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4 years ago

14

Describe how total resistance is determined for series and parallel circuits

1 answer:

antiseptic1488 [7]4 years ago

3 0

-- When two or more resistance are in series, the net effective resistance
of the group is the sum of the individual resistances.

-- When two or more resistances are in parallel, the net effective resistance
of the group is

1 / (the sum of the reciprocals of the individual resistances) .

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Name two specific contributions bohr made to our understanding of atomic structure

MAXImum [283]

Https://answers.yahoo.com/question/index?qid=20120227184717AAzEq8g

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4 years ago

Faiyez wrote the problem below

Ilya [14]

445/100 - 5/4 = 445/100 - 125/100 = 320/100 = 16/5 = 3 1/5.

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4 years ago

You have four fixed-volume containers at STP . Container A has 0.5 mol of gas in 11.2 L. Container B has 2 mol of gas in 22.4 L.

Colt1911 [192]

Answer:

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3 years ago

A 51.0 kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 225 N. For the first 10.0 m th

shepuryov [24]

Answer:

The final speed of the crate is 12.07 m/s.

Explanation:

For the first 10.0 meters, the only force acting on the crate is 225 N, so we can calculate the acceleration as follows:

$F = ma$

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Now, we can calculate the final speed of the crate at the end of 10.0 m:

$v_{f}^{2} = v_{0}^{2} + 2ad_{1}$

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For the next 10.5 meters we have frictional force:

$F - F_{\mu} = ma$

$F - \mu mg = ma$

So, the acceleration is:

$a = \frac{F - \mu mg}{m} = \frac{225 N - 0.17*51.0 kg*9.81 m/s^{2}}{51.0 kg} = 2.74 m/s^{2}$

The final speed of the crate at the end of 10.0 m will be the initial speed of the following 10.5 meters, so:

$v_{f}^{2} = v_{0}^{2} + 2ad_{2}$

$v_{f} = \sqrt{(9.39 m/s)^{2} + 2*2.74 m/s^{2}*10.5 m} = 12.07 m/s$

Therefore, the final speed of the crate after being pulled these 20.5 meters is 12.07 m/s.

I hope it helps you!

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3 years ago

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igor_vitrenko [27]

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3 years ago