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3 years ago

When electrical energy is being used by an electric light, what really happens to the energy?

Physics

2 answers:

vitfil [10]3 years ago

7 0

Energy is not created and not destroyed it will only change form

So heres your answer ; It is given off as other forms of energy/light and heat !!

=answer 2nd one

Ronch [10]3 years ago

5 0

A. it is given off as other forms of energy

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3 years ago

These capacitors are then disconnected from their batteries, and the positive plates are now connected to each other and the neg

Kay [80]

Answer:

Following are the solution to the given question:

Explanation:

For charging plates that are connected in a similar manner:

Calculating the total charge:

$\to q =q_1 + q_2 = C_1V_1 +C_2V_2 =1320 + 2714 = 4034 \mu C$

Calculating the common potential:

$\to V = \frac{q}{C}= \frac{q}{(C_1 + C_2)} =\frac{4034}{6.8} = 593 \ V\\\\$

Calculating the charge after redistribution:

$When: \\\\q = q_{1}' + q_{2}' = q_1 + q_2$

$\to q_{1}' = C_1V = 2.2 \times 593 = 1305\ \mu C\\ \\ \to q_{2}' = C_2V = 4.6 \times 593 = 2729 \ \mu C$

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3 years ago

A laser beam is incident on two slits with a separation of 0.230 mm, and a screen is placed 4.75 m from the slits. If the bright

Mariulka [41]

Answer:

$7.55\times 10^{-7} m$

Explanation:

We are given that

d=0.23 mm= $0.23\times 10^{-3} m$

$1mm=10^{-3} m$

Screen is placed from the slits at distance ,L=4.75 m

The bright interference fringes on the screen are separated by 1.56 cm.

$\Delta y=1.56 cm=1.56\times 10^{-2} m$

1 m=100 cm

We have to find the wavelength of laser light.

We know that

$\Delta y=\frac{\lambda L}{d}$

Substitute the values

$1.56\times 10^{-2}=\frac{\lambda\times 4.75}{0.23\times 10^{-3}}$

$\lambda=\frac{1.56\times 10^{-2}\times 0.23\times 10^{-3}}{4.75}$

$\lambda=7.55\times 10^{-7} m$

4 0

3 years ago