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3 years ago

A circuit with several current paths, whose total current equals the sum of the current in its branches is a _____ circuit.

2 answers:

6 0

<span>A circuit with several current paths, whose total current equals the sum of the current in its branches is a parallel circuit.</span>

disa [49]3 years ago

4 0

Answer:

Parallel

Explanation:

A circuit with several current paths can be connected in two basic forms; series and parallel.

When a circuit is connected in series, the same current flows through out the circuit.

However where the circuit is connected in parallel, the sum of the currents flowing in the branches gives the total current flowing in the circuit.

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1=A.0.83m/s

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A mini rail gun made of a copper rod of mass m and radius r rests on two parallel copper rails that are a distance L apart and o

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Answer:

Explanation:

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Work done by this force = F X d

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3 years ago

In an experiment performed in a space station, a force of 74 N causes an object to have an acceleration equal to 9 m/s2. What is

Stells [14]

Answer:

<h3>The answer is 8.2 kg</h3>

Explanation:

The mass of the object can be found by using the formula

$m = \frac{f}{a} \\$

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a is the acceleration

From the question we have

$m = \frac{74}{9} \\ = 8.222222...$

We have the final answer as

Hope this helps you

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3 years ago

How much work would be needed to raise the payload from the surface of the moon (i.e., x = r) to the "end of the universe"?

meriva

Moon was rotation Earth or sun
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3 years ago

Read 2 more answers

I attach a 4.1 kg block to a spring that obeys Hooke's law and supply 3.8 J of energy to stretch the spring. I release the block

borishaifa [10]

Answer:

The amplitude of the oscillation is 2.82 cm

Explanation:

Given;

mass of attached block, m = 4.1 kg

energy of the stretched spring, E = 3.8 J

period of oscillation, T = 0.13 s

First, determine the spring constant, k;

$T = 2\pi \sqrt{\frac{m}{k} }$

where;

T is the period oscillation

m is mass of the spring

k is the spring constant

$T = 2\pi \sqrt{\frac{m}{k} } \\\\k = \frac{m*4\pi ^2}{T^2} \\\\k = \frac{4.1*4*(3.142^2)}{(0.13^2)} \\\\k = 9580.088 \ N/m\\\\$

Now, determine the amplitude of oscillation, A;

$E = \frac{1}{2} kA^2$

where;

E is the energy of the spring

k is the spring constant

A is the amplitude of the oscillation

$E = \frac{1}{2} kA^2\\\\2E = kA^2\\\\A^2 = \frac{2E}{k} \\\\A = \sqrt{\frac{2E}{k} } \\\\A = \sqrt{\frac{2*3.8}{9580.088} }\\\\A = 0.0282 \ m\\\\A = 2.82 \ cm$

Therefore, the amplitude of the oscillation is 2.82 cm

8 0

2 years ago