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vichka [17]
3 years ago
12

A cubic box has a 0.6 cubic decimetre side. Find the mass of air in the box in gram. (1 Liter of air has a mass of 1.3 g).

Physics
1 answer:
morpeh [17]3 years ago
4 0

Answer:

The mass of air in the box is 0.78 grams.

Explanation:

A litre equals a cubic decimetre. By definition of density (\rho), in grams per litre, we calculate the mass of air inside the cube (m), in grams:

m = \rho \cdot V (1)

Where V is the volume occupied by air within the cube, in litres.

If we know that V = 0.6\,L and \rho = 1.3\,\frac{g}{L}, then the mass of air is:

m = \left(1.3\,\frac{g}{L} \right)\cdot (0.6\,L)

m = 0.78\,g

The mass of air in the box is 0.78 grams.

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Answer:

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Explanation:

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3 years ago
The rubber band contains .......potential energy as it is stretched.
kodGreya [7K]

Answer:

elastic potential energy

You input potential (stored) energy into the rubber band system when you stretched the rubber band back. Because it is an elastic system, this kind of potential energy is specifically called elastic potential energy.

Explanation:

4 0
2 years ago
Find the energy in Joules required to lift a 55.0 Megagram object a distance of 500 cm.
fredd [130]

Energy to lift something =

               (mass of the object) x (gravity) x (height of the lift).

BUT ...

This simple formula only works if you use the right units.

Mass . . . kilograms
Gravity . . . meters/second²
Height . . . meters

For this question . . .

Mass = 55 megagram = 5.5 x 10⁷ grams = 5.5 x 10⁴ kilograms

Gravity (on Earth) = 9.8 m/second²

Height = 500 cm  =  5.0 meters

So we have ...

Energy = (5.5 x 10⁴ kilogram) x (9.8 m/s²) x (5 m)

            =  2,696,925 joules .

That's quite a large amount of energy ... equivalent to
straining at the rate of 1 horsepower for almost exactly an
hour, or burning a 100 watt light bulb for about 7-1/2 hours.

The reason is the large mass that's being lifted.
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7 0
3 years ago
50 grams of ice cubes at -15°C are used to chill a water at 30°C with mass mH20 = 200 g. Assume that the water is kept in a foam
Arada [10]

Answer : The final temperature is, 25.0^oC

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

c_1 = specific heat of ice = 2.09J/g^oC

c_2 = specific heat of water = 4.18J/g^oC

m_1 = mass of ice = 50 g

m_2 = mass of water = 200 g

T_f = final temperature = ?

T_1 = initial temperature of ice = -15^oC

T_2 = initial temperature of water = 30^oC

Now put all the given values in the above formula, we get:

50g\times 2.09J/g^oC\times (T_f-(-15))^oC=-200g\times 4.184J/g^oC\times (T_f-30)^oC

T_f=25.0^oC

Therefore, the final temperature is, 25.0^oC

5 0
3 years ago
What renewable source of energy would be suited to chicago IL?
Kay [80]

Answer:

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Explanation:

7 0
3 years ago
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