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Paladinen [302]

3 years ago

14

What help in reversing direction of current of current

1 answer:

Novay_Z [31]3 years ago

3 0

Answer:

<em>To reverse the direction of an electric current, we simply reverse the voltage either automatically with the help of some switching circuitry or manually by changing the voltage source terminals connection. </em>

Explanation:

For electric current to flow, there must be a potential difference, usually referred to as the voltage. The electric current flow is analogous to the flow of water under the action of a pump, through a series of pipe connections. The voltage is similar to the driving action of the pump, and current flows the same way water flows. The resistance due to drag on the pipe wall is equivalent to electric resistance. For current to flow in the reverse direction, the voltage or rather, the potential difference is changed, causing the current to flow in the opposite direction. This can be done by switching the terminals of the voltage source, or by automatic means. The automatic switching can be done with a transistor based circuitry.

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The x vector component of a displacement vector has a magnitude of 146 m and points along the negative x axis. The y vector comp

larisa86 [58]

Answer:

a) the magnitude of r is 184.62

b) the direction is 37.74° south of the negative x-axis

Explanation:

Given the data in the question;

as illustrated in the image blow;

To find the the magnitude of r, we will use the Pythagoras theorem

r² = y² + x²

r = √( y² + x²)

we substitute

r = √((-113)² + (-146)²)

r = √(12769 + 21316 )

r = √(34085 )

r = 184.62

Therefore, the magnitude of r is 184.62

To find its direction, we need to find ∅

from SOH CAH TOA

tan = opposite / adjacent

tan∅ = -113 / -146

tan∅ = 0.77397

∅ = tan⁻¹( 0.77397 )

∅ = 37.74°

Therefore, the direction is 37.74° south of the negative x-axis

7 0

2 years ago

postnew [5]

Explanation:

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5 0

3 years ago

How much money would be saved by turning off one 100.0-W lightbulb 3.0 h/day for 365 days if the

Pavlova-9 [17]

Answer:

the money that would be saved is $13.14.

Explanation:

Given;

power consumed by the light bulb, P = 100 W = 0.1 kW

time of running the bulb, t = 3 hours for 365 days = 1,095 hours

cost rate of power consumption, C = $0.12 per kWh

Energy consumed by the light bulb for the given days;

E = Pt

E = 0.1 kW x 1,095 hr

E = 109.5 kWh

Cost of energy consumed = 109.5 kWh x $0.12 / kWh

= $13.14

Therefore, the money that would be saved is $13.14.

3 0

3 years ago

If the tension of the cable is 25.0 N what is the mass of the ball

marysya [2.9K]

Answer:576

Explanation:

5 0

3 years ago

By what factor must the amplitude of a sound wave be decreased in order to decrease the intensity by a factor of 3?

s344n2d4d5 [400]

Answer:

Amplitude is decreased by a factor of $\sqrt3$ if intensity is decreased by a factor of 3.

Explanation:

Intensity of a sound wave is directly proportional to the square of its amplitude.

Therefore, if intensity is $I$ and amplitude is $A$ , then

$I=kA^2$ , where, $k$ is constant of proportionality.

Now, if intensity of sound wave is decreased by a factor of 3. So,

New intensity is, $I_{new}=\frac{I}{3}$

$I_{new}=kA_{new}^2\\\frac{I}{3}=kA_{new}^2$

Plug in $kA^2$ for $I$ . This gives,

$\frac{kA^2}{3}=kA_{new}^2\\A_{new}^2=\frac{A^2}{3}\\A_{new}=\sqrt{\frac{A^2}{3}}=\frac{A}{\sqrt{3}}$

Therefore, amplitude is decreased by a factor of $\sqrt3$ .

4 0

3 years ago