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Genrish500 [490]
3 years ago
14

When a voltage difference is applied to a piece of metal wire, a 10.0 mA current flows through it. If this metal wire is now rep

laced with a silver wire having twice the diameter of the original wire, how much current will flow through the silver wire?
Physics
1 answer:
Mrrafil [7]3 years ago
6 0

Answer:

10 mA

Explanation:

We have given when a potential difference is applied then a current of 10 mA is flows so current i = 10 mA

According to ohms law voltage V =i R

And resistance is given as R= \frac{\rho l}{A}

From the relation it is clear that resistance is inversely proportional to the area

As in new case the wire have twice diameter means 4 times the area so the resistance will decrease by 4 times from the previous resistance

As current i=\frac{V}{R} current is inversely proportional to resistance so current will become 4 times of previous value

So current = 10×4=40 mA

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200J

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K.E=½×16kg×5m/s²

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A cat with a mass of 4.50 kilograms sits on a ledge 0.800 meters above the ground. If it jumps to the ground, how much kinetic e
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5 0
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Read 2 more answers
1. Simplify the following expression<br> 8-6/4-12+3^2
DanielleElmas [232]

Answer:

-5/2

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8-6/4-12+3^2

8-6/4-12+9

2/4-12+9

LCM =4

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5 0
2 years ago
You are comparing two diffraction gratings using two different lasers: a green laser and a red laser. You do these two experimen
Nikolay [14]

Answer:

a. (a) grating A has more lines/mm; (b) the first maximum less than 1 meter away from the center

Explanation:

Let  n₁ and n₂ be no of lines per unit length  of grating A and B respectively.

λ₁ and λ₂ be wave lengths of green and red respectively , D be distance of screen and d₁ and d₂ be distance between two slits of grating A and B ,

Distance of first maxima for green light

= λ₁ D/ d₁

Distance of first maxima for red light

= λ₂ D/ d₂

Given that

λ₁ D/ d₁ = λ₂ D/ d₂

λ₁ / d₁ = λ₂ / d₂

λ₁ / λ₂  = d₁ / d₂

But

λ₁  <  λ₂

d₁ < d₂

Therefore no of lines per unit length of grating A will be more because

no of lines per unit length  ∝ 1 / d

If grating B is illuminated with green light first maxima will be at distance

λ₁ D/ d₂

As λ₁ < λ₂

λ₁ D/ d₂ < λ₂ D/ d₂

λ₁ D/ d₂ < 1 m

In this case position of first maxima will be less than 1 meter.

Option a is correct .

5 0
3 years ago
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