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Doss [256]
3 years ago
7

Rod A and rod B are cylindrical rods made of the same metal. amd they differ only in size. Rod B has double the length and doubl

e the width of rod a. In an experiment, rod A is found to comduct heat at a rate of 25 W between a hot reservoir and a cold reservoir. At what width of rod A. rate will rod B conduct heat between the same two reservoirs?
A.12.5 W
B.100 W
C. 25 W
D.50 W
E. 200 W
Physics
1 answer:
Mice21 [21]3 years ago
5 0

Answer:

it would take rod B twice as much time

Explanation:

it would take rod B twice as much time as it is twice as thick and twice as long. Due to this reason it would take the electric charge not only more time but even more voltage to travel through the rod

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Read 2 more answers
A copper wire and a tungsten wire of the same length have the same resistance. What is the ratio of the diameter of the copper w
spayn [35]

Answer:

Therefore the ratio of diameter of the copper to that of the tungsten is

\sqrt{3} :\sqrt{10}

Explanation:

Resistance: Resistance is defined to the ratio of voltage to the electricity.

The resistance of a wire is

  1. directly proportional to its length i.eR\propto l
  2. inversely proportional to its cross section area i.eR\propto \frac{1}{A}

Therefore

R=\rho\frac{l}{A}

ρ is the resistivity.

The unit of resistance is ohm (Ω).

The resistivity of copper(ρ₁) is 1.68×10⁻⁸ ohm-m

The resistivity of tungsten(ρ₂) is 5.6×10⁻⁸ ohm-m

For copper:

A=\pi r_1^2 =\pi (\frac{d_1}{2} )^2

R_1=\rho_1\frac{l_1}{\pi(\frac{d_1}{2})^2 }

\Rightarrow (\frac{d_1}{2})^2=\rho_1\frac{l_1}{\pi R_1 }......(1)

Again for tungsten:

R_2=\rho_2\frac{l_2}{\pi(\frac{d_2}{2})^2 }

\Rightarrow (\frac{d_2}{2})^2=\rho_2\frac{l_2}{\pi R_2 }........(2)

Given that R_1=R_2   and    l_1=l_2

Dividing the equation (1) and (2)

\Rightarrow\frac{ (\frac{d_1}{2})^2}{ (\frac{d_2}{2})^2}=\frac{\rho_1\frac{l_1}{\pi R_1 }}{\rho_2\frac{l_2}{\pi R_2 }}

\Rightarrow( \frac{d_1}{d_2} )^2=\frac{1.68\times 10^{-8}}{5.6\times 10^{-8}}   [since R_1=R_2   and    l_1=l_2]

\Rightarrow( \frac{d_1}{d_2} )=\sqrt{\frac{1.68\times 10^{-8}}{5.6\times 10^{-8}}}

\Rightarrow( \frac{d_1}{d_2} )=\sqrt{\frac{3}{10}}

\Rightarrow d_1:d_2=\sqrt{3} :\sqrt{10}

Therefore the ratio of diameter of the copper to that of the tungsten is

\sqrt{3} :\sqrt{10}

8 0
4 years ago
in the game of tetherball, the struck ball whirls around the pole. in what direction does the net force on the ball point?
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In the  tetherball game, struck ball whirls around a pole. The vertical component of the tension balances the weight and the horizontal component of tension provides the centripetal force that points toward the center of the circle.

There are two forces acting on a tether ball as it is rotating in uniform circular motion. The force of tension acts along the rope while the weight acts in downward.

The ball will move in a circular motion due to centripetal force and the tether which keeps the ball linked to the pole pulling to the center. For centripetal force, the tether can be an actual object, like a chain, or it may be a force, like gravity that acts like a tether.

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