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Leto [7]
3 years ago
15

How does approaching a temperature of absolute zero affect kinetic energy.?

Physics
1 answer:
Alexus [3.1K]3 years ago
7 0

First of all the kinetic energy is when the particles move in continuous random motion.

If the temperature is high the colliding particles will collide more. and if the temperature is low the colliding particles will collide less.

Low temperature result in low kinetic energy 
High temperature result in high kinetic energy

Absolute zero is the point where where all molecules have no kinetic energy. It is a theoretical value (it has never been reached).

The Kelvin temperature scale is based on absolute zero being the lowest possible temperature that could theoretically be reached. That is why there is no such thing as a negative Kelvin temperature value.

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A boy flies a kite with the string at a 30 degree angle to the horizontal. The tension in the string is 4.5N .
sp2606 [1]
How much work in J does the string do on the boy if the boy stands still? 

<span>answer: None. The equation for work is W = force x distance. Since the boy isn't moving, the distance is zero. Anything times zero is zero </span>
<span>--------------------------------------... </span>
<span>How much work does the string do on the boy if the boy walks a horizontal distance of 11m away from the kite? </span>

<span>answer: might be a trick question since his direction away from the kite and his velocity weren't noted. Perhaps he just set the string down and walked away 11m from the kite. If he did this, it is the same as the first one...no work was done by the sting on the boy. </span>

<span>If he did walk backwards with no velocity indicated, and held the string and it stayed at 30 deg the answer would be: </span>
<span>4.5N + (boys negative acceleration * mass) = total force1 </span>
<span>work = total force1 x 11 meters </span>
<span>--------------------------------------... </span>

<span>How much work does the string do on the boy if the boy walks a horizontal distance of 11m toward the kite? </span>

<span>answer: same as above only reversed: </span>
<span>4.5N - (boys negative acceleration * mass) = total force2 </span>
<span>work = total force2 x 11 meters</span>
6 0
3 years ago
An office window has dimensions 3.1 m by 2.1 m. As a result of the passage of a storm, the outside air pressure drops to 0.954 a
Virty [35]

Answer:

Net forces which pushes the window is 30342.78 N.

Explanation:

Given:

Dimension of the office window.

Length of the window = 3.1 m

Width of the window = 2.1 m

Area of the window = (3.1\times 2.1) = 6.51\ m^2

Difference in air pressure = Inside pressure - Outside pressure

                                           = (1.0-0.954) atm = 0.046 atm

Conversion of the pressure in its SI unit.

⇒  1 atm = 101325 Pa

⇒ 0.046 atm = 0.046\times 101325 =4660.95 Pa

We have to find the net force.

We know,

⇒ Pressure = Force/Area

⇒ Pressure=\frac{Force }{Area}

⇒ Force =Pressure\times Area

⇒ Plugging the values.

⇒ Force =4660.95\times 6.51

⇒ Force=30342.78 Newton (N)

So,

The net forces which pushes the window is 30342.78 N.

3 0
3 years ago
All substances are built from​
Greeley [361]
An atom. Hope this helped
6 0
3 years ago
a person dives off the edge of aa cliff 33m above the surface of the sea. assuming that air resistance is negligible, how long d
USPshnik [31]

Answer:

The time is t = 2.595 \  s

The speed is v = 25.43 \ m/s

Explanation:

From the question we are told that

    The height of the cliff is  h =  33 \  m

 Generally from kinematic equation we have that

       h  =  ut + \frac{1}{2} gt^2

before the jump the persons initial velocity is  u =  0 m/s

 So

        33   =  0 * t + \frac{1}{2} 9.8 * t^2

=>      t = 2.595 \  s

Generally from kinematic equation

     v= u + gt

=>  v= 0 + 9.8 * 2.595

=>  v = 25.43 \ m/s

3 0
3 years ago
a chamber with a fixed volume of 1.0 meters cubed contains a monatomic gas at 3.00 *10^K. The chamber is heated to a temperature
igomit [66]

Answer:

Explanation:

Given

Volume of fixed chamber V=1 m^3

Initial Temperature T_1=300 K

Final Temperature T_2=400 K

Heat Supplied Q=10 J

From First law of thermodynamics

Change in internal energy of the system is equal to heat added minus work done by the system

\Delta U=Q-W

as the volume is fixed therefore work

W=\int PdV=0

thus \Delta U=mc_v\Delta T=Q

c_v for mono-atomic gas is 12.471 J/K-mol

n\times 12.471\times (400-300)=10

n=0.008018 mol

and 1 mole contains 6.022\times 10^{23} molecules

thus  No of molecules=0.008018\times 6.022\times 10^{23}

No of molecules=4.82\times 10^{21} molecules

3 0
3 years ago
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