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Stells [14]
3 years ago
12

Find the ratio of the diameter of aluminium to copper wire, if they have the same

Physics
1 answer:
kicyunya [14]3 years ago
3 0

Answer:

1.24

Explanation:

The resistivity of copper\rho_1=2.65\times 10^{-8}\ \Omega-m

The resistivity of Aluminum,\rho_2=1.72\times 10^{-8}\ \Omega-m

The wires have same resistance per unit length.

The resistance of a wire is given by :

R=\rho \dfrac{l}{A}\\\\R=\rho \dfrac{l}{\pi (\dfrac{d}{2})^2}\\\\\dfrac{R}{l}=\rho \dfrac{1}{\pi (\dfrac{d}{2})^2}

According to given condition,

\rho_1 \dfrac{1}{\pi (\dfrac{d_1}{2})^2}=\rho_2 \dfrac{1}{\pi (\dfrac{d_2}{2})^2}\\\\\rho_1 \dfrac{1}{{d_1}^2}=\rho_2 \dfrac{1}{{d_2}^2}\\\\(\dfrac{d_2}{d_1})^2=\dfrac{\rho_1}{\rho_2}\\\\\dfrac{d_2}{d_1}=\sqrt{\dfrac{\rho_1}{\rho_2}}\\\\\dfrac{d_2}{d_1}=\sqrt{\dfrac{2.65\times 10^{-8}}{1.72\times 10^{-8}}}\\\\=1.24

So, the required ratio of the diameter of Aluminum to Copper wire is 1.24.

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constant force F=5i+5j−1kF=5i+5j−1k is applied to an object that is moving along a straight line from the point (−5,−3,−4)(−5,−3
meriva

Answer:

W = 71J

Explanation:

Given force F = (5i+5j−1k)N

d = Δr

r1 = (−5,−3,−4)m

r2 = (2,5,0)m

Δr = r2 – r1 = (2-(-5), 5-(-3), 0-(-4))

Δr = (2+5, 5+3, 0+4) = (7i+ 8j +4k)m

W = F•d = (5i+5j−1k)•(7i+ 8j +4k)

W = 5×7 + 5×8 +-1×4 = 35 + 40 - 4

W = 71J

6 0
3 years ago
Hey guys, is this question for a series circuit or for parallel?
Mandarinka [93]

Answer:

I need help with the same question

Explanation:

3 0
3 years ago
Suppose there was a star with a parallax angle of 1 arcsecond. How far away would it be? Select all that apply.
lesya692 [45]

Answer:

option E

Explanation:

given,                          

Parallax angle(d) = 1 arcsecond

using Parallax formula                  

      d = \dfrac{1}{p}

 p is the parsecs angle which is measured in 1 arcsecond

 d is the distance in parsec

now,                                            

      P = \dfrac{1}{d}

      P = \dfrac{1}{1}

      P = 1 \ parsec

we know,                                  

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hence, the answer will be option E

6 0
4 years ago
PLEASE HELP (100)
morpeh [17]

Answer:

The work done will be W=55.27\: J

Explanation:

The work equation is given by:

W=F\cdot x

Where:

F is the force due to gravity (weight = mg)

x is the length of the ramp (3 m)

Now, the force acting here is the component of weight in the ramp direction, so it will be:

F_{x-direction}=mgsin(28)

Therefore, the work done will be:

W=mgsin(28)*3

W=4*9.81*sin(28)*3

W=55.27\: J

I hope it helps you!

3 0
3 years ago
A 565 N rightward force pulls a large box across the floor with a constant velocity of 0.75 m/s. If the coefficient of friction
alukav5142 [94]
The first thing you should know is that the friction force is equal to the coefficient of friction due to normal force.
 Therefore, clearing the normal force we have:
 The friction is 565N.
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6 0
4 years ago
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