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1 year ago

8

What is the equivalent resistance of a circuit that contains four 75.02

1 answer:

viktelen [127]1 year ago

3 0

Answer:

option A is the correct answer

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How to find the maximum amount of a substance that will dissolve using a graph?

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Solubility indicates the maximum amount of a substance that can be dissolved in a solvent at a given temperature. Such a solution is called saturated. Divide the mass of the compound by the mass of the solvent and then multiply by 100 g to calculate the solubility in g/100g .

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A rock is dropped from the top of a tall tower. Half a second later another rock, twice as massive as the first, is dropped. Ign

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B- the acceleration is greater for the more massive rock

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1 year ago

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Consider a 2250-lb automobile clocked by law-enforcement radar at a speed of 85.5 mph (miles/hour). if the position of the car i

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As per law of Heisenberg uncertainty law

product of uncertainty in position and uncertainty in momentum will be constant

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2 years ago

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PLEASE HELP. NOTICE HOW THEY PUT FUTURE

MrMuchimi

I think its <u>B</u> because it looks like it might indicate a future rain storm.

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3 years ago

A +71 nC charge is positioned 1.9 m from a +42 nC charge. What is the magnitude of the electric field at the midpoint of these c

viva [34]

Answer:

The net Electric field at the mid point is 289.19 N/C

Given:

Q = + 71 nC = $71\times 10^{- 9} C$

Q' = + 42 nC = $42\times 10^{- 9} C$

Separation distance, d = 1.9 m

Solution:

To find the magnitude of electric field at the mid point,

Electric field at the mid-point due to charge Q is given by:

$\vec{E} = \frac{Q}{4\pi\epsilon_{o}(\frac{d}{2})^{2}}$

$\vec{E} = \frac{71\times 10^{- 9}}{4\pi\8.85\times 10^{- 12}(\frac{1.9}{2})^{2}}$

$\vec{E} = 708.03 N/C$

Now,

Electric field at the mid-point due to charge Q' is given by:

$\vec{E'} = \frac{Q'}{4\pi\epsilon_{o}(\frac{d}{2})^{2}}$

$\vec{E'} = \frac{42\times 10^{- 9}}{4\pi\8.85\times 10^{- 12}(\frac{1.9}{2})^{2}}$

$\vec{E'} = 418.84 N/C$

Now,

The net Electric field is given by:

$\vec{E_{net}} = \vec{E} - \vec{E'}$

$\vec{E_{net}} = 708.03 - 418.84 = 289.19 N/C$

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2 years ago