All categories

3 years ago

Pressure is a function of force and area on which the force is exerted: P

Physics

1 answer:

sammy [17]3 years ago

7 0

Answer:

The change in volume decreases the area over which collisions of

the container and gas particles occur, so the pressure increases.

Explanation:

P = F/A = F ÷ V/h = Fh/V

You might be interested in

What information about elements can be collected from the periodic table?

Dafna1 [17]

Answer:

Using the data in the table scientists, students, and others that are familiar with the periodic table can extract information concerning individual elements. For instance, a scientist can use carbon's atomic mass to determine how many carbon atoms there are in a 1 kilogram block of carbon.

Explanation:

HOPE THIS HELPS LIKE AN RATE PLZ

3 0

3 years ago

Read 2 more answers

Which of the following statements is true for ideal gases, but is not always true for real gases?

xxMikexx [17]

Answer:

C. Replacing one gas by another under the same conditions, has no effect on pressure.

Explanation:

Ideal gas:

A gas is treated as an ideal gas if temperature is high and pressure is low.

Kinetic energy for ideal gas given as

$K.E.=\dfrac{3}{2}KT$

So when temperature of gas is increases then Average molecular kinetic energy will also increases.

The size of molecule is negligible as compare to the dimension of container. It mean that volume occupied by molecule is less as compare to the volume of container.

The between molecules is perfectly elastic.

Ideal gas equation

P V = m R T

So the option C is not always true.

4 0

3 years ago

A series of pulses, each of amplitude 0.1m , is sent down a string that is attached to a post at one end. The pulses are reflect

sp2606 [1]

The net displacement at a point on the string where the pulses cross is 0.2 m.

The term "displacement" refers to a shift in an object's position. It has a magnitude and a direction, making it a vector quantity. An arrow pointing from the starting point to the finishing point serves as its symbol.

A string that is connected to a post at one end is used to transmit a sequence of pulses, each measuring 0.1 meters in amplitude.

At the post, the pulses are reflected and return along the string without losing any of their amplitude.

Now, let's say the ends are free.

There is no inversion on reflection if the end is free. The amplitude at their intersection is 2A.

Now, since A = 0.1 m

Then, 2A = 2(0.1) = 0.2 m

As a result, the net displacement at the string's intersection of two pulses is 0.2 m.

The correct option is (c).

Learn more about amplitude here:

brainly.com/question/3613222

#SPJ4

4 0

2 years ago

How can a nuke (Nuclear bomb) be hotter than the sun?

Art [367]

Cause it can take so much heat then when the sun beams down on it then the process gets alot more complicated it gets a little hotter than it starts to get more and more hotter .

6 0

3 years ago

A ray of light is moving from a material having a high indexof refraction into a material with a lower index of refraction.

luda_lava [24]

(a) Away from the normal

We can find the direction of bending of the ray of light by using Snell's equation:

$n_1 sin \theta_1 = n_2 sin \theta_2$

where we have:

n1, n2: index of refraction of the first and second medium

$\theta_1, \theta_2$ ; angle that the incident and the refracted ray form with the normal to the surface

Here, the light ray moves from a material with high index of refraction to a material with lower index, so we have

$n_1 > n_2$

Re-arranging Snell's law we find

$sin \theta_2 = \frac{n_1}{n_2} sin \theta_1$

since we have

$\frac{n_1}{n_2}>1$

this implies

$sin \theta_2 > sin \theta_1\\\theta_2 > \theta_1$

so the ray of light bends away from the normal.

(b) The wavelength is greater in the second material (the one with lower index of refraction)

The wavelength of the light in a medium is given by

$\lambda=\frac{\lambda_0}{n}$

where

$\lambda_0$ is the wavelength of the light in a vacuum

n is the refractive index

The equation can be rewritten as

$\lambda_0 = \lambda_1 n_1 = \lambda_2 n_2$

and again it can be rewritten as

$\lambda_2 = \frac{n_1}{n_2} \lambda_1$

where

$\lambda_1 = 600 nm\\\frac{n_1}{n_2}>1$

Therefore, we have that the wavelength in the second medium (the one with lower index of refraction) is longer than the wavelength in the first medium.

(c) The frequency remains the same

Wavelength and speed of a light ray depend on the medium in which the wave is travelling through, however the frequency does not depend on that, so it remains the same in the two mediums.

8 0

3 years ago