A burning candle provides :

The cylinder valve is open and the gas is collected and atmospheric pressure

jok3333 [9.3K]

Less gas will be collected because some of the gases will escape from the open cylinder valve.

Cylinders used to store carbon dioxide will have thicker walls than those of butane because of higher pressures.

<h3>What are compressed gases?</h3>

Compressed gases are gases which are compressed under high pressure in gas cylinders.

Cylinder valves are used to reduce the pressure of the compressed gases and in the process, some of the gas molecules escape.

Since the cylinder valve is open and the gas is collected at atmospheric pressure, less gas will be collected because some of the gases will escape.

Since, the carbon dioxide not liquefy under pressure compared to butane, the cylinders used to store carbon dioxide will have thicker walls than those of butane.

Learn more about compressed gases at: brainly.com/question/518065

4 0

2 years ago

A girl (mass M) standing on the edge of a frictionless merry-go-round (radius R, rotational inertia I) that is not moving. She t

vladimir1956 [14]

a) $\omega=\frac{-mvR}{I+MR^2}$

b) $v=\frac{-mvR^2}{I+MR^2}$

Explanation:

a)

Since there are no external torques acting on the system, the total angular momentum must remain constant.

At the beginning, the merry-go-round and the girl are at rest, so the initial angular momentum is zero:

$L_1=0$

Later, after the girl throws the rock, the angular momentum will be:

$L_2=(I_M+I_g)\omega +L_r$

where:

$I$ is the moment of inertia of the merry-go-round

$I_g=MR^2$ is the moment of inertia of the girl, where

M is the mass of the girl

R is the distance of the girl from the axis of rotation

$\omega$ is the angular speed of the merry-go-round and the girl

$L_r=mvR$ is the angular momentum of the rock, where

m is the mass of the rock

v is its velocity

Since the total angular momentum is conserved,

$L_1=L_2$

So we find:

$0=(I+I_g)\omega +mvR\\\omega=\frac{-mvR}{I+MR^2}$

And the negative sign indicates that the disk rotates in the direction opposite to the motion of the rock.

b)

The linear speed of a body in rotational motion is given by

$v=\omega r$

where

$\omega$ is the angular speed

r is the distance of the body from the axis of rotation

In this problem, for the girl, we have:

$\omega=\frac{-mvR}{I+MR^2}$ is the angular speed

$r=R$ is the distance of the girl from the axis of rotation

Therefore, her linear speed is:

$v=\omega R=\frac{-mvR^2}{I+MR^2}$

5 0

3 years ago

Um elétron é lançado entre duas placas eletrizadas como mostra a figura. Sejam v= 6x10^6 m/s, ângulo 45°, E= 2x10^3 N/C, d= 3 cm

Svetlanka [38]

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3 0

3 years ago

Express the kinetic energy K in terms of the potential energy U. K=GMm/2R

max2010maxim [7]

Answer:

K = -½U

Explanation:

From Newton's law of gravitation, the formula for gravitational potential energy is;

U = -GMm/R

Where,

G is gravitational constant

M and m are the two masses exerting the forces

R is the distance between the two objects

Now, in the question, we are given that kinetic energy is;

K = GMm/2R

Re-rranging, we have;

K = ½(GMm/R)

Comparing the equation of kinetic energy to that of potential energy, we can derive that gravitational kinetic energy can be expressed in terms of potential energy as;

K = -½U

7 0

3 years ago

Una prenda de 320gramos de ropa gira en el interior de una lavadora si dicha lavadora tiene 40 cm y gira con una frecuencia de 4

Nitella [24]

Answer:

Período del tambor: $T = 0.25\,s$ , fuerza sobre la prenda: $F \approx 80.852\,N$ , velocidad lineal del tambor: $v \approx 10.053\,\frac{m}{s}$ , velocidad angular del tambor: $\omega \approx 25.133\,\frac{rad}{s}$ .

Explanation:

La expresión tiene un error por omisión, su forma correcta queda descrita a continuación:

"Una prenda de 320 gramos de ropa gira en el interior de una lavadora si dicha lavadora tiene un radio de 40 centímetros y gira con una frecuencia de 4 hertz. Halle a) el período, b) la velocidad angular, c) la fuerza con la que gira la prenda y d) la velocidad lineal de la lavadora."

El tambor gira a velocidad angular constante ( $\omega$ ), en radianes por segundo, lo cual significa que la prenda experimenta una aceleración centrífuga ( $a$ ), en metros por segundo al cuadrado. En primer lugar, calculamos el período de rotación del tambor ( $T$ ), en segundos: