

Explanation:

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3 years ago

A sailor strikes the side of his ship just below the waterline. He hears the echo of the sound reflected from the ocean floor di

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Answer:

2625 m deep

Explanation:

Let the sound speed in sea water be 1500 m/s. If he hears the echo 3.5s after the strike, then the sound would have traveled a distance of 1500 * 3.5 = 5250 m to the bottom and back. This would mean the ocean is 5250 / 2 = 2625 m deep.

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3 years ago