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3 years ago

A 30 ohm resistor and a 20 ohm resistor are

Physics

1 answer:

Reptile [31]3 years ago

8 0

The current that would pass through the 30 ohms resistor is 2 A.

<h3>What is electric current?</h3>

Electric current is the rate of flow of electric charge round a conductor.

To calculate the electric current that would pass through the 30 ohms resistor, we use the formula below

Formula:

I = V/Rt........... Equation 1

Where:

I = Electric current passing through the 30 ohms resistor
V = Voltage
Rt = Total or effective resistance of the resistors.

From the question,

Given:

V = 100 volts
Rt = (30+20) ohms (since both resistors are connected in series)

Rt = 50 ohms

Substitute these values into equation 1

I = 100/50
I = 2 A

Hence, The current that would pass through the 30 ohms resistor is 2 A.

Learn more about electric current here: brainly.com/question/1100341

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3 years ago

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Answer:

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3 years ago

The frequency of the musical note F#3 is 1.85x102 hertz. What is its period?

g100num [7]

The period of the musical note is $\text { 5. } 4 \times 10^{-3}$ seconds.

Answer: Option A

<u>Explanation:</u>

The frequency is defined as the number of oscillations or a complete cycle of wave occurred in a given time interval. So the frequency is inversely proportional to the time period. Thus the mathematical representation of frequency with time period is

$\text {Frequency}=\frac{1}{\text {Time period }}$

As the frequency is given as $1.85 \times 10^{2} \mathrm{Hz}$ , the time period can be found as

$\text { Time period }=\frac{1}{\text { Frequency }}$

Thus,

$\text { Time period }=\frac{1}{1.85 \times 10^{2}}=5.4 \times 10^{-3} \mathrm{s}$

Thus the time period for the frequency of the musical note is $\text { 5. } 4 \times 10^{-3}$ seconds.

5 0

3 years ago

As shown in the figure below, a bullet is fired at and passes through a piece of target paper suspended by a massless string. Th

NikAS [45]

Answer:

M = 0.730*m

V = 0.663*v

Explanation:

Data Given:

$v_{bullet, initial} = v\\v_{bullet, final} = 0.516*v\\v_{paper, initial} = 0\\v_{paper, final} = V\\mass_{bullet} = m\\mass_{paper} = M\\Loss Ek = 0.413 Ek$

Conservation of Momentum:

$P_{initial} = P_{final}\\m*v_{i} = m*0.516v_{i} + M*V\\0.484m*v_{i} = M*V .... Eq1$

Energy Balance:

$\frac{1}{2}*m*v^2_{i} = \frac{1}{2}*m*(0.516v_{i})^2 + \frac{1}{2}*M*V^2 + 0.413*\frac{1}{2}*m*v^2_{i}\\\\0.320744*m*v^2_{i} = M*V^2\\\\M = \frac{0.320744*m*v^2_{i} }{V^2} ....... Eq 2$