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Vlad1618 [11]
3 years ago
13

Assume the work done compressing the He gas is -63 kJ and the internal energy change of the gas is 79 kJ. What is the heat loss

or heat gain of this process?
Engineering
1 answer:
klemol [59]3 years ago
3 0

Answer:

Heat gain of 142 kJ

Explanation:

We can see that job done by compressing the He gas is negative, it means that the sign convention we are going to use is negative for all the work done by the gas and positive for all the job done to the gas. With that being said, the first law of thermodynamics equation will help us to solve this problem.

ΔU = Q + W ⇒ Q = ΔU -W

Q = 79 - (-63) = 142 kJ

Therefore, the gas gained heat by an amount of 142 kJ.

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7 0
3 years ago
2 samples of water of equal volume are put into dishes and kept at room temp for several days. the water in the first dish is co
Over [174]

Answer:

Vaporization is the process by which a substance changes from its solid or liquid state to a gaseous state.

Since both liquids are of the same volume and are placed under the same temperature condition, for them to not to vaporize at the same time, they must have been in different containers.

For vaporization to take place, the volume of liquid, amount of air exposure and area of the surface must be considered.

Maybe the first liquid was in a dish which has a large opening, thereby exposing a large amount which can make water to evaporate faster, whereas the second liquid was somehow enclosed (in a deeper dish).

5 0
3 years ago
What pounds per square inch is required by a bubbler system to produce bubbles at a depth of 4 ft 7 in water?
Brut [27]
2.31 ft/psig

1.98 psig would equalize feet of head 4’7”

4.583/2.31 = 1.98. .583 is 7”/12”


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4 0
3 years ago
Which dimension of wellness focuses on finding enjoyment and satisfaction in daily endeavours?
UNO [17]

Answer:

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Explanation:

5 0
3 years ago
An object is supported by a crane through a steel cable of 0.02m diameter. If the natural swinging of the equivalent pendulum is
devlian [24]

Answer:

22.90 × 10⁸ kg

Explanation:

Given:

Diameter, d = 0.02 m

ωₙ = 0.95 rad/sec

Time period, T = 0.35 sec

Now, we know

T= 2\pi\sqrt{\frac{L}{g}}

where, L is the length of the steel cable

g is the acceleration due to gravity

0.35= 2\pi\sqrt{\frac{L}{9.81}}

or

L = 0.0304 m

Now,

The stiffness, K is given as:

K = \frac{\textup{AE}}{\textup{L}}

Where, A is the area

E is the elastic modulus of the steel = 2 × 10¹¹ N/m²

or

K = \frac{\frac{\pi}{4}d^2\times2\times10^11}{0.0304}

or

K = 20.66 × 10⁸ N

Also,

Natural frequency, ωₙ = \sqrt{\frac{K}{m}}

or

mass, m = \sqrt{\frac{K}{\omega_n^2}}

or

mass, m = \sqrt{\frac{20.66\times10^8}{0.95^2}}

mass, m = 22.90 × 10⁸ kg

4 0
3 years ago
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