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3 years ago

a current of 5 ampere is passed for 2 hours in an electric iron having a resistance of 100 ohms calculate the heat produced

Physics

1 answer:

Alina [70]3 years ago

3 0

Answer:

$\boxed{\sf Heat \ produced \ (H) = 5 \ kWh}$

Given:

Resistance (R) = 100 Ω

Current (I) = 5 A

Time (t) = 2 hours

To Find:

Heat developed (H) in the electric iron

Explanation:

Formula:

$\boxed{ \bold{ \sf H = I^2Rt}}$

Substituting values of I, R & t in the equation:

$\sf \implies H = {5}^{2} \times 100 \times 2 \\ \\ \sf \implies H = 25 \times 100 \times 2 \\ \\ \sf \implies H = 5000 \\ \\ \sf \implies H = 5 \: kWh$

$\therefore$

Heat developed (H) in the electric iron = 15 kWh

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F = m × a

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a baseball is traveling (=20m/s) and us hit by a bat. it leaves the bat traveling (-30 m/s). What is the change in the velocity?

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Suppose you have three identical metal spheres, AA, BB, and CC. Initially sphere AA carries a charge qq and the others are uncha

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Complete Question

Suppose you have three identical metal spheres, A, B, and C. Initially sphere A carries a charge q and the others are uncharged. Sphere A is brought in contact with sphere B, and then the two are separated. Spheres CC and BB are then brought in contact and separated. Finally spheres AA and CC are brought in contact and then separated. What is the final charge on the sphere B, in terms of q?

a. 3/8q

b. 1/4q

c. 3/4q

d. q

e. 5/8q

f. 1/3q

g.1/2q

h. 0

Answer:

The correct option is b

Explanation:

From the question we are told that

The charge carried by A is q C

The charge carried by B is 0 C

The charge carried by C is 0 C

When A and B are brought close and then separated the charge carried by A and B is mathematically evaluated as

$\frac{ 0 + q}{2} = \frac{q}{2}$

When C and B are brought close and then separated the charge carried by C and B is mathematically evaluated as

$\frac{0 + \frac{q}{2} }{2} = \frac{q}{4}$

When C and A are brought close and then separated the charge carried by C and A is mathematically evaluated as

$\frac{\frac{q}{4} + \frac{q}{2} }{2} = \frac{3q}{8}$

Looking at these calculation we can see that the charge carried by B is

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