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3 years ago

-Which of the following wires will have the least resistance?

2 answers:

7 0

The wires that will have the least resistance is :
C. A short thick wire
in order to get the least resistence, you need the wire to be the lowest in length and the highest in Area

hope this helps

Pani-rosa [81]3 years ago

5 0

Answer:

Option (C)

Explanation:

The obstruction offered by the conductor in the path of flow of electrons is called the resistance of the conductor.

The reistance of the conductor is directly proportional to the length of the conductor and inversely proportional to the area of crossection of the conductor.

To minimize the resistance we need a wire of short length and more crossection area.

So, wire should be short and thick.

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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

$PE = -\frac{GMm}{R}$

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

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

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

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

$v = 1.83*10^7m/s$

Therefore the velocity when they are about to collide is $1.83*10^7m/s$

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3 years ago

You toss a walnut at a speed of 15.0 m/s at an angle of 50.0∘ above the horizontal. The launch point is on the roof of a buildin

kaheart [24]

Answer:

a.3.51s

b.33.8m

c. 9.64,-22.9

Explanation:

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2 years ago

A round object of mass 10 kg and radius 0.5 m rolls without slipping down a hill from a height of 4.5 m. If its speed at the bot

Mice21 [21]

Answer:

moment of inertia is 2.72 kg m²

Explanation:

given data

mass m = 10kg

height h = 4.5 m

radius r = 0.5 m

speed v = 6.5 m/s

to find out

moment of inertia

solution

we apply here conservation of energy

that is

mgh = 1/2 ×mv² + 1/2 × Iω²

here I is moment of inertia we find and

we know ω = Velocity / radius = 6.5 / 0.5 = 13

and g = 9.8

so put here all these value

10 (9.8) 4.5 = 1/2 ×(10)(6.5)² + 1/2 × I(13)²

441 = 211.25 + 1/2 × I( 169 )

I = 2.72

so moment of inertia is 2.72 kg m²

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3 years ago

Essam is abseiling down a steep cliff. How much gravitational potential energy does he lose for every metre he descends? His mas

Dafna11 [192]

Answer:

720 J

Explanation:

The gravitational potential energy that Essam loses for every metre is given by:

$\Delta U=mg \Delta h$

where

m=72 kg is Essam's mass

$g=10 N/kg$ is the gravitational field strength

$\Delta h=1 m$ is the difference in height

By substituting the numbers into the formula, we find

$\Delta U=(72 kg)(10 N/kg)(1 m)=720 J$

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3 years ago

Read 2 more answers

Compared to the inertia of a 1-kilogram mass, the inertia of a 4-kilogram mass is

Mumz [18]

Explanation:

The inertia of a 4 kg mass is four times as great as a 1 kg mass.

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3 years ago