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xz_007 [3.2K]
3 years ago
12

Suppose that there are N=108 two-state systems, each with energy difference E=6 × 10-21 J between the two states. The environmen

t temperature is 300 K. The number of systems in their high energy states is M. Therefore, the free energy is a function of M: F(M).What is dU/dM?
Physics
1 answer:
kati45 [8]3 years ago
5 0

Answer:

The value of \dfrac{dU}{dM} is 6\times10^{-29}\ J/unit.

Explanation:

Given that,

Number N=10^{8}

Energy difference = 6\times10^{-21}\ J[/tex]

Temperature T =300 K

We need to calculate the value of \dfrac{dU}{dM}

We know that,

\dfrac{dU}{dM}=\dfrac{energy\ difference}{Change\ in\ number\ of\ system}

\dfrac{dU}{dM}=\dfrac{6\times10^{-21}}{10^{8}}

\dfrac{dU}{dM}=6\times10^{-29}\ J/unit

Hence, The value of \dfrac{dU}{dM} is 6\times10^{-29}\ J/unit.

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What does it mean to say that some elements are reactive and form ions easily whereas others do not?
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4 0
2 years ago
A spring on a horizontal surface can be stretched and held 0.5 m from its equilibrium position with a force of 60 N. a. How much
Lostsunrise [7]

Answer:

a)1815Joules b) 185Joules

Explanation:

Hooke's law states that the extension of a material is directly proportional to the applied force provided that the elastic limit is not exceeded. Mathematically;

F = ke where;

F is the applied force

k is the elastic constant

e is the extension of the material

From the formula, k = F/e

F1/e1 = F2/e2

If a force of 60N causes an extension of 0.5m of the string from its equilibrium position, the elastic constant of the spring will be ;

k = 60/0.5

k = 120N/m

a) To get the work done in stretching the spring 5.5m from its position,

Work done by the spring = 1/2ke²

Given k = 120N/m, e = 5.5m

Work done = 1/2×120×5.5²

Work done = 60× 5.5²

Work done = 1815Joules

b) work done in compressing the spring 1.5m from its equilibrium position will be gotten using the same formula;

Work done = 1/2ke²

Work done =1/2× 120×1.5²

Works done = 60×1.5²

Work done = 135Joules

8 0
3 years ago
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You wad up a piece of paper and throw it into the wastebasket. How far will
vitfil [10]

The range of the piece of paper is C) 1.4 m

Explanation:

The motion of the piece of paper is the motion of a projectile, which consists of two separate motions:

- A uniform motion along the horizontal direction, with constant velocity

- A uniformly accelerated motion along the vertical direction, with constant acceleration (the acceleration of gravity, g=9.8 m/s^2)

From the equation of motion, it is possible to find an expression for the range (the total horizontal distance covered) of a projectile, which is given by:

d=\frac{u^2 sin 2\theta}{g}

where

u is the initial velocity

\theta is the angle of projection

g is the acceleration of gravity

For the piece of paper in this problem,

u = 4.3 m/s

\theta=65^{\circ}

Substituting,

d=\frac{(4.3)^2 sin(2\cdot 65^{\circ})}{9.8}=1.45 m \sim 1.4 m

Learn more about projectile motion:

brainly.com/question/8751410

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6 0
3 years ago
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A soft tennis ball is dropped onto a hard floor from a height of 1.50 m and rebounds to a height of 1.10 m. (a) Calculate its ve
Gemiola [76]

Answer:

(a)   v = 5.42m/s

(b)   vo = 4.64m/s

(c)   a = 2874.28m/s^2

(d)   Δy = 5.11*10^-3m

Explanation:

(a) The velocity of the ball before it hits the floor is given by:

v=\sqrt{2gh}        (1)

g: gravitational acceleration = 9.8m/s^2

h: height where the ball falls down = 1.50m

v=\sqrt{2(9.8m/s^2)(1.50m)}=5.42\frac{m}{s}

The speed of the ball is 5.42m/s

(b) To calculate the velocity of the ball, after it leaves the floor, you use the information of the maximum height reached by the ball after it leaves the floor.

You use the following formula:

h_{max}=\frac{v_o^2}{2g}       (2)

vo: velocity of the ball where it starts its motion upward

You solve for vo and replace the values of the parameters:

v_o=\sqrt{2gh_{max}}=\sqrt{2(9.8m/s^2)(1.10m)}=4.64\frac{m}{s}

The velocity of the ball is 4.64m/s

(c) The acceleration is given by:

a=\frac{\Delta v}{\Delta t}=\frac{v_o-v}{3.50*10^{-3}s}=\frac{4.64m/s-(-5.42m)/s}{3.50*10^{-3}s}=2874.285\frac{m}{s^2}

a=\frac{\Delta v}{\Delta t}=\frac{v_o-v}{3.50*10^{-3}s}=\frac{4.64m/s-5.42m/s}{3.50*10^{-3}s}=-222.85\frac{m}{s^2}

The acceleration of the ball is 2874.28/s^2

(d) The compression of the ball is:

\Deta y=\frac{v^2}{2(a)}=\frac{(5.42m/s)^2}{2(2874.28m/s^2)}=5.11*10^{-3}m

THe compression of the ball when it strikes the floor is 5.11*10^-3m

4 0
3 years ago
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