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dusya [7]
3 years ago
12

How much current must be applied across a 60 Ω light bulb filament in order for it to consume 55 W of power? Unserious answers w

ill be deleted, and reported. (Please show some work)
Physics
1 answer:
sergij07 [2.7K]3 years ago
4 0

Answer: current I = 0.96 Ampere

Explanation:

Given that the

Resistance R = 60 Ω 

Power = 55 W

Power is the product of current and voltage. That is

P = IV ...... (1)

But voltage V = IR. From ohms law.

Substitutes V in equation (1) power is now

P = I^2R

Substitute the above parameters into the formula to get current I

55 = 60 × I^2

Make I^2 the subject of formula

I^2 = 55/60

I^2 = 0.92

I = sqr(0.92)

I = 0.957 A

Therefore, 0.96 A current must be applied.

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Calculate the energy of a photon emitted when an electron in a hydrogen atom undergoes a transition from =7 to =1.
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1.549×10-19lJ is the energy of a photon emitted when an electron in a hydrogen atom undergoes a transition from =7 to =1.

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The Rydberg formula is used to determine the energy change.

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2 years ago
What is the equivalent resistance of a circuit that contains three 10.0  resistors connected in series to a 6.0 V battery
disa [49]
The equivalent resistance of n resistors in series is given by:
R_{eq} = R_1 + R_2 + ... + R_n
In our circuit, we have three resistors of 10.0 \Omega each, therefore the equivalent resistance of the circuit is
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4 0
3 years ago
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3 years ago
Who exerts more pressure? a) A girl of 50 kg, wearing heels with an area of 1 cm2. b) An elephant of 4000 kg with foot area of 2
Mrrafil [7]

Answer:

The girl exerts more pressure.

Explanation:

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i.e P = F/A

<u>Girls</u>

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Force = mg = 50 × 10 = 500N

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\frac{500}{10 ^{ - 4} }

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<u>Elephant</u>

<u>Area</u><u> </u><u>=</u><u> </u><u>2</u><u>5</u><u>0</u><u>cm</u><u>²</u><u> </u><u>=</u><u> </u><u>2</u><u>.</u><u>5</u><u> </u><u>x</u><u> </u><u>1</u><u>0</u><u>^</u><u>(</u><u>-</u><u>2</u><u>)</u><u>b</u><u> </u><u>m</u><u>²</u>

<u>Force</u><u> </u><u>=</u><u> </u><u>mg</u><u> </u><u>=</u><u> </u><u>4</u><u>0</u><u>0</u><u>0</u><u>0</u><u>N</u>

<u>Pressure</u><u> </u><u>=</u><u> </u>

<u>\frac{40000}{2.5 \times  {10}^{ - 2} }</u>

<u>= 1.6 \times  {10}^{6}</u>

5 0
3 years ago
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