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3 years ago

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

2 answers:

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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Which of the following statements is true?

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3 years ago

Which best describes a major characteristic of both volcanoes and earthquakes? they are centered at the poles?

Arturiano [62]

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3 years ago

A 460 W heating unit is designed to operate with an applied potential difference of 120 V (a) By what percentage will its heat o

dybincka [34]

Answer:

(a) = -0.16%

(b) = smaller

Explanation:

given

power = 460 W

potential difference = 120 V

(a) what percentage will its heat output drop if the applied potential difference drops to 110 V ?

we know $p = \frac{v^2}{R}$ .....................(i)

we need to find change in power

$\Delta P = \frac{\Delta (V^2)}{R}$

$\Delta P = \frac{2 V \Delta V}{R}$ ..............(ii)

from equations we get

$\frac{\Delta P}{P} = \frac{2 \Delta V}{V}$

$\frac{\Delta P}{P} = 2 \frac{110 -120}{120}$

$\frac{\Delta P}{P} = -2(\frac{10}{120})$

$\frac{\Delta P}{P} = - 0.16 %$

(b)

if we increase temperature resistance will increase and decrease with decrease in temperature and we know power is inversely proportional to resistance so if potential decrease and it would cause drop in power

and due to this increment of heating power resistance will decrease so actual drop in the power would be smaller

7 0

3 years ago

~~~~~NEED HELP ASAP~~~~~

Romashka-Z-Leto [24]

Answer:

Centripetal Acceleration 18.75 m/s^2, Rotational Kinetic Energy 843.75 J

Explanation:

a Linear acceleration (we cant find tangential acceleration with the givens so we will find centripetal)

a= ω^2*r

ω= 300rev/min

convert into rev/s

300/60= 5rev/s

a= 18.75m/s^2

b) use Krot= 1/2 Iω^2

plug in gives

1/2(30*2.25)(25)= 843.75 J

8 0

2 years ago