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aivan3 [116]
4 years ago
14

¿que es la energia interna?

Physics
2 answers:
solong [7]4 years ago
6 0
La energía interna es el resultado de la contribución de la energía cinética
Vladimir [108]4 years ago
4 0

Answer:

La energía interna es el resultado de la contribución de la energía cinética de las moléculas o átomos que lo constituyen, de sus energías de rotación, traslación y vibración, además de la energía potencial intermolecular debida a las fuerzas de tipo gravitatorio, electromagnético y nuclear.

Explanation:

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What is measured in watts that starts with a p
Nesterboy [21]
Power is measured in watts
6 0
3 years ago
How much work (in J) is required to expand the volume of a pump from 0.0 L to 2.7 L against an external pressure of 1.0 atm?
spin [16.1K]

Answer:

-2.79 x 10²J

Explanation:

using the pressure volume work formula which states that

work = -PΔV

           = -(1.0 atm) (2.7-0.0)L = -2.7L . atm

Convert litre.atmosphere to Joules.

1 L . atm = 101.325 joules

-2.75 L .atm = -2.75 x 101.325 = -278.64375 =

Work = -2.79 x 10²J

5 0
4 years ago
How many times larger than a centigram is a dekagram
Umnica [9.8K]

Answer:

A dekagram is thousand (1000) times larger than a centigram.

Explanation:

→ [1 dekagram = 1,000 centigrams]

→ 1 dekagram = 10 grams

→ 10 grams = 100 decigrams

→ 100 decigrams = 1,000 centigrams

3 0
3 years ago
W
ludmilkaskok [199]

Answer: W represents an element Y represents a mixture

Explanation: I just took a quiz on this and got it right. Good Luck

4 0
3 years ago
The combined focal length of two thin lens is 24 cm and the focal length of one converging lens is 8
qwelly [4]

Answer: f = -12 cm

Explanation: <u>Combined</u> <u>lenses</u> is an array of  simple lenses with a common axis. The combination is useful for correction of optical aberrations which cannot be corrected by simple lenses.

When two lenses are in contact and are thin, focal lengths are related as:

\frac{1}{F} =\frac{1}{f_{1}} +\frac{1}{f_{2}}

If there is a distance between the lenses, the focal length will be:

\frac{1}{F} =\frac{1}{f_{1}} +\frac{1}{f_{2}} -\frac{d}{f_{1}f_{2}}

Since the lenses in the question above are thin and in contact, the focal length of one of them will be:

\frac{1}{F} =\frac{1}{f_{1}} +\frac{1}{f_{2}}

\frac{1}{f_{2}} =\frac{1}{f_{1}} -\frac{1}{F}

\frac{1}{f_{2}} =\frac{1}{8} -\frac{1}{24}

\frac{1}{f_{2}} =\frac{-2}{24}

f_{2}= -12

The focal length of the other lens is -12 cm, with the negative sign meaning it's a converging lens.

7 0
3 years ago
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