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Verizon [17]
3 years ago
8

A wire of length L and cross-sectional area A has resistance R.

Physics
2 answers:
lara31 [8.8K]3 years ago
4 0

Answer:

The resistance will be twice the original resistance

Explanation:

This is a fairly simple question, the formular for the resistance of a wire is given as

R = rho *length / area

Where R = resistance

Rho = resistivity

L = length

A = area

Since the density and area are constant I.e they do not change

R/ length = rho/ area

The initial length is given as L, when this length is stretched to twice its original length ,it becomes 2×L = 2L

Let x represent the resistance when the length is doubled

R/ L = x / 2L

x = 2LR / L ; dividing by L

We have that x = 2R ; twice the resistance

soldi70 [24.7K]3 years ago
4 0

Answer:

Explanation:

Given:

L2 = 2 × L1

Using the formula,

Resistivity, d = (R × A)/L

Where,

R = resistance

A = area

L = length

1. d1 = (R1 × A1)/L1

2. d2 = (R2 × A2)/L2

Equating both 1 and 2 together,

(R2 × A2)/L2 = (R1 × A1)/L1

(R2 × A2)/(2 × L1) = (R1 × A1)/L1

Assume A1 = A2,

R2 = [(R1 × A1) × 2 × L1]/(A1 × L1)

R2 = 2 × R1

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Answer:

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Explanation:

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8 0
3 years ago
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For the image of the overhead projector to be in focus, the distance from the projector lens to the image, <img src="https://tex
rjkz [21]
Given:
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projector lens focal length, f
distance from the transparency to the projector lens, do

thin lens equation: 1/f = 1/di + 1/do
do = 4 inches
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convert feet to inches, for uniformity.
1 foot = 12 inches
8 feet * 12 inches/ft = 96 inches
 
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Adding fractions, denominator must be the same.

1/f = (1/96 * 1/1) + (1/4 * 24/24)
1/f = 1/96 + 24/96
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4 0
3 years ago
Please help!! I will give brainliest!
grandymaker [24]

Answer:

Depends.

Explanation:

Whether the object is going left or right, the speed will stay the same until friction eventually stops it. <em>However, </em>if, for example, we're talking about an object going straight before veering right, then yes, speed <em>does</em> matter. An object will normally have to speed up or slow down momentarily when changing direction to keep itself sustained on the ground.

So, honestly? It really depends on what we're talking about!

Hope this helped!

Source(s) used: None.

7 0
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<span>Hope this helps.</span>
7 0
3 years ago
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