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Tamiku [17]
3 years ago
6

A battery with a potential difference of V connects to a resistor, and the resulting current is measured. You remove the battery

, and a new battery with a potential difference of 2V is put in its place. As a result,a. the current will double.b. the resistance will double.c. both the current and the resistance will double.d. battery replacement will have no effect on the circuit.
Physics
2 answers:
dybincka [34]3 years ago
8 0

Answer:

a. the current will doubled.

Explanation:

IN CASE 1:

Let , R be the resistance of resistor and it is given that the potential difference produced by battery is V.

Therefore, current through circuit is , I=\dfrac{V}{R}.

Now,

IN CASE 2:

The potential difference now is doubled i.e. 2 V.

Since, resistance of a body depends on its nature.

Therefore, Temperature remains constant.

So, current through circuit is , I_2=\dfrac{2V}{R}=2\times I.

Therefore, potion a) is correct.

Oksi-84 [34.3K]3 years ago
5 0

Answer:

a. the current will double

Explanation:

I = Current

V = Voltage

R = Resistance

From Ohm's law we have the relation

V=IR\\\Rightarrow I=\dfrac{V}{R}

It can be seen that the current is proportional to the voltage.

This means that with increase in current there will be increase in voltage and vice versa.

So, if the voltage doubles the current will double

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Two concentric conducting spherical shells produce a radially outward electric field of magnitude49,000 N/C at a point 4.10 m fr
Korolek [52]

To solve this problem it is necessary to apply the concepts related to the electric field according to the definition of Coulomb's law.

 The electric field is defined mathematically as a vector field that associates to each point in space the (electrostatic or Coulomb) force per unit of charge exerted on an infinitesimal positive test charge at rest at that point.

Mathematically this can be described as:

E = \frac{1}{4\pi \epsilon_0}\frac{q}{r^2}

Where,

\epsilon_0 = permittivity of free space

r = Distance

q = Charge

E = Electric Field

Our values are given as,

E= 49.000N/C

r= 4.1m

\epsilon_0=8.854*10{-12}C^2N^{-1}m^{-2}

Replacing we have,

E = \frac{1}{4\pi \epsilon_0}\frac{q}{r^2}

49000 = \frac{1}{4\pi (8.854*10{-12})}\frac{q}{(4.1)^2}

q= 9.16*10^{-5} C

q= 91.6\mu C

Therefore the amoun of charge on the outer surface of the larger shell is 91.6 \mu C

6 0
4 years ago
PLEASE HELP ME IM TIMED
dusya [7]

Answer:

imma just get lol its d

Explanation:

6 0
3 years ago
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A bag of blood with a density of 1050 kg/m3 is raised about 1.00 m higher than the level of a patient's arm. How much greater is
SSSSS [86.1K]

Answer:

A bag of blood with a density of 1050 kg/m4 is raised about 1.00 m higher than the level of a patient's arm. How much greater is the blood pressure at the patient's arm than it would be if the bag were at the same height as the army is the solisan of kind

8 0
3 years ago
A ray of light crosses a boundary between two transparent materials. The medium the ray enters has a larger optical density. Whi
Tpy6a [65]

Answer:

The wavelength of the light decreases as it enters into the medium with the greater optical density.

The frequency of the light remains constant as it transitions between materials.

Explanation:

- When a ray of light crosses a boundary between two different materials, it undergoes refraction: the ray changes direction, and it also changes speed, according to the relationship:

v=\frac{c}{n}

where v is the speed of the ray of light in the material, c is the speed of light in a vacuum, n is the index of refraction of the material, which is larger for a medium with larger optical density. So, from the equation, we see that the larger the optical density, the smaller the speed of the wave.

- The frequency of the light does not depend on the properties of the medium, so it remains unchanged: therefore the statement

The frequency of the light remains constant as it transitions between materials.

is correct.

- Moreover, the wavelength of the ray of light is related to the speed and the frequency by the equation

\lambda=\frac{v}{f}

where v is the speed and f the frequency. Since we have seen that v decreases and f remains constant, this means that the wavelength decreases as well, so the statement

The wavelength of the light decreases as it enters into the medium with the greater optical density.

is also correct.

5 0
3 years ago
Lila is a track and field athlete. She must complete four laps around a circular track. The track itself is a 400 meter track an
fgiga [73]

Answer: 4.45m/s

Explanation:

I assume we're solving for her speed.

400m x 4 laps = 1600m

1600m / 6 minutes = 266.67m/min

(266.67m/min) x (1 minute/60 seconds) = 4.45m/s

266.67/60 = 4.45m/s

6 0
4 years ago
Read 2 more answers
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