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mixer [17]
2 years ago
12

In a certain process, the energy of the system decreases by 250 kJkJ. The process involves 480 kJkJ of work done on the system.

Find the amount of heat QQQ transferred in this process
Physics
1 answer:
BigorU [14]2 years ago
5 0

Answer:

-730KJ

Explanation:

According to the first law of thermodynamics;

Let the total energy of the system be ∆E

Let heat be q and let work the w

Since the energy decreases ∆E is negative

Since work is done on the system w= positive

So;

- 250 = 480 + q

q = -250 - 480

q=-730KJ

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A = 200/100

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Summarize: Based on what you have learned, how will the sound that the observer hears
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Read 2 more answers
A +71 nC charge is positioned 1.9 m from a +42 nC charge. What is the magnitude of the electric field at the midpoint of these c
viva [34]

Answer:

The net Electric field at the mid point is 289.19 N/C

Given:

Q = + 71 nC = 71\times 10^{- 9} C

Q' = + 42 nC = 42\times 10^{- 9} C

Separation distance, d = 1.9 m

Solution:

To find the magnitude of electric field at the mid point,

Electric field at the mid-point due to charge Q is given by:

\vec{E} = \frac{Q}{4\pi\epsilon_{o}(\frac{d}{2})^{2}}

\vec{E} = \frac{71\times 10^{- 9}}{4\pi\8.85\times 10^{- 12}(\frac{1.9}{2})^{2}}

\vec{E} = 708.03 N/C

Now,

Electric field at the mid-point due to charge Q' is given by:

\vec{E'} = \frac{Q'}{4\pi\epsilon_{o}(\frac{d}{2})^{2}}

\vec{E'} = \frac{42\times 10^{- 9}}{4\pi\8.85\times 10^{- 12}(\frac{1.9}{2})^{2}}

\vec{E'} = 418.84 N/C

Now,

The net Electric field is given by:

\vec{E_{net}} = \vec{E} - \vec{E'}

\vec{E_{net}} = 708.03 - 418.84 = 289.19 N/C

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