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3 years ago

The current through a 10 ohm resistor is 1.2 amperes.What is the potential difference across the resistor?

Physics

1 answer:

Natalija [7]3 years ago

6 0

<u>Answer</u>: The potential difference across the resistor is 12 volts.

<u>Explanation:</u>

To calculate the potential difference cross the resistor, we use Ohm's Law. This law states that the potential difference across two wires is directly proportional to the current flowing through that wire.

Mathematically,

$V\propto I\\V=IR$

Where,

V = potential difference = ?V

I = Current flowing = 1.2 A

R = Resistor = $10\Omega$

Putting values in above equation, we get:

$V=1.2\times 10=12V$

Hence, the potential difference across the resistor is 12 volts

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A charge of 12 c passes through an electroplating apparatus in 2. 0 min. what is the average current?

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A charge of 12 c passes through an electroplating apparatus in 2.0 min, then the average current is 0.1 ampere.

<h3>What is an electric charge?</h3>

Charged material experiences a force when it is exposed to an electromagnetic field due to the physical property of electric charge. You can have a positive or negative electric charge.

Electric current is defined as the charge per unit of time.

The mathematical relation between current and the electric charge

I =Q/T

where I is the current flowing

Q is the total electric charge

T is the time period for which the current is flowing

As given in the problem A charge of 12 c passes through an electroplating apparatus in 2.0 min

Let us first convert the time period of minutes into seconds

1 min = 60 seconds

2 min = 2*60 seconds

=120 seconds

By using the above relation between electric current and electric charge

and by substituting the respective values of the charge and the time period

I =Q/T

I = 12c/120 seconds

I = 0.1 Ampere

Thus, the average current flowing through the apparatus would be 0.1 Ampere.

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5 0

1 year ago

Visible light passes through a diffraction grating that has 900 slits per centimeter, and the interference pattern is observed o

kobusy [5.1K]

Answer:

$\Delta \lambda=14.3\ nm$

Explanation:

It is given that,

The number of lines per unit length, N = 900 slits per cm

Distance between the formed pattern and the grating, l = 2.3 m

n the first-order spectrum, maxima for two different wavelengths are separated on the screen by 2.98 mm, $\Delta Y=2.98\ mm = 0.00298\ m$

Let d is the slit width of the grating,

$d=\dfrac{1}{N}$

$d=\dfrac{1}{900\ cm}$

$d=1.11\times 10^{-5}\ m$

For the first wavelength, the position of maxima is given by :

$y_1=\dfrac{L\lambda_1}{d}$

For the other wavelength, the position of maxima is given by :

$y_2=\dfrac{L\lambda_2}{d}$

So,

$\Delta \lambda=\dfrac{\Delta y d}{l}$

$\Delta \lambda=\dfrac{0.00298\times 1.11\times 10^{-5}}{2.3}$

$\Delta \lambda=1.43\times 10^{-8}\ m$

$\Delta \lambda=14.3\ nm$

So, the difference between these wavelengths is 14.3 nm. Hence, this is the required solution.

3 0

2 years ago

A proton and an alpha particle are momentarily at rest at adistance r from each other. They then begin to move apart.Find the sp

Arte-miy333 [17]

Answer:

The unknown quantities are:

E and F

The final velocity of the proton is:

√(8/3) k e^2/(m*r)

Explanation:

Hello!

We can solve this problem using conservation of energy and momentum.

Since both particles are at rest at the beginning, the initial energy and momentum are:

Ei = k (q1q2)/r

pi = 0

where k is the coulomb constant (= 8.987×10⁹ N·m²/C²)

and q1 = e and q2 = 2e

When the distance between the particles doubles, the energy and momentum are:

Ef = k (q1q2)/2r + (1/2)m1v1^2 + (1/2)m2v2^2

pf = m1v1 + m2v2

with m1 = m, m2 = 4m, v1=vf_p, v2 = vf_alpha

The conservation momentum states that:

pi = pf

Therefore:

m1v1 + m2v2 = 0

That is:

v2 = (1/4) v1

The conservation of energy states that:

Ei = Ef

Therefore:

k (q1q2)/r = k (q1q2)/2r + (1/2)m1v1^2 + (1/2)m2v2^2

Replacing

m1 = m, m2 = 4m, q1 = e, q2 = 2e

and v2 = (1/4)v1

We get:

(1/2)mv1^2 = k e^2/r + (1/2)4m(v1/4)^2 = k e^2/r + (1/8)mv1^2

(3/8) mv1^2 = k e^2/r

v1^2 = (8/3) k e^2/(m*r)

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3 years ago