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Lady_Fox [76]
3 years ago
11

In an electrical circuit, what happens to the current flowing through the wire if the initial voltage of 18 V is doubled, and th

e initial resistance of 35 Ohms is reduced by a factor of 4?
ANSWER QUICKLY PLEASE! ITS URGENT!
Physics
2 answers:
Evgen [1.6K]3 years ago
5 0

Answer:

I hope this helps you :)

Explanation:

The current is directly proportional to the voltage and inversely proportional to the resistance. ... So doubling or tripling the resistance will cause the current to be one-half or one-third of the original value.

Genrish500 [490]3 years ago
3 0

The current is increased by a factor of 8.

From 18/35 A. to 144/35 A.

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Liquid pools of methane are found on the surface of Titan, one of Saturn's moons. The temperature on the surface of Titan is -18
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Answer:

-292 degrees F

Explanation:

C = 5/9( F -32)

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F =  -292 degrees

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21. If the Sun's rays were at 45° to a vertical pillar, how would
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Answer:

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Answer:

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Explanation:

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Two neutron stars are separated by a distance of 1.0 x 1012 m. They each have a mass of 1.0 x 1028 kg and a radius of 1.0 x 103
son4ous [18]

To develop this problem it is necessary to apply the concepts related to Gravitational Potential Energy.

Gravitational potential energy can be defined as

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As M=m, then

PE = -\frac{Gm^2}{R}

Where,

m = Mass

G =Gravitational Universal Constant

R = Distance /Radius

PART A) As half its initial value is u'=2u, then

U = -\frac{2Gm^2}{R}

dU = -\frac{2Gm^2}{R}

dKE = -dU

Therefore replacing we have that,

\frac{1}{2}mv^2 =\frac{Gm^2}{2R}

Re-arrange to find v,

v= \sqrt{\frac{Gm}{R}}

v = \sqrt{\frac{6.67*10^{-11}*1*10^{28}}{1*10^{12}}}

v = 816.7m/s

Therefore the  velocity when the separation has decreased to one-half its initial value is 816m/s

PART B) With a final separation distance of 2r, we have that

2r = 2*10^3m

Therefore

dU = Gm^2(\frac{1}{R}-\frac{1}{2r})

v = \sqrt{Gm(\frac{1}{2r}-\frac{1}{R})}

v = \sqrt{6.67*10^{-11}*10^{28}(\frac{1}{2*10^3}-\frac{1}{10^{12}})}

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7 0
3 years ago
A 110 kg hoop rolls along a horizontal floor so that the hoop's center of mass has a speed of 0.220 m/s. How much work must be d
Softa [21]

Answer:

the work that must be done to stop the hoop is 2.662 J

Explanation:

Given;

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W = ΔK.E

W = ¹/₂mv²

W = ¹/₂ x 110 x 0.22²

W = 2.662 J

Therefore, the work that must be done to stop the hoop is 2.662 J

6 0
3 years ago
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