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Olegator [25]
3 years ago
8

Several light bulbs, each of resistance 1.5 Ω, are connected in a series across a 120 V source of emf. If the current through th

e circuit is 2 A, how many light bulbs are there in the circuit?
Physics
1 answer:
Leni [432]3 years ago
7 0
<h3><u>Answer;</u></h3>

40 light bulbs

<h3><u>Explanation</u>;</h3>

The total resistance of components or bulbs in series is given as the sum of resistance of all the components.

Thus; if there are bulbs in series each with a resistance of 1.5 Ω, the the total resistance will be; 1.5nΩ

From the ohms law;

V = IR , where V is the voltage, I is the current and R is the resistor.

Thus; R = V/i

         R = 120/2

            = 60 Ω

But, there are n bulbs each with 1.5 Ω; thus there are;

n = 60/1.5

<u>  = 40 Bulbs </u>

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Practice questions, will mark brainliest!
andrew-mc [135]

Answer:

266.67Watts

Explanation:

Time = 2.5hr to seconds

3600s = 1hr

2.5hrs = 3600×2.5= 9000s

Force = 32N

Distance = 75km  to m

1000m = 1km

75km = 1000×75 = 75000m

Power = workdone / time

Work = force × distance

Therefore work = 32N × 75000m

Work = 2400000Nm

Power = work ➗ time

Power = 2400000Nm ➗ 9000s

Power = 266.67Watts

Watts is the S. i unit of power

I hope this was helpful, please mark as brainliest

4 0
3 years ago
A measure of how easily current will pass through a material ?
Jlenok [28]

Answer:

It is Conductivity because it is the measure of the ease.

6 0
3 years ago
Why metals have thermoconductivity higher than ceramic?
Pie

Answer:

Thermal Conductivity Easily Transmits Heat Among Fine Ceramics

4 0
2 years ago
Water is flowing in a pipe with a circular cross section but with varying cross-sectional area, and at all points the water comp
slamgirl [31]

(a) 5.66 m/s

The flow rate of the water in the pipe is given by

Q=Av

where

Q is the flow rate

A is the cross-sectional area of the pipe

v is the speed of the water

Here we have

Q=1.20 m^3/s

the radius of the pipe is

r = 0.260 m

So the cross-sectional area is

A=\pi r^2 = \pi (0.260 m)^2=0.212 m^2

So we can re-arrange the equation to find the speed of the water:

v=\frac{Q}{A}=\frac{1.20 m^3/s}{0.212 m^2}=5.66 m/s

(b) 0.326 m

The flow rate along the pipe is conserved, so we can write:

Q_1 = Q_2\\A_1 v_1 = A_2 v_2

where we have

A_1 = 0.212 m^2\\v_1 = 5.66 m/s\\v_2 = 3.60 m/s

and where A_2 is the cross-sectional area of the pipe at the second point.

Solving for A2,

A_2 = \frac{A_1 v_1}{v_2}=\frac{(0.212 m^2)(5.66 m/s)}{3.60 m/s}=0.333 m^2

And finally we can find the radius of the pipe at that point:

A_2 = \pi r_2^2\\r_2 = \sqrt{\frac{A_2}{\pi}}=\sqrt{\frac{0.333 m^2}{\pi}}=0.326 m

6 0
3 years ago
Rock at the top of a 20 m tall hill the rock has a mass of 10 kg how much potential energy does it have
Blizzard [7]
PE = 10 * 10 * 20 = 2000 Joule
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3 years ago
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