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grigory [225]
4 years ago
11

48 grams 12cm^3, what would the density of the material be

Physics
1 answer:
Licemer1 [7]4 years ago
7 0
Density is mass over volume:

D =  \frac{m}{V}

In your case, mass is 48 grams and volume is 12cm^3

If you put that into the equation, you will get your density:

D =  \frac{m}{V} =  \frac{48g}{12cm^{3} } =4g/ cm^{3}
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Describes the first law of thermodynamics
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Thermodynamics is usually defined as a branch of physics that deals with the study of the heat and various form of energy, and their interaction between the.

The first law says that heat appears as energy, and it cannot be produced and also cannot be demolished. It can only change from one form to another. This signifies that the total amount of energy present in the universe remains constant.

This first law can be mathematically represented as:

ΔU = Q - W

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The speed of a wave is 65 m/sec. if the wavelength os the wave is 0.8 meters. what is the spedd of this wave?
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The time of a wave is 65 m/sec. if the wavelength of the wave is 0.8 meters. what is the speed of this wave?

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Who was first person who envisaged a telescope to see the objects far away​
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3 years ago
TUL<br> Which of the following terms refers to the exactness of a measurement?
Oksana_A [137]

Accuracy

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Accuracy is a term that refers to the exactness of a measurement.

Accuracy is the nearness of measured value to the true value.

The difference between the measured value and the true value is the uncertainty in the measurement.

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5 0
3 years ago
Determine the ratio of Earth's gravitational force exerted on an 80-kg person when at Earth's surface and when 1400 km above Ear
postnew [5]

m = mass of the person = 80 kg

M = mass of earth = 5.98 x 10²⁴ kg

R = radius of earth = 6.37 x 10⁶ m

h = height above the earth's surface = 1400 km = 1.4 x 10⁶ m

r₁ = initial distance of the person from the center of earth when on surface = R =  6.37 x 10⁶ m

r₂ = final distance of the person from the center of earth when at some height = R + h =  6.37 x 10⁶ + 1.4 x 10⁶ = 7.77 x 10⁶ m


F₁ = Gravitational force of earth on the person when at surface

Gravitational force of earth on the person when at surface is given as

F₁ = G M m/r₁²                                             eq-1

F₂ = Gravitational force of earth on the person when at some height

Gravitational force of earth on the person when at some height is given as

F₂ = G M m/r₂²                                             eq-2

dividing eq-1 by eq-2

F₁ /F₂ = (G M m/r₁² )/(G M m/r₂²)

F₁ /F₂ = r₂²/r₁²

inserting the values

F₁ /F₂ = (7.77 x 10⁶)²/(6.37 x 10⁶)²

F₁ /F₂ = 1.49

3 0
4 years ago
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