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hram777 [196]
2 years ago
5

If the temperature of a gas is increased by 10x, what will happen to the volume?

Physics
1 answer:
xxMikexx [17]2 years ago
8 0

The volume of a gas will increase by ten times if the temperature is increased by ten times.

<h3>Relationship between the volume of a gas and temperature</h3>

The relationship between the volume of a gas and its temperature is explained in Charles' law of gases which states that:

  • The volume of a fixed mass of gas is directly proportional to its temperature provided the pressure of the gas is kept constant.

This means that if the temperature of a gas is increased by any given factor, the volume increases by the same factor proportionally.

Therefore, if the volume of a gas will increase by ten times if the temperature is increased by ten times.

Learn more about gas volume and temperature at: brainly.com/question/18706379

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2 years ago
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What is 3600Hz has in rpm?​
Setler [38]

Answer:

Explanation:

N=Rotor Speed in Revolution per minute(rpm)

for P=4 and N=3600, f comes out to be 120 Hz.

So frequency of voltage produced is 120 Hz. But this is not practical. Generally 4-Pole generator has N=1500rpm(for 50 Hz) or 1800rpm for 60 Hz. Two pole generator can have N=3600rpm(f=60Hz).

The most practical situation is generator having N=3600Hz with 2 Poles.

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8 0
3 years ago
Example 3 :
kherson [118]

The friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 × 10^8 respectively. Also, the friction factor and head loss when velocity is 3m/s is 0.096 and 5.3 × 10^8 respectively.

<h3>How to determine the friction factor</h3>

Using the formula

μ = viscosity = 0. 06 Pas

d =  diameter = 120mm = 0. 12m

V =  velocity = 1m/s and 3m/s

ρ = density = 0.9

a. Velocity = 1m/s

friction factor = 0. 52 × \frac{0. 06}{0. 12* 1* 0. 9}

friction factor = 0. 52 × \frac{0. 06}{0. 108}

friction factor = 0. 52 × 0. 55

friction factor = 0. 289

b. When V = 3mls

Friction factor = 0. 52 × \frac{0. 06}{0. 12 * 3* 0. 9}

Friction factor = 0. 52 × \frac{0. 06}{0. 324}

Friction factor = 0. 52 × 0. 185

Friction factor = 0.096

Loss When V = 1m/s

Head loss/ length = friction factor × 1/ 2g × velocity^2/ diameter

Head loss = 0. 289 × \frac{1}{2*6. 6743 * 10^-11} × \frac{1^2}{0. 120} × \frac{1}{100}

Head loss =  1. 80 × 10^8

Head loss When V = 3m/s

Head loss = 0. 096 × \frac{1}{1. 334 *10^-10} × \frac{3^2}{0. 120} × \frac{1}{100}

Head loss = 5. 3× 10^8

Thus, the friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 ×10^8 respectively also, the friction factor and head loss  when velocity is 3m/s is 0.096 and 5.3 ×10^8 respectively.

Learn more about friction here:

brainly.com/question/24338873

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4 0
1 year ago
Give reasons for the following:
Juliette [100K]

Answer:

a. lower surface area, less resistence

b. more surface area, the load is split so no single tire overstrained

c. more surface area, more resistance against the sand. human steps sink down in the sand.

d. rapid change in air pressure on eardrums lead to somewhat-painful tension

e. air would always find its way in so no pressure difference can be achieved

(would indeed appreciate the brainliest if you appreciate the work)

4 0
3 years ago
What is the molar mass of an ideal gas if a 0.800 g sample of this gas occupies a volume of 200. mL at 50.0 oC and 720. mm Hg?
Paladinen [302]
PV=nRT
(720/760)(0.200)=(0.800/x)(0.08206)(323.15)
(0.1894736842)=(0.800/x)(0.08206)(323.15)
.0071451809=(0.800/x)
x=MM=111.9635758 g/mol
7 0
3 years ago
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