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GuDViN [60]
3 years ago
15

What happens to the current supplied by the battery when you add an identical bulb in parallel to the original bulb?

Physics
1 answer:
Marysya12 [62]3 years ago
6 0

Answer:

b. The current stays the same.

Explanation:

In the case given current is supplied by the battery to a bulb . Here, we should know that bulb also apply resistance to the flow of current .

Now, when an identical bulb is connected in parallel to the original bulb .

Therefore, both the resistance( bulb) are in parallel.

We know, when two resistance are in parallel , current through them is same and voltage is divided between them.

Therefore, in this case current stays same in the original bulb.

Hence, this is the required solution.

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qwelly [4]

Answer:

B

Explanation:

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3 years ago
True or false Weight is constant every place in the universe.
ycow [4]
False. it's depend on g -constant.

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4 years ago
How does heat energy move by convection when a cook heats a pot of water?
laiz [17]

The correct choice is

B. Particles at the bottom of the water carry heat energy to the top of the water.

when pot of water is heater, the bottom of pot gets heated. the particles of water in contact with the bottom of the pot gets heat through conduction. after getting heat, these particles of water near the bottom, move away towards top and their position is taken by cooler particles from top. that way heat travels

3 0
3 years ago
Read 2 more answers
A 1.5 kg cart is attached to a spring with spring constant of 5 N/m. The cart & spring is pulled to stretch the spring by 3
vaieri [72.5K]

22.5 J

Explanation:

Given:

x = 3 m

k = 5\:\text{N/m}

The spring potential energy PE_s is

PE_s = \frac{1}{2}kx^2 = \frac{1}{2}(5\:\text{N/m})(3\:\text{m})^2

\:\:\:\:\:\:\:=22.5\:\text{J}

3 0
3 years ago
Calculate the kinetic energy that the earth has because of (a) its rotation about its own axis and (b) its motion around the sun
Dominik [7]

Answer:

a. K_{Axis}=2.574x10^{29}J

b. K_{Orbit}=2.6577x10^{33}J

Explanation:

K_{Axis}=\frac{1}{2}I*w^2

I_{Sphere}=\frac{2}{5}*m*r^2

w=\frac{2\pi }{T} , T=24hrs*\frac{3600s}{1hr} =86400s

radius earth = 6371 km

mass earth = 5,972*10^24 kg

a.

K_{Axis}=\frac{1}{2}*\frac{2}{5}*m*r^2*(\frac{2\pi}{T})^2

K_{axis}=\frac{4\pi^2}{5}*5.98x10^{24}kg*(6.38x10^6m)^2*(\frac{1}{86400s})^2

K_{Axis}=2.574x10^{29}J

b.

T=1year*\frac{365day}{1year}*\frac{24hr}{1day}*\frac{3600s}{1hr}=31536000s

K_{Orbit}=\frac{1}{2}*I*w

I=m*r^2

K_{Orbit}=\frac{1}{2}*m*r^2*(\frac{2\pi}{T})^2

K_{Orbit}=\frac{4\pi^2}{5}*5.98x10^{24}*6.38x10^6m*(\frac{1}{31.536x10^6s})^2

K_{Orbit}=2.6577x10^{33}J

4 0
4 years ago
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