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3 years ago

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Physics

1 answer:

fomenos3 years ago

4 0

Answer:

The bulbs should be connected in parallel.

Explanation:

We want to find out a way to hook up 2 light bulbs and a battery so that when one bulb burns out or is disconnected the other bulbs stays lit.

We must connect the two bulbs in parallel so that even when one bulb is burns out, it will have no effect on the other bulb and the 2nd bulb will keep on working. The current flowing in each bulb will depend upon the resistance of each bulb and the voltage will be same across each bulb.

On the other hand, if we use a series circuit then if one bulb burns out then the there is no flow of current in the circuit and therefore, the second bulb will not be operational.

The current flowing through each bulb is given by

I = V/R

The voltage across each bulb is given by

V = IReq

Where I is the current and Req is the equivalent resistance of the two bulbs connected in parallel and is given by

Req = (R₁*R₂)/(R₁+R₂)

The connection diagram is attached where two bulbs are connected in parallel and are power with a battery.

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siniylev [52]

Answer:

The size (diameter) of the basketball's shadow on the wall is approximately 53.38 cm

Explanation:

The given parameters of the basketball are;

The diameter of the basketball = 23 cm (9 inches)

The distance of the light bulb from the closest side of the basketball = 3 m

The distance from the ball to the wall = 4 m

The distance from the light source to the center of the ball, d = 3 m + 0.23/2 m = 3.115 m

The angle the light ray makes with the edge of the ball, θ = arctan(0.115/3.115)

Therefore, the ratio of the shadow width divided by 2 to the distance from the light from the wall = 0.115/3.115

The distance from the light from the wall = 3 m + 4 m + 0.23 m = 7.23 m

Therefore;

((The width of the shadow)/2)/(The distance from the light from the wall) = 0.115/3.115

∴ ((The width of the shadow)/2)/(7.23 m) = 0.115/3.115

((The width of the shadow)/2) = 7.23 m × 0.115/3.115 = 16629/62300 m ≈ 0.2669 m = 26.69 cm

The width (diameter) of the shadow on the wall = 2 × 16629/62300 m ≈ 0.5338 m = 53.38 cm

The size (diameter) of the basketball's shadow on the wall ≈ 53.38 cm

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3 years ago

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MakcuM [25]

Answer:

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