At which point is potential energy greatest?

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:

1)

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,

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

2)

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

What do nebulae have the most in common with?

alexdok [17]

Most commonly the answer is A, the sun. A nebula is a nursery for stars. The Sun itself was created in a nebula. Also,you do not need college physics to understand this question

3 0

3 years ago

A force of 10 lb is required to hold a spring stretched 4 in. beyond its natural length. How much work W is done in stretching i

Sunny_sXe [5.5K]

Answer:

114.44 J

Explanation:

From Hook's Law,

F = ke................. Equation 1

Where F = Force required to stretch the spring, k = spring constant, e = extension.

make k the subject of the equation

k = F/e.............. Equation 2

Given: F = 10 lb = (10×4.45) N = 44.5 N, e = 4 in = (4×0.254) = 1.016 m.

Substitute into equation 2

k = 44.5/1.016

k = 43.799 N/m

Work done in stretching the 9 in beyond its natural length

W = 1/2ke²................. Equation 3

Given: e = 9 in = (9×0.254) = 2.286 m, k = 43.799 N/m

Substitute into equation 3

W = 1/2×43.799×2.286²

W = 114.44 J

3 0

3 years ago

Read 2 more answers

How to calculate time using power and energy

Troyanec [42]

Answer:

The formula that links energy and power is: Energy = Power x Time. The unit of energy is the joule, the unit of power is the watt, and the unit of time is the second.

Explanation:

7 0

3 years ago

The acceleration vector of a particle in projectile motion ________.

Alex73 [517]

Answer:

Points downward, and its magnitude is 9.8 m/s^2

Explanation:

The motion of a projectile consists of two independent motions:

- A uniform horizontal motion, with constant velocity and zero acceleration. In fact, there are no forces acting on the projectile along the horizontal direction (if we neglect air resistance), so the acceleration along this direction is zero.

- A vertical motion, with constant acceleration g = 9.8 m/s^2 towards the ground (downward), due to the presence of gravity wich "pulls" the projectile downward.

The total acceleration of the projectile is given by the resultant of the horizontal and vertical components of the acceleration. But we said that the horizontal component is zero, therefore the total acceleration corresponds just to its vertical component, therefore it is a vector with magnitude 9.8 m/s^2 which points downward.

4 0

3 years ago