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3 years ago

What is the difference between static friction and kinetic friction

Physics

2 answers:

alekssr [168]3 years ago

6 0

Static friction keeps stationary objects calm
Kinetic friction slows down a moving thing

sergejj [24]3 years ago

3 0

The Force of Static Friction keeps a stationary object at rest! Once the Force ofStatic Friction is overcome, the Force of Kinetic Friction is what slows down a moving object.

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When doing numerical calculations involving temperature, you need to pay particular attention to the temperature scale you are u

Mrac [35]

1) $293 ^{\circ}C$

2) $859^{\circ}C$

Explanation:

The average kinetic energy of the molecules of an ideal gas is directly related to the Kelvin temperature of the gas, by the formula

$KE=\frac{3}{2}kT$

where

KE is the kinetic energy

k is the Boltzmann constant

T is the Kelvin temperature

We can say therefore that the average kinetic energy of the particles is directly proportional to the absolute temperature of the gas; so, we can write:

$KE\propto T$

And therefore

$\frac{KE_1}{KE_2}=\frac{T_1}{T_2}$ (1)

In this problem, we have:

$KE_1 = K_{10}$ is the initial kinetic energy of the molecules when the temperature of the gas is

$T_1=10^{\circ}+273=283 K$

Here we want to find the temperature $T_2$ at which the average kinetic energy of the particles is

$KE_2=2K_{10}$

So, twice the initial value. Substituting into eq.(1) and solving for T2, we find:

$T_2=\frac{T_1 KE_2}{KE_1}=\frac{(283)(2K_{10})}{K_{10}}=566 K$

Converting into Celsius degrees,

$T_2=566-273=293 ^{\circ}C$

The root-mean-square (rms) speed of the molecules in a gas is given by the equation

$v=\sqrt{\frac{3kT}{m}}$

where

k is the Boltzmann constant

T is the Kelvin temperature of the gas

m is the mass of each molecule

Therefore, from the equation we can say that the rms speed is proportional to the square root of the temperature:

$v\propto \sqrt{T}$

So we can write:

$\frac{v_1}{v_2}=\frac{\sqrt{T_1}}{\sqrt{T_2}}$ (2)

where in this problem:

$v_1 = v_{rms}$ is the rms speed of the molecules when the temperature is

$T_1=10^{\circ}C+273=283 K$

$v_2=2v_{rms}$ is the final rms speed of the molecules

Solving eq.(2), we find the temperature at which the rms speed is twice the initial value:

$T_2=T_1 (\frac{v_2}{v_1})^2=(283)(\frac{2v_{rms}}{v_{rms}})^2=1132 K$

Converting into Celsius degrees,

$T_2=1132-273=859^{\circ}C$

8 0

3 years ago

Area is found by multiplying the length of a surface times the width. If a floor measures 5.28 m2, how many square centimeters d

Scilla [17]

62,500 cm^2.........

8 0

3 years ago

Help me please???!!!!

Mariulka [41]

I got you gurl so if your looking for speed kts speed =distance +time and when you do all your steps correctly its speed=320 m/s....... Hope this helped

6 0

3 years ago

A crate of mass 10.0 kg is pulled up a rough incline with an initial speed of 1.50m/s . The pulling force is 100 N parallel to t

sergey [27]

167.67 Joules of work are done by the gravitational force on the crate.

Force is defined as the product of the mass of the object and acceleration. The SI unit of the force of Newton.

Gravitational force is the force applied by gravity on an object. The gravity of the earth is 9.8 m/s².

Force = Mass × Acceleration

F = m × a

Mass of the crate = 10 kg

The initial speed of the crate = 1.50 m/s

The pulling force = 100 N