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strojnjashka [21]

3 years ago

8

A wire with a resistance R is lengthened to 3.07 times its original length by pulling it through a small hole. Find the new resi

stance of the wire after it was stretched.

1 answer:

charle [14.2K]3 years ago

4 0

Explanation:

Below is an attachment containing the solution.

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Which law states that the pressure and absolute temperature of a fixed quantity of gas are directly proportional under constant

Brrunno [24]

<u>Gay Lussac’s law</u> state that the pressure and absolute temperature of a fixed quantity of a gas are directly proportional under constant volume conditions.

<h2>Further Explanation </h2><h3>Gay-Lussac’s law </h3>

It states that at constant volume, the pressure of an ideal gas I directly proportional to its absolute temperature.
Thus, an increase in pressure of an ideal gas at constant volume will result to an increase in the absolute temperature.

<h3>Boyles’s law </h3>

This gas law states that the volume of a fixed mass of a gas is inversely proportional to its pressure at constant absolute temperature.
Therefore, when the volume of an ideal gas is increased at constant temperature then the pressure of the gas will also increase.

<h3>Charles’s law </h3>

It states that the volume of a fixed mass of a gas is directly proportional to absolute temperature at constant pressure.
Therefore, an increase in volume of an ideal gas causes a corresponding increase in its absolute temperature and vice versa while the pressure is held constant.

<h3>Dalton’s law </h3>

It is also known as the Dalton’s law of partial pressure. It states that the total pressure of a mixture of gases is always equivalent to the total sum of the partial pressures of individual component gases.
Partial pressure refers to the pressure of an individual gas if it occupies the same volume as the mixture of gases.

Keywords: Gas law, Gay-Lussac’s law, pressure, volume, absolute temperature, ideal gas

<h3>Learn more about: </h3>

Gay-Lussac’s law: brainly.com/question/2644981
Charles’s law: brainly.com/question/5016068
Boyles’s law: brainly.com/question/5016068
Dalton’s law: brainly.com/question/6491675

Level: High school

Subject: Chemistry

Topic: Gas laws

Sub-topic: Gay-Lussac’s law

8 0

3 years ago

Read 2 more answers

Show that rigid body rotation near the Galactic center is consistent with a spherically symmetric mass distribution of constant

irakobra [83]

To solve this problem we will use the concepts related to gravitational acceleration and centripetal acceleration. The equality between these two forces that maintains the balance will allow to determine how the rigid body is consistent with a spherically symmetric mass distribution of constant density. Let's start with the gravitational acceleration of the Star, which is

$a_g = \frac{GM}{R^2}$

Here

$M = \text{Mass inside the Orbit of the star}$

$R = \text{Orbital radius}$

$G = \text{Universal Gravitational Constant}$

Mass inside the orbit in terms of Volume and Density is

$M =V \rho$

Where,

V = Volume

$\rho =$ Density

Now considering the volume of the star as a Sphere we have

$V = \frac{4}{3} \pi R^3$

Replacing at the previous equation we have,

$M = (\frac{4}{3}\pi R^3)\rho$

Now replacing the mass at the gravitational acceleration formula we have that

$a_g = \frac{G}{R^2}(\frac{4}{3}\pi R^3)\rho$

$a_g = \frac{4}{3} G\pi R\rho$

For a rotating star, the centripetal acceleration is caused by this gravitational acceleration. So centripetal acceleration of the star is

$a_c = \frac{4}{3} G\pi R\rho$

At the same time the general expression for the centripetal acceleration is

$a_c = \frac{\Theta^2}{R}$

Where $\Theta$ is the orbital velocity

Using this expression in the left hand side of the equation we have that

$\frac{\Theta^2}{R} = \frac{4}{3}G\pi \rho R^2$

$\Theta = (\frac{4}{3}G\pi \rho R^2)^{1/2}$

$\Theta = (\frac{4}{3}G\pi \rho)^{1/2}R$

Considering the constant values we have that

$\Theta = \text{Constant} \times R$

$\Theta \propto R$

As the orbital velocity is proportional to the orbital radius, it shows the rigid body rotation of stars near the galactic center.

So the rigid-body rotation near the galactic center is consistent with a spherically symmetric mass distribution of constant density

6 0

3 years ago

The larger the push, the larger the change in velocity. This is an example of Newton's Second Law of Motion which states that th

Paul [167]

According to Newton, an object will only accelerate if there is a net or unbalanced forceacting upon it. The presence of an unbalanced force will accelerate an object - changing its speed, its direction, or both its speed and direction.

6 0

3 years ago

Read 2 more answers

Which type of physical activity is being performed in the picture?

mafiozo [28]

B strength training I think that’s the answer

3 0

3 years ago

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Consider a 20 cm thick granite wall with a thermal conductivity of 2.79 W/m·K. The temperature of the left surface is held const

kozerog [31]

Answer:

The right wall surface temperature and heat flux through the wall is 35.5°C and 202.3W/m²

Explanation:

Thickness of the wall is L= 20cm = 0.2m

Thermal conductivity of the wall is K = 2.79 W/m·K

Temperature at the left side surface is T₁ = 50°C

Temperature of the air is T = 22°C

Convection heat transfer coefficient is h = 15 W/m2·K

Heat conduction process through wall is equal to the heat convection process so

$Q_{conduction} = Q_{convection}$

Expression for the heat conduction process is

$Q_{conduction} = \frac{K(T_1 - T)}{L}$

Expression for the heat convection process is

$Q_{convection} = h(T_2 - T)$

Substitute the expressions of conduction and convection in equation above

$Q_{conduction} = Q_{convection}$

$\frac{K(T_1 - T_2)}{L} = h(T_2 - T)$

Substitute the values in above equation

$\frac{2.79(50- T_2)}{0.2} = 15(T_2 - 22)\\\\T_2 = 35.5^\circC$

Now heat flux through the wall can be calculated as

$q_{flux} = Q_{conduction} \\\\q_{flux} = \frac{K(T_1 - T_2)}{L}\\\\q_{flux} = \frac{2.79(50 - 35.5)}{0.2}\\\\q_{flux} = 202.3W/m^2$

Thus, the right wall surface temperature and heat flux through the wall is 35.5°C and 202.3W/m²

6 0

3 years ago