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3 years ago

4. In a object filled with a type of gas, which of the following actions would decrease the pressure

Physics

1 answer:

ExtremeBDS [4]3 years ago

7 0

Answer:

Decrease the number of gas particles, increase the object's area, and reduce the temperature of the gas

Explanation:

TEMPERATURE:

Decreasing the temperature will slow down the molecules. Hence, less no. of collisions will take place between walls of object and molecules. This will result in decrease of pressure.

Therefore, the pressure of a gas can be decreased by increasing its temperature.

NUMBER OF GAS PARTICLES:

Decreasing the number of particles will result in less no. of collisions, hence decreasing the pressure.

Therefore, the pressure of a gas can be decreased by decreasing its no. of molecules or no. of particles.

AREA OF OBJECT:

The pressure is given by the formula:

$P = \frac{F}{A}\\\\For constant Force (F):\\P\ \alpha\ \frac{1}{A}$

where,

A = Area of Object

Therefore, the pressure of a gas can be decreased by increasing area of object.

So, the correct option is:

Decrease the number of gas particles, increase the object's area, and reduce the temperature of the gas

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A trick shot archer shoots an arrow with a velocity of 30.0 m/s at an angle of 20.0 degrees with respect to the horizontal. An a

svlad2 [7]

Answer:

$u'=10.259\ m.s^{-1}$ is the initial velocity of tossing the apple.

the apple should be tossed after $\Delta t=0.0173\ s$

Explanation:

Given:

velocity of arrow in projectile, $v=30\ m.s^{-1}$

angle of projectile from the horizontal, $\theta=20^{\circ}$

distance of the point of tossing up of an apple, $d=30\ m$

Now the horizontal component of velocity:

$v_x=v\ cos\ \theta$

$v_x=30\times cos\ 20^{\circ}$

$v_x=28.191\ m.s^{-1}$

The vertical component of the velocity:

$v_y=v.sin\ \theta$

$v_y=30\times sin\ 20^{\circ}$

$v_y=10.261\ m.s^{-1}$

Time taken by the projectile to travel the distance of 30 m:

$t=\frac{d}{v_x}$

$t=\frac{30}{28.191}$

$t=1.0642\ s$

Vertical position of the projectile at this time:

$h=v_y.t-\frac{1}{2}g.t^2$

$h=10.261\times 1.0642-\frac{1}{2} \times 9.8\times 1.0642^2$

$h=5.3701\ m$

Now this height should be the maximum height of the tossed apple where its velocity becomes zero.

$v'^2=u'^2-2g.h$

$0^2=u'^2-2\times 9.8\times 5.3701$

$u'=10.259\ m.s^{-1}$ is the initial velocity of tossing the apple.

Time taken to reach this height:

$v'=u'-g.t'$

$0=10.259-9.8\times t'$

$t'=1.0469\ s$

We observe that $t>t'$ hence the time after the launch of the projectile after which the apple should be tossed is:

$\Delta t=t-t'$

$\Delta t=1.0642-1.0469$

$\Delta t=0.0173\ s$

8 0

3 years ago

If a soundproof door is almost shut but still leaves a small opening, what wave phenomenon is responsible for someone being able

lana66690 [7]

The phenomenon which is responsible for this effect is called diffraction.

Diffraction is the ability of a wave to propagate when it meets an obstacle or a slit. When the wave encounters the obstacle or the slit, it 'bends' around it and it continues propagate beyond it. A classical example of this phenomenon is when a sound wave propagates through a wall where there is a small aperture (as in the example of this problem)

5 0

3 years ago

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A rope is vibrating so as to form the standing wave pattern shown. How many antinodes are present in the rope?a- 5b- 4c- 8d- 10e

denpristay [2]

c) 8

Explanation

When you shake a rope, the particles in the rope move up and down, and the wave moves forward or away from the source of energy. The rope moves in a direction that is perpendicular

Nodes are the places where the rope doesn't move at all; antinodes occur where the motion is greatest.

Step 1

let's check the graph:

a) Nodes

and Antinodes

so, in the ripe there are 8 antinodes

therefore, the answer is

c) 8

4 0

1 year ago

In photosynthesis, carbon dioxide is converted to oxygen and then released.

3241004551 [841]

What is the question yes it is converted to and then released

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3 years ago

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An object is 16.0cm to the left of a lens. The lens forms an image 36.0cm to the right of the lens.

CaHeK987 [17]

A) 11.1 cm

We can find the focal length of the lens by using the lens equation:

$\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$

where

f is the focal length

p = 16.0 cm is the distance of the object from the lens

q = 36.0 cm is the distance of the image from the lens (taken with positive sign since it is on the opposide side to the image, so it is a real image)

Solving the equation for f:

$\frac{1}{f}=\frac{1}{16.0 cm}+\frac{1}{36.0 cm}=0.09 cm^{-1}\\f=\frac{1}{0.09 cm^{-1}}=11.1 cm$