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3 years ago

Q. Calculate the annual energy

Physics

1 answer:

melomori [17]3 years ago

3 0

Answer:

60 kWh

Explanation:

The computation of the annual energy consumption in KW-h is shown below:

As we know that

1 kw = 1000 w

So, for 1400 W it would be

= 1,400 ÷ 1,000

= 1.4 kW

Now the number of hours it used in a year

= 7 minutes × 365 days ÷ 60 minutes

= 42.58333 hours

So in one year it used

= 1.4 kW × 42.58333

= 59.61 kWh

= 60 kWh

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The protective device that switches off if the circuit overloads is a _______.

Kitty [74]

Answer:

E.) Circuit Breaker is the correct answer.

Explanation:

Circuit Breaker is an automated electrical device that is designed to control and stop the excessive current flowing if the crosses a particular level and it protects the electrical circuit from the damage that can be caused due to overload of current that is the reason Circuit Breaker is called protective device.

There are different types of circuit breakers such as oil circuit breaker, vacuum circuit breaker, outdoor circuit breaker, Spring operated circuit breaker, Hydraulic circuit breaker and Medium-voltage circuit breaker, etc.

7 0

3 years ago

Find the currents flowing in the circuit in the figure below. (Assume the resistances are

Nitella [24]

By Kirchoff's law in left side loop

$E_1 + E_2 = {r_1 + R_1 + R_4)I_1 + (R_2 + r_2) I_3$

similarly kirchoff's law in right side loop

$E_3 - E_4 - E_2 = (r_3 + r_4 + R_3)I_2 - (R_2 + r_2)I_3$

also by junction law we know that

$I_1 = I_2 + I_3$

now by plug in all values we have

$18 + 3 = (0.5 + 8 + 15)I_1 + (10 + 0.25)I_3$

$21 = 23.5I_1 + 10.25I_3$

$12 - 24 - 3 = (0.75 + 0.25 + 12)I_2 - (10+ 0.25)I_3$

$-15 = 13I_2 - 10.25I_3$

So by solving above equations we have

$I_1 = 0.492 A$

$I_2 = -0.428 A$

$I_3 = 0.920 A$

6 0

3 years ago

A solid sphere of radius 40.0cm has a total positive charge of 26.0μC uniformly distributed throughout its volume. Calculate the

Rudiy27

The magnitude of the electric field for 60 cm is 6.49 × 10^5 N/C

R(radius of the solid sphere)=(60cm)( 1m /100cm)=0.6m

$Q\;(\text{total charge of the solid sphere})=(26\;\mathrm{\mu C})\left(\dfrac{1\;\mathrm{C}}{10^6\;\mathrm{\mu C}} \right)={26\times 10^{-6}\;\mathrm{C}}$

Since the Gaussian sphere of radius r>R encloses all the charge of the sphere similar to the situation in part (c), we can use Equation (6) to find the magnitude of the electric field:

$E=\dfrac{Q}{4\pi\epsilon_0 r^2}$

Substitute numerical values:

$E&=\dfrac{24\times 10^{-6}}{4\pi (8.8542\times 10^{-12})(0.6)}\\ &={6.49\times 10^5\;\mathrm{N/C}\;\text{directed radially outward}}}$

The spherical Gaussian surface is chosen so that it is concentric with the charge distribution.

As an example, consider a charged spherical shell S of negligible thickness, with a uniformly distributed charge Q and radius R. We can use Gauss's law to find the magnitude of the resultant electric field E at a distance r from the center of the charged shell. It is immediately apparent that for a spherical Gaussian surface of radius r < R the enclosed charge is zero: hence the net flux is zero and the magnitude of the electric field on the Gaussian surface is also 0 (by letting QA = 0 in Gauss's law, where QA is the charge enclosed by the Gaussian surface).

Learn more about Gaussian sphere here:

brainly.com/question/2004529

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