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3 years ago

Calculate the total current in parallel circuit using Kirchhoff’s Current Law? anyone give example and solution...

Physics

1 answer:

Ad libitum [116K]3 years ago

4 0

Answer:

please check attachment and follow the steps.

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Two blocks can collide in a one-dimensional collision. The block on the left hass a mass of 0.30 kg and is initially moving to t

snow_lady [41]

Answer:

a) 3.632 m/s

b) 0.462 m/s

Explanation:

Using the law of conservation of momentum:

$m_{1} u_{1} + m_{2} u_{2}= m_{1} V_{1} + m_{2} V_{2}$ ..........(1)

$m_{1} = 0.30 kg\\u_{1} = 2.4 m/s\\m_{2} = 0.80 kg\\u_{2} = 0 m/s$

Substituting the above values into equation (1) and make V2 the subject of the formula:

$0.3(2.4) + 0.80(0)= 0.3 V_{1} + 0.8 V_{2}\\$

$V_{2} = \frac{0.72 - 0.3 V_{1}}{0.8}$ ..................(2)

Using the law of conservation of kinetic energy:

$0.5m_{1} u_{1} ^{2} + 1.2 = 0.5m_{1} V_{1} ^{2} + 0.5m_{2} V_{2} ^{2}\\0.5(0.3) (2.4) ^{2} + 1.2 = 0.5(0.3) V_{1} ^{2} + 0.5(0.8)V_{2} ^{2}\\$

$2.064 = 0.15 V_{1} ^{2} + 0.4V_{2} ^{2}$ .......(3)

Substitute equation (2) into equation (3)

$2.064 = 0.15 V_{1} ^{2} + 0.4(\frac{0.72 - 0.3V_{1} }{0.8}) ^{2}\\2.064 = 0.15 V_{1} ^{2} + 0.4(\frac{0.5184 - 0.432V_{1} + 0.09V_{1} ^{2} }{0.64}) \\1.32096 = 0.096 V_{1} ^{2} + 0.20736 - 0.1728V_{1} + 0.036V_{1} ^{2} \\0.132 V_{1} ^{2} - 0.1728V_{1} - 1.1136 = 0\\V_{1} = 3.632 m/s$

Substituting $V_{1}$ into equation(2)

$V_{2} = \frac{0.72 - 0.3 *3.632}{0.8}\\V_{2} = \frac{0.72 - 0.3 *(3.632)}{0.8}\\V_{2} = 0.462 m/s$

8 0

3 years ago

Select the correct answer.

svet-max [94.6K]

Heya!!!

Answer to your question:
C.
Energy transferred to a machine equals energy transferred from it.

Law of conservation of energy states that total energy of an isolated system remains constant
Energy can neither be created nor be destroyed. Machine follows the same rule.

Hope it helps ^_^

6 0

3 years ago

Read 2 more answers

How much work is done on 10.0C of charge to move it through a potential difference of 9V in 10s?

Norma-Jean [14]

Here is the link with ans on it
https://moorsscience.wikispaces.com/file/view/chapter+12+answers.pdf
hope it helps

5 0

3 years ago

Consider the electric field lines drawn below for a configuration of two charges. Several locations are labeled on the diagram.

geniusboy [140]

Answer:

The magnitude or strength of an electric field in the space surrounding a source charge is related directly to the quantity of charge on the source charge and inversely to the distance from the source charge. The direction of the electric field is always directed in the direction that a positive test charge would be pushed or pulled if placed in the space surrounding the source charge.

Explanation:

I really hope this helps you!

6 0

2 years ago

What happens to centripetal acceleration as the radius of curvature decreases and the speed is constant, and why

lutik1710 [3]

It increases, because the centripetal acceleration is inversely proportional to the radius of the curvature.

Hopr it helps :)

4 0

3 years ago