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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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A disk of radius R = 11 cm is pulled along a frictionless surface with a force of F = 16 N by a string wrapped around the edge.A
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Explanation:

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Consider the interference pattern produced by two parallel slits of width a and separation d, in which d = 3a. The slits are ill
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Answer:

a)   m =1  θ = sin⁻¹  λ  / d,  m = 2        θ = sin⁻¹ ( λ  / 2d) ,   c)     m = 3

Explanation:

a) In the interference phenomenon the maxima are given by the expression

         d sin θ = m λ

the maximum for m = 1 is at the angle

          θ = sin⁻¹  λ  / d

the second maximum m = 2

          θ = sin⁻¹ ( λ  / 2d)

the third maximum m = 3

        θ = sin⁻¹ ( λ  / 3d)

the fourth maximum m = 4

       θ = sin⁻¹ ( λ  / 4d)

b) If we take into account the effect of diffraction, the intensity of the maximums is modulated by the envelope of the diffraction of each slit.

       I = I₀ cos² (Ф) (sin x / x)²

       Ф = π d sin θ /λ

       x = pi a sin θ /λ

where a is the width of the slits

with the values ​​of part a are introduced in the expression and we can calculate intensity of each maximum

c) The interference phenomenon gives us maximums of equal intensity and is modulated by the diffraction phenomenon that presents a minimum, when the interference reaches this minimum and is no longer present

maximum interference       d sin θ = m λ

first diffraction minimum    a sin θ = λ

we divide the two expressions

                       d / a = m

In our case

                   3a / a = m

                    m = 3

order three is no longer visible

7 0
2 years ago
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