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Lelu [443]
3 years ago
10

What is the most likely reason it is dangerous to break open a compressed gas tank?

Physics
2 answers:
erastova [34]3 years ago
8 0
The main reason why it's very dangerous to open compress gas tank is it can cause major disaster not only in human but also in plants and animals. Compress gas is flammable. And it's like a bomb the damage of gas tank is very catastrophic. It is very harmful if mishandled.
svp [43]3 years ago
7 0

Answer:

Extreme pressure of the gas exerted on the walls of tank or cylinder is most likely the reason

Explanation:

Normally, compressed gas is filled in a container at a high pressure, because there is no other method you can fill a gas in this container. You need a greater quantity of gas filled in container and then transported to other places for various purposes. Now usually a valve is placed at the opening of tank or cylinder to control the flow of gas when using it. If you break open the container, the gas molecules will break out at a rapid velocity, since, according to Newton's third law of motion, which is,

Every action has an equal but opposite reaction

the force exerted by gas to flow out into atmosphere will equally act on container as well but in opposite direction. It can cause container to move haphazardly in very fast speed and container can collide with you or any other person standing nearby.

Secondly, it depend upon the nature of gas that it is might be poisonous or inflammable  and when breaks out in environment, it can cause damage to the people who inhale it or it might cause fire in the surroundings. That is why, it is dangerous to break open a container of compressed gas.

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Answer:

The magnitude of the induced e.m.f is 571 mV

Explanation:

Given;

magnetic field strength, B = 0.4T

area of the loop, A = 1m²

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Since the angle of inclination of the magnetic field, then the direction of the induced e.m.f, θ = 0

According to faraday's law; the induced emf is directly proportional to the rate of change of flux.

Thus, magnitude of the induced e.m.f is given as

E.M.F = \frac{\delta \phi}{\delta t} \\\\\phi = BAcos \theta =  BAcos0 = BA\\\\E.M.F = \frac{0.4 X1}{0.7} = 0.571 V

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2 years ago
9.00 V is applied to a wire with a resistance of 52.0 ohm. At what distance from the wire is the magnetic field 2.22x 10^-8 T?
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