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Lana71 [14]
3 years ago
8

The coefficient of linear expansion of copper is 17 x 10^-6 K-1. A block of copper 30 cm wide, 45 cm long, and 10 cm thick is he

ated from 0°C to 100°C What is the change in the volume of the block?
Physics
1 answer:
schepotkina [342]3 years ago
7 0

Answer:

The change in volume is 6.885\times 10^{- 5}\

Solution:

As per the question:

Coefficient of linear expansion of Copper, \alpha = 17\times 10^{- 6}\ K^{- 1}

Initial Temperature, T = 0^{\circ} = 273 K

Final Temperature, T' = 100^{\circ} = 273 + 100 = 373 K

Now,

Initial Volume of the block, V = 30\times 45\times 10\times 10^{- 6}\ m^{3} = 0.0135\ m^{3}

V' = V(1 + \gamma \Delta T)

\gamma = 3\alpha

V' = V(1 + 3\alpha \Delta T)

where

V' = Final volume

V' - V= 0.0135\times 17\times 10^{- 6} \times (T' - T))

\Delta V= 0.0135\times 3\times 17\times 10^{- 6} \times (373 - 273)) = 6.885\times 10^{- 5}\

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9806.65 Joules

Explanation:

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3 years ago
What are the largest and smallest resistances you can obtain by connecting a 36.0-Ω , a 50.0-Ω , and a 700-Ω resistor together?
myrzilka [38]
<h2>Answer:</h2>

786Ω and 20.32Ω respectively.

<h2>Explanation:</h2>

(a) Given a number of resistors each with its own resistance, the largest resistance can be obtained when these resistors are connected in series.

From the question, the resistors have the following resistances;

36.0-Ω, 50.0-Ω , and 700-Ω

Now, when they are connected in series, the total resistance (R) obtainable is given by the sum of these individual resistances as follows;

R = 36.0-Ω + 50.0-Ω + 700-Ω

R = 786Ω

Therefore, the largest resistance that can be obtained by connecting  a 36.0-Ω , a 50.0-Ω , and a 700-Ω resistor together is 786Ω

(b) Similarly, given a number of resistors each with its own resistance, the smallest resistance can be obtained when these resistors are connected in parallel.

From the question, the resistors have the following resistances;

36.0-Ω, 50.0-Ω , and 700-Ω

Now, when they are connected in parallel, the total resistance (R) obtainable is given by using the relation as follows;

\frac{1}{R} = \frac{1}{36.0} + \frac{1}{50.0} + \frac{1}{700.0}

\frac{1}{R} = \frac{35000+25200+1800}{1260000}

\frac{1}{R} = \frac{62000}{1260000}

\frac{1}{R} = \frac{62}{1260}

R = \frac{1260}{62}

R = 20.32Ω

Therefore, the smallest resistance that can be obtained by connecting  a 36.0-Ω , a 50.0-Ω , and a 700-Ω resistor together is 20.32Ω

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3 years ago
The radius of an atom is closest in size to a
ioda

the radius of the entire atom was 0.00000001 cm.

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nlexa [21]

Answer:

chris gets 960 and molly gets 1440

Explanation:

add the ratio up and divide

2+3=5

2400/5=480

480x2=960

480x3=1440

960+1440= 2400

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What is the charge on a hypothetical ion with 35 protons and 33 electrons?
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