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Mice21 [21]
3 years ago
6

Select the correct answer.

Physics
1 answer:
Aloiza [94]3 years ago
7 0

Answer:

B

Explanation:

B

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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
sergij07 [2.7K]

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.

4 0
3 years ago
Definition: A potential path for the flow of electricity with at least one gap in the path that prevents this flow. The gap is u
Andreas93 [3]

Answer:

Open circuit

Explanation:

An open circuit is simply an electrical circuit that is not complete. In such a circuit, there is a gap and this will not allow the electric current to pass through.

Despite all the elements being complete in the circuit, an open circuit will halt the flow of electric current and will not do deliver the necessary energy it is supposed to.

In such a circuit, the wires are cut of and not connected properly.

The reverse is a closed circuit.

7 0
3 years ago
Read 2 more answers
While spinning down from 500 rpm to rest, a flywheel does 3.9 kj of work. this flywheel is in the shape of a solid uniform disk
Ksivusya [100]

 

The answer is 4.0 kg since the flywheel comes to rest the kinetic energy of the wheel in motion is spent doing the work. Using the formula KE = (1/2) I w².

Given the following:

I =  the moment of inertia about the axis passing through the center of the wheel; w = angular velocity ; for the solid disk as I = mr² / 2 so KE = (1/4) mr²w². Now initially, the wheel is spinning at 500 rpm so w = 500 * (2*pi / 60) rad / sec = 52.36 rad / sec.

The radius = 1.2 m and KE = 3900 J

3900 J = (1/4) m (1.2)² (52.36)²

m = 3900 J / (0.25) (1.2)² (52.36)²

m = 3.95151 ≈ 4.00 kg

6 0
4 years ago
The low-_____ areas in a sound wave are called rarefactions. A. period B. frequency C. energy D. density
Delicious77 [7]
You can picture a sound wave a lot like a Slinky wave . . . a
thicker, compressed blob moving along the path, with thinner,
stretched-out places before and after it.

The thicker parts of a sound wave, where the air is more dense,
are called compressions.

The thinner parts of a sound wave, where the air is less dense,
are called rarefactions.
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3 years ago
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A very long, straight solenoid with a cross-sectional area of 2.00 cm2 is wound with 94.1 turns of wire per centimeter. Starting
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Yeah I have no clue what you saw me on lol yeah I
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