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3 years ago

What is the connection between temperature and kinetic energy

Physics

1 answer:

joja [24]3 years ago

8 0

Answer:

The relationship between temperature and kinetic energy is directly proportional.

Explanation:

The temperature is directly proportional to the kinetic energy of the gas molecules. We know that kinetic energy is the energy due to the motion and when gas molecules absorbs heat, they move with more speed and the kinetic energy of particles increases as a result.

According to the kinetic theory of gases

K = 3/2RT/Na

Where K is the average kinetic energy of gas molecules, T is the temperature of the gas, Na is Avogadro's number and R is the universal gas constant.

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When you stretch a spring 13 cm past its natural length, it exerts a force of 21

zloy xaker [14]

Answer:

A. 1.6 N/cm

Explanation:

spring constant = 21/13 = 1.6 N/cm

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3 years ago

An electric hoist does 500 joules of work lifting a crate 2 meters. How

LenKa [72]

Hope this answer helps

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3 years ago

A hollow cylinder with an inner radius of 5 mm and an outer radius of 26 mm conducts a 4-A current flowing parallel to the axis

bearhunter [10]

Answer:

B = 38.2μT

Explanation:

By the Ampere's law you have that the magnetic field generated by a current, in a wire, is given by:

$B=\frac{\mu_o I_r}{2\pi r}$ (1)

μo: magnetic permeability of vacuum = 4π*10^-7 T/A

r: distance from the center of the cylinder, in which B is calculated

Ir: current for the distance r

In this case, you first calculate the current Ir, by using the following relation:

$I_r=JA_r$

J: current density

Ar: cross sectional area for r in the hollow cylinder

Ar is given by $A_r=\pi(r^2-R_1^2)$

The current density is given by the total area and the total current:

$J=\frac{I_T}{A_T}=\frac{I_T}{\pi(R_2^2-R_1^2)}$

R2: outer radius = 26mm = 26*10^-3 m

R1: inner radius = 5 mm = 5*10^-3 m

IT: total current = 4 A

Then, the current in the wire for a distance r is:

$I_r=JA_r=\frac{I_T}{\pi(R_2^2-R_1^2)}\pi(r^2-R_1^2)\\\\I_r=I_T\frac{r^2-R_1^2}{R_2^2-R_1^2}$ (2)

You replace the last result of equation (2) into the equation (1):

$B=\frac{\mu_oI_T}{2\pi r}(\frac{r^2-R_1^2}{R_2^2-R_1^2})$

Finally. you replace the values of all parameters:

$B=\frac{(4\pi*10^{-7}T/A)(4A)}{2\PI (12*10^{-3}m)}(\frac{(12*10^{-3})^2-(5*10^{-3}m)^2}{(26*10^{-3}m)^2-(5*10^{-3}m)^2})\\\\B=3.82*10^{-5}T=38.2\mu T$

hence, the magnitude of the magnetic field at a point 12 mm from the center of the hollow cylinder, is 38.2μT

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3 years ago

An element becomes a positively charged ion when it

Anuta_ua [19.1K]

An element becomes a + ion when it...loses electrons

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3 years ago

Read 2 more answers

alexgriva [62]

Answer:

Get your answer for $1 from www.gotit-pro.com

Explanation:

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3 years ago