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Softa [21]
3 years ago
8

A low resistance light bulb and a high resistance light bulb are connected in parallel with each other. Which bulb is brighter i

f the parallel combination is in series with a battery?
Physics
1 answer:
sweet [91]3 years ago
6 0
<h2>Answer:</h2>

The bulb with low resistance will be brighter.

<h2>Explanation:</h2>

The brightness of a bulb is a function of both the voltage across the bulb and current flowing through the bulb. The higher the voltage, the higher the current. Hence the brighter the bulb.

Now, according to the question, the bulbs (the high resistance bulb and the low resistance bulb) are connected in parallel with each other. This means that the same voltage passes across them.

Also, we know that according to Ohm's law, the voltage (V) and current (I) through a conductor are related by the following equation;

V =  I x R                -------------------(i)

Where;

R is the resistance of the conductor.

We can re-write equation (i) as follows;

I = V / R               -----------------------(ii)

According to equation (ii), at fixed voltage (V), the current (I) will increase as the resistance (R) decreases.

Now, since the two bulbs have the same voltage, the bulb with the low resistance will allow a larger flow of current than the bulb with high resistance.  Therefore, as said earlier that brightness is dependent on voltage and current, the bulb with the low resistance (and having larger current at some voltage) will be brighter than the bulb with the high resistance (having smaller current at same voltage).

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dolphi86 [110]

Answer:

a = 10.07m/s^2

Their acceleration in meters per second squared is 10.07m/s^2

Explanation:

Acceleration is the change in velocity per unit time

a = ∆v/t

Given;

∆v = 50.0miles/hour - 0

∆v = 50.0miles/hours × 1609.344 metres/mile × 1/3600 seconds/hour

∆v = 22.352m/s

t = 2.22 s

So,

Acceleration a = ∆v/t = 22.352m/s ÷ 2.22s

a = 10.07m/s^2

Their acceleration in meters per second squared is 10.07m/s^2

7 0
2 years ago
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In an electricity demonstration at the Deutsches Museum in Munich, Germany, a person sits inside a metal sphere of radius 0.90 m
vladimir2022 [97]

Answer:

 E_interior = 0

Explanation:

As the sphere is metallic, the electrical charges are distributed on its surface, as far away as possible from each other.

If we apply Gauss's law, as the charge is on the surface, when drawing a spherical Gaussian surface, we see that there is no charge inside, therefore there is no electric field inside the metallic sphere.

          E_interior = 0

6 0
3 years ago
1) Andrea and Chuck are riding on a merry-go-round. Andrea rides on a horse at the outer rim of the circular platform, twice as
telo118 [61]

Explanation:

The tangential speed of Andrea is given by :

v=r\omega

Where

r is radius of the circular path

ω is angular speed

The merry-go-round is rotating at a constant angular speed. Let the new distance from the center of the circular platform is r'

r' = 2r

New angular speed,

v'=r'\omega'\\\\v'=(2r)\omega\\\\v'=2r\omega\\\\v'=2v

New angular speed is twice that of the Chuck's speed.

8 0
3 years ago
Does someone know how to do math with that equation
mestny [16]

Answer:

f = 5 cm

Explanation:

using the thin lens equation, given as follows:

\frac{1}{f} = \frac{1}{d_{o}}+\frac{1}{d_{i}}\\

where,

f = focal length = ?

do = the distance of object from lens = 20 cm

di = the distance of image from lens = 6.6667 cm

Therefore,

\frac{1}{f} = \frac{1}{20\ cm}+\frac{1}{6.6667\ cm}\\\\\frac{1}{f} =  0.199999\ cm^{-1}\\\\f = \frac{1}{0.199999\ cm^{-1}}\\\\

<u>f = 5 cm</u>

8 0
3 years ago
The strength of the electric field at a certain distance from a point charge is represented by E. What is the strength of the el
LUCKY_DIMON [66]

Answer:

e.)At twice the distance, the strength of the field is E/4.

Explanation:

The strength of the electric field at a certain distance from a point charge is given by:

E=k\frac{Q}{r^2}

where

k is the Coulomb's constant

Q is the charge

r is the distance from the point charge

In this problem, the distance from the point charge is doubled:

r' = 2r

So the new electric field strength is

E'=k\frac{Q}{(2r)^2}=k \frac{Q}{4 r^2}=\frac{1}{4} (k\frac{Q}{r^2})=\frac{E}{4}

so, at twice the distance the strength of the field is E/4.

4 0
2 years ago
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