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Strike441 [17]
2 years ago
10

1. Calculate the resistivity of the given material whose resistance is 2Ω, length and area of cross section are 15 cm and 25cm²

respectedly
Chemistry
1 answer:
Katena32 [7]2 years ago
6 0

Answer:

0.033 Ω . m

Explanation:

This is a typical problem that can be solved with the resistivity formula.

ρ = R . A / l

where R is the resistance, A is the area and l, length of the object.

We have all the data in the statement but firstly, we need to convert the area and length in m² and m, respectively because units of ρ are Ω . m

15 cm . 1 m/100 cm = 0.15 m

25 cm² .  1m² / 10000 cm² = 0.0025 m²

We replace → R = 2Ω . (0.0025 m² / 0.15) ⇒ 0.033 Ω . m

A conductive object will be more or less resistant according to the resistivity, which increases with length and decreases with area. Furthermore, resistivity is also conditioned to temperature, where as T ° increases, resistivity increases too.

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A man heats a balloon in the oven. If the balloon initially has a volume of 0.4 liters and a temperature of 20 degrees celsius,
Airida [17]

Answer:

0.714 liter.

Explanation:

Given:

The balloon initially has a volume of 0.4 liters and a temperature of 20 degrees Celsius.

It is heated to a temperature of 250 degrees Celsius.

Question asked:

What will be the volume of the balloon after he heats it to a temperature of 250 degrees Celsius ?

Solution:

By using:

PV=nRT

Assuming pressure as constant,

V∝ T

Now, let  K is the constant.

V = KT

Let initial volume of balloon , V_{1} = 0.4 liter

1000 liter = 1 meter cube

1 liter = \frac{1}{1000} m^{3} = 10^{-3} m^{3

0.4 liter = 0.4\times10^{-3}=4\times10^{-4} m^{3}

And initial temperature of balloon, T_{1} = 20°C = (273 + 20)K

                                                                          = 293 K

Let the final volume of balloon is V_{2}

And a given, final temperature of balloon, T_{2} is 250°C = (273 + 250)K

                                                                                          = 523 K

Now, V_{1} = KT_{1}

          4\times10^{-4}=K\times293\ (equation\ 1 )

V_{2} = KT_{2}

    =K\times523\ (equation 2)

Dividing equation 1 and 2,

 \frac{4\times10^{-4}}{V_{2} } =\frac{K\times293}{K\times523}

K cancelled by K.

By cross multiplication:

293V_{2} =4\times10^{-4} \times523\\V_{2} =\frac{ 4\times10^{-4} \times523\\}{293} \\          = \frac{2092\times10^{-4}}{293} \\          =7.14\times10^{-4}m^{3}

Now convert it into liter with the help of calculation done above.

7.14\times10^{-4} \times1000\\7.14\times10^{-4} \times10^{3} \\0.714\ liter

Therefore, the volume of the balloon be after he heats it to a temperature of 250 degrees Celsius is 0.714 liter.

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When a heavy atom is split in a fission reaction, the daughter nuclei that form have higher binding energies than the parent nuc
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