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viktelen [127]
3 years ago
6

The copper wire and bulb is connected in a series with 220 V electric supply. Why only an electric bulb glows where as the coppe

r wire remains the same.Give REASON to support the answer.
Physics
1 answer:
soldi70 [24.7K]3 years ago
3 0
<span>-- In a series circuit, the current ( I ) is the same at every point.

-- The power dissipated by any section of the circuit is I² x Resistance.

-- The wire has very low resistance, so I²R is very low dissipated power.

-- The filament in the bulb has most all of the resistance in the circuit,
so it dissipates virtually all the power of the circuit, and certainly much
more than the wires do.
</span>
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C) Explain relative velocity with examples.​
Mnenie [13.5K]

Answer:

we encounter ocassion where one or more object move in the which is not stationary with respect to another example a boat is cross a river that is flowing at some rate of aeroplane encountring wind durning it motion

5 0
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The rate at which a candle burns in millimeters per minute is:
anzhelika [568]
since both components, length and time, are measurable 
<span>since Rate = length ÷ time </span>
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You have a nightlight plugged into an outlet in the hallway, which uses 3.5 watts when plugged in. If the house circuit provides
nekit [7.7K]
Base on the said question or problem that state and ask to calculate the current of the said light bulb and in my further calculation and further analysis, I would say that the current of the light bulb would be 0.0292. I hope you are satisfied with my answer and feel free to ask for more 
7 0
3 years ago
Read 2 more answers
Calculate the power of an engine which can supply 60kg of energy per minute​
Nutka1998 [239]

Answer:

1 Watt

Explanation:

Power (P) = w / Δt

w = work done

Δt = change in time

Power (P) = 60kg / 60 seconds

Power (P) = 1 Watt

6 0
4 years ago
A horizontal 649 N merry-go-round of radius 1.05 m is started from rest by a constant horizontal force of 61.3 N applied tangent
nexus9112 [7]

Answer:

The kinetic energy of the merry-go-round is 632.82 J

Explanation:

Given;

weight of the merry-go-round, W = 649 N

radius of the merry-go-round, r = 1.05 m

applied horizontal force, F =  61.3 N

acceleration due to gravity, g = 9.8 m/s²

mass of  merry-go-round, m = W/g

                                               = 649/9.8  = 66.225 kg

moment of inertia of merry-go-round, I = ¹/₂mr²

                                                                 = ¹/₂ x 66.225 x (1.05)²

                                                                 = 36.507 kg.m²

Angular acceleration of the merry-go-round, α

τ = Iα = Fr

α = Fr / I

Where;

α is angular acceleration

α = (61.3 x 1.05) / 36.507

α = 1.763 rad/s²

Angular velocity of the merry-go-round, ω

ω = αt

ω = 1.763 x 3.34

ω = 5.888 rad/s

Finally, the kinetic energy of the merry-go-round, K.E

K.E = ¹/₂Iω²

K.E = ¹/₂ x 36.507 x (5.888)²

K.E = 632.82 J

4 0
3 years ago
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