An ideal gas in a cubical box having sides of length L exerts a pressure p on the walls of the box. If all of this gas is put into a box having sides of length 0.5L without changing its temperature, the pressure it exerts on the walls of the larger box will be...

p.

2p.

4p.

8p.

12p.

Answers

Answer 1

Answer:

2P

Explanation:

See attached file

An Ideal Gas In A Cubical Box Having Sides Of Length L Exerts A Pressure P On The Walls Of The Box. If

Related Questions

HELP ILL MARK BRAINLIEST PLS!!!!

A patch of mud has stuck to the surface of a bicycle tire as shown. The stickiness of
the mud is the centripetal or tension force that keeps the mud on the tire as it spins.
Has work been done on the mud as the tire makes one revolution, if the mud stays
on the tire? Explain.

Answers

Answer:

Yes, work has been done on the mud.

Explanation:

Work is done on a body, when a force is applied on the body to move it through a certain distance. In the case of the mud, the tire exerts a centripetal force on the mud. The centripetal force moves the mud along a path that follows the circle formed by the tire in one revolution of the tire. The total distance traveled is the circumference of the circle formed. The work done on the mud is therefore the product of the centripetal force on the mud from the tire, and the circumference of the circle formed by the tire, usually expressed in radian.

A load of 223,000 N is placed on an aluminum column 10.2 cm in diameter. If the column was originally 1.22 m high find the amount that the column has shrunk.

Answers

Answer:

0.4757 mm

Explanation:

Given that:

Load P = 223,000 N

the length of the height of the aluminium column = 1.22 m

the diameter of the aluminum column = 10.2 cm = 0.102 m

The amount that the column has shrunk ΔL can be determined by using the formula:

[tex]\Delta L = \dfrac{PL}{AE_{Al}}[/tex]

where;

A = πr²

2r = D

r = D/2

r = 0.102/2

r = 0.051

A = π(0.051)²

A = 0.00817

Also; the young modulus of aluminium [tex]E_{Al}[/tex] is:

[tex]E_{Al}= 7*10^{10} \Nm^{-2}[/tex]

[tex]\Delta L = \dfrac{PL}{AE_{Al}}[/tex]

[tex]\Delta L = \dfrac{223000* 1.22}{0.00817* 7*10^{10}}[/tex]

ΔL = 4.757 × 10⁻⁴ m

ΔL =  0.4757 mm

Hence; the amount that the column has shrunk is 0.4757 mm

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