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3 years ago

13

A steel bridge is 1000 m long at -20°C in winter. What is the change in length when the temperature rises to 40°C in summer? The

average coefficient of linear expansion of this steel is 11 × 10-6 C-1.

1 answer:

xenn [34]3 years ago

6 0

Answer:

ΔL = 0.66 m

Explanation:

The change in length on an object due to rise in temperature is given by the following equation of linear thermal expansion:

ΔL = αLΔT

where,

ΔL = Change in Length of the bridge = ?

α = Coefficient of linear thermal expansion = 11 x 10⁻⁶ °C⁻¹

L = Original Length of the Bridge = 1000 m

ΔT = Change in Temperature = Final Temperature - Initial Temperature

ΔT = 40°C - (-20°C) = 60°C

Therefore,

ΔL = (11 x 10⁻⁶ °C⁻¹)(1000 m)(60°C)

<u>ΔL = 0.66 m</u>

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2 years ago

A circuit consists of a battery connected to three resistors (65 ω, 25ω, and 170ω) in parallel. the total current through the re

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A. To find the total emf of the battery, just remember that in a parallel circuit, the voltage is the same throughout the circuit. So you can get the total voltage of the circuit by using Ohm's Law.

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Where:
I = current (A)
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Now you can derive the formula of Voltage by transposing the Resistance to the other side of the equation to isolate Voltage. The formula you will now use will be:
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However, you cannot solve this yet because the resistance you need is the total resistance in the circuit. To do this, you need to get the total resistance in this parallel circuit and the formula would be:

$\frac{1}{R_{T}} = \frac{1}{R_{1}}+ \frac{1}{R_{2}}+ \frac{1}{R_{3}}...+ \frac{1}{R_{n}}$

You have three resistors with the following resistance:
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$\frac{1}{R_{T}} = \frac{1}{R_{65}}+ \frac{1}{R_{25}}+ \frac{1}{R_{170}}$

$\frac{1}{R_{T}} =0.0153+0.04+0.006+0.0059$
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Get the reciprocal of both sides and divide:

$R_{T} = \frac{1}{0.0613} =16.32$

The total resistance then is 16.32Ω

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$V = 29.376V$

The emf of the battery is 29.376V

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The current flowing through resistor 1 with a resistance of 65Ω is 0.45A.

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The current flowing through resistor 2 with a resistance of 25Ω is 1.18A.

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$I= \frac{V}{R}$
$I= \frac{29.376}{170}$
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The current flowing through resistor 3 with a resistance of 170Ω is 0.17A.

If you add up all their current it confirms the given that the total current running through all of them is 1.8A.

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2 years ago

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Answer:

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Explanation:

See attached picture.

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