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2 years ago

According to the law of conservation of energy, mechanical energy can be changed to heat energy.

Physics

1 answer:

Leno4ka [110]2 years ago

8 0

Answer:

The answer to this question is given below in this explanation section.

Explanation:

" law of conservation of energy"

The law of conservation of energy states that energy can neither be created nor destroyed only converted from one form of energy into another.This mean that a system always has a same account of a energy,unless it is added from the outside.This is particularly confusing in the case of non conversation forces,where energy is converted from ,mechanical energy into thermal energy.but the overall energy does remain the same.The only way to use energy is to transform energy from one form to another.

The amount of energy in any system than it is determined by the following equation.

Ut=Ui +W+Q

Ut is the total internal energy of a system.
Ui is the initial internal energy of a system.
W is the work done by or on the system.
Q is the heat added to or removed by the system.

It is also possible to determined the change in internal energy of the system using the equation.

ΔU=W+Q

The mechanical energy of a system increases provided their is no loss of energy due to friction.The energy would transform to kinetic energy when the speed is increasing.Te mechanical energy of a system remain constant provided their is no loss of energy due to friction.

The law of conversation of energy which say that in a closed system total energy is conserved that is it constant.

KE1 + PE1=KE2+PE2

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Answer:

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2 years ago

Dos cargas puntuales q1 = −50μC y q2 = +30μC se encuentran

alina1380 [7]

Answer:

Datos:

q1 = -50 μC = $50*10^{-6}$

q2 = +30 μC = $30*10^{-6}$

F = 10 N

a) x si la F = 10N

Aplicando la Ley de Coulomb:

x = $\sqrt\frac{k0 * q1 *q2}{F}$ = $\sqrt\frac{(9*10^{9} )*(50*10^{-6})*(30*10^{-6})}{10}$ = 1,162m

b) x si la F = 20 N

x= $\sqrt\frac{(9*10^{9} )*(50*10^{-6})*(30*10^{-6})}{20}$ = 0,822m

c)x si la F = 50 N

x = $\sqrt\frac{(9*10^{9} )*(50*10^{-6})*(30*10^{-6})}{50}$ = 0,520m

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2 years ago

You are observing the radiation from a distant active galaxy and you notice that the amplitude of the signal varies in strength

GREYUIT [131]

Answer:

Period of the signal.

Explanation:

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When You take a look at Quasars, we get the to know that the more you think you can see, the more they move away from us.

Thus, when "You are observing the radiation from a distant active galaxy and you notice that the amplitude of the signal varies in strength regularly over a certain period. The maximum possible size for the source of this radiation can now be calculated from the "PERIOD OF THE SIGNAL.

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3 years ago

T=2pi square root 1/g solve for g. Explanation would be really helpful.

Natalija [7]

I added individual steps for clarity. Note that g must be positive if the solution is to be real.

$T=2\pi \sqrt{\frac{1}{g}}=2\pi g^{-\frac{1}{2}}\\g^{-\frac{1}{2}} = \frac{T}{2\pi}\\(g^{-\frac{1}{2}})^{-2} = (\frac{T}{2\pi})^{-2}\\g = \frac{4\pi^2}{T^2}\,\,\,, g>0}$

Let me know if you have any questions.

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2 years ago

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Alexus [3.1K]

Answer:

Explanation:

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3 years ago