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3 years ago

A gas is stored in a sealed container of constant volume. The temperature of the gas increases.

Physics

1 answer:

Alexus [3.1K]3 years ago

6 0

Answer:

If the gas volume is decreased, the container wall area decreases and the molecule-wall collision frequency increases, both of which increase the pressure exerted by the gas. Avogadro's law. At constant pressure and temperature, the frequency and force of molecule-wall collisions are constant.

Explanation:

As the temperature increases, the average kinetic energy increases as does the velocity of the gas particles hitting the walls of the container. The force exerted by the particles per unit of area on the container is the pressure, so as the temperature increases the pressure must also increase.

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A block whose weight is 45 N rests on a horizontal table. A horizontal force of 36 N is applied to the block. The coefficient of

riadik2000 [5.3K]

Answer:

1.5 m/s²

Explanation:

For the block to move, it must first overcome the static friction.

Fs = N μs

Fs = (45 N) (0.42)

Fs = 18.9 N

This is less than the 36 N applied, so the block will move. Since the block is moving, kinetic friction takes over. To find the block's acceleration, use Newton's second law:

∑F = ma

F − N μk = ma

36 N − (45 N) (0.65) = (45 N / 9.8 m/s²) a

6.75 N = 4.59 kg a

a = 1.47 m/s²

Rounded to two significant figures, the block's acceleration is 1.5 m/s².

Usually the coefficient of static friction is greater than the coefficient of kinetic friction. You might want to double check the problem statement, just to be sure.

6 0

3 years ago

A 124-kg balloon carrying a 22-kg basket is descending with a constant downward velocity of 20.0 m/ s. A I.O-kg stone is thrown

nadya68 [22]

(a) 296.6 m

The motion of the stone is the motion of a projectile, thrown with a horizontal speed of

$v_x = 15.0 m/s$

and with an initial vertical velocity of

$v_{y0} = -20.0 m/s$

where we have put a negative sign to indicate that the direction is downward.

The vertical position of the stone at time t is given by

$y(t) = h + v_{0y} t + \frac{1}{2}gt^2$ (1)

where

h is the initial height

g = -9.81 m/s^2 is the acceleration due to gravity

The stone hits the ground after a time t = 6.00 s, so at this time the vertical position is zero:

y(6.00 s) = 0

Substituting into eq.(1), we can solve to find the initial height of the stone, h:

$0 = h + v_{0y} y + \frac{1}{2}gt^2\\h = -v_{0y} y - \frac{1}{2}gt^2=-(-20.0 m/s)(6.00 s) - \frac{1}{2}(9.81 m/s^2)(6.00 s)^2=296.6 m$

(b) 176.6 m

The balloon is moving downward with a constant vertical speed of

$v_y = -20 m/s$

So the vertical position of the balloon after a time t is

$y(t) = h + v_y t$

and substituting t = 6.0 s and h = 296.6 m, we find the height of the balloon when the rock hits the ground:

$y(t) = 296.6 m + (-20.0 m)(6.00 s)=176.6 m$

(c) 198.2 m

In order to find how far is the rock from the balloon when it hits the ground, we need to find the horizontal distance covered by the rock during the time of the fall.

The horizontal speed of the rock is

$v_x = 15.0 m/s$

So the horizontal distance travelled in t = 6.00 s is

$d_x = v_x t = (15.0 m/s)(6.00 s)=90 m$

Considering also that the vertical height of the balloon after t=6.00 s is

$d_y = 176.6 m$

The distance between the balloon and the rock can be found by using Pythagorean theorem:

$d=\sqrt{(90 m)^2+(176.6 m)^2}=198.2 m$

(di) 15.0 m/s, -58.8 m/s

For an observer at rest in the basket, the rock is moving horizontally with a velocity of

$v_x = 15.0 m/s$

Instead, the vertical velocity of the rock for an observer at rest in the basket is

$v_y (t) = gt$

Substituting time t=6.00 s, we find

$v_y = (-9.8 m/s)(6.00 s)=-58.8 m/s$

(dii) 15.0 m/s, -78.8 m/s

For an observer at rest on the ground, the rock is still moving horizontally with a velocity of

$v_x = 15.0 m/s$

Instead, the vertical velocity of the rock for an observer on the ground is now given by

$v_y (t) = v_{0y} + gt$

Substituting time t=6.00 s, we find

$v_y = (-20.0 m/s)+(-9.8 m/s)(6.00 s)=-78.8 m/s$

6 0

2 years ago

At a particular instant, a hot air balloon is 100 m in the air and descending at a constant speed of 2.0 m/s. at this exact inst

rewona [7]

Answer:

86.4 m horizontal from landing spot

Explanation:

Find out how long before the ball hits the ground

vertical speed of ball = -2 m/s gravity = - 9.81 m/s^2

find time to hit ground from 100 m

( height will be<u> zero</u> when it hits the ground)

<u>0 </u>= 100 - 2 t - 1/2 ( 9.81) t^2

use Quadratic Formula to find t = 4.32 seconds

horizontal speed of ball = 20 m/s

in 4.32 seconds it will travel horizontally 20 m/s * 4.32 s = 86.4 m

3 0

1 year ago

What relationship between the sun and earth did copernicus formulate?

Dmitrij [34]

Answer:

The astronomical model created and published by Nicholas Copernicus in the year 1543 is called Copernican heliocentrism. The model set the Sun in immobile position near the center of the solar system with Earth, as well as the other planets, spherical, epicycled and at consistent frequencies around it.

5 0

3 years ago

To a stationary observer, a bus moves north with a speed of 10 m/s. A man inside walks toward the back of the bus with a speed o

Rudiy27

9.6m/s - apex .........................

3 0

3 years ago

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