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mixer [17]
2 years ago
13

Energy in inductors: you need an inductor that will store 20 j of energy when a 3. 0-a current flows through it. what should be

its self-inductance?
Physics
1 answer:
Diano4ka-milaya [45]2 years ago
3 0

The self inductance can be calculated as 4.444 Henry.

The inductor will store energy = 20 joule,

current = 3 ampere.

<h3>What is Self inductance?</h3>

     The current carrying conductor which allows or opposes the change of flow of current is known as self-inductance. The rate of change of current can also be given through self inductance.

Formula can be given  for the energy in the inductor as,

                      E = 1 / 2 × L × I² joule.

where as,    E - Energy

                    L - Inductance

                    I - Current flowing through the coil.

The equation can be rearranged as,

                      L =  2 E/ I²

                         =  2 × 20 / 3²

                         =  40 / 9

Hence,           L =  4.4444 Henry

Learn more about energy in inductors,

brainly.com/question/28030901

#SPJ4

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A woman walked 115 m. As she did so, her speed increased from 4.20 m/s to 5.00 m/s. How long did it take her to walk this distan
ASHA 777 [7]

Answer:

25 seconds

Explanation:

Assuming the woman is accelerating at a constant rate of a \;\;m/s^2 from the initial velocity, u=4.20 m/s, to the final velocity, v=5.00 m/s.

Let she takes t seconds to cover the distance, s=115 m.

As acceleration, a=\frac{v-u}{t}=\frac{5-4.2}{t}

\Rightarrow at=0.8\cdots(i)

Now, from the equation of motion

s=ut+\frac 12 at^2

\Rightarrow s=ut+\frac 12 at(t)

\Rightarrow 115=4.2t+\frac 12 \times 0.8 t [ from equation (i)]

\Rightarrow 115=(4.2+0.4)t

\Rightarrow t= 115/4.6 = 25 seconds.

Hence, she takes 25 seconds to walk the distance.

5 0
3 years ago
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
what is the potential difference across the headlight bulbs when the starter motor is operated, requiring an additional 39 a fro
frosja888 [35]

The starter motor's potential difference across the headlight bulbs is 38.45V, requiring an additional 39 a from the battery. Voltage, also known as potential difference.

It is sometimes described as the amount of work needed to move a test charge between two sites, expressed as a unit of charge. Volt is the potential difference's SI unit (V). We only take into account the charge between the locations P and Q when current moves between them in an electric circuit. Electric potential difference between two sites is referred to as voltage, also known as electric pressure, electric tension, or (electric) potential difference. an electric field that is static.

Vh = I*Rn

Vh = 39/5.476*5.40v

Vh = 38.45v

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5 0
1 year ago
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