All categories

3 years ago

Using the statistical definition of entropy, what is the entropy of a system where W = 4?

Physics

1 answer:

Oliga [24]3 years ago

5 0

<span> S = k ln(W)
= 1.38*10^-23 J/K * ln(4)
= 1.91*10^-23 J/K</span>
so it will be c

You might be interested in

The half-life of an anticancer drug is 170 days at 25 oC. Degradation follows the first-order kinetics. When the drug is stored

julsineya [31]

Answer:

1694 days

Explanation:

In first-order kinetics, the rate is proportional to the amount.

dA/dt = kA

For first-order kinetics, the rate k can be found using the half-life:

t₁,₂ = (ln 2) / k

In other words, the half-life is inversely proportional with the rate.

At the lower temperature, the rate is reduced to a third, so the half-life increases by a factor of 3. Meaning that the new half-life is 170 × 3 = 510 days.

The "shelf life" is the time it takes to reduce the initial amount to 10%. We can solve for this using the half-life equation.

A = A₀ (½)^(t / t₁,₂)

A₀/10 = A₀ (½)^(t / 510)

1/10 = (½)^(t / 510)

ln(1/10) = (t / 510) ln(½)

ln(10) = (t / 510) ln(2)

ln(10) / ln(2) = t / 510

t = 510 ln(10) / ln(2)

t ≈ 1694

5 0

2 years ago

Two 13 cm -long thin glass rods uniformly charged to +11nC are placed side by side, 4.0 cm apart. What are the electric field st

DedPeter [7]

Answer:

E1 = 10.15 * 10^4 N/C

E2 = 0

E3 = 10.15 *10^4 N/C

Explanation:

Given data:

Two 13 cm-long thin glass rods ( L ) = 0.13 m

charge (Q) = +11nC

distance between thin glass rods = 4 cm .

<u>Calculate the electric field strengths </u>

electric charge due to a single glass rod in the question ( E ) = $\frac{Q}{2\pi e_{0}rL }$

equation 1 can be used to determine E1, E2 and E3 because the points lie within the two rods hence the net electric field produced will be equal to the difference in electric fields produced

applying equation 1 to determine E1

E1 = $\frac{Q}{2\pi e_{0}rL }$ $( \frac{1}{0.01} - \frac{1}{0.03} )$ ( distance from 1 rod is 0.01 m and from the other rod is 0.03 )

= $\frac{11*10^{-9} }{2*3.14*8.85*10^{-12}*0.13 } ( 66.67 )$

= 10.15 * 10^4 N/C

applying equation 1 to determine E2

E2 = $\frac{Q}{2\pi e_{0}rL }$ $( \frac{1}{0.02} - \frac{1}{0.02} )$

therefore E2 = 0

E1 = E3

hence E3 = 10.15*10^4 N/C

4 0

2 years ago

A 6 kg block is sliding down a horizontal frictionless surface with a constant speed of 5 m/s. It then slides down a frictionles

BARSIC [14]

Answer:

5in

Explanation:

4 0

3 years ago

Sodium chloride, NaCl, is formed when a sodium atom transfers its electron to a chlorine atom. The difference in charge between

victus00 [196]

Sodium chloride, NaCl, is formed when a sodium atom transfers its electron to a chlorine atom. The difference in charge between the two atoms creates a(n) electrostatic attraction that bonds them together.

8 0

2 years ago

Read 2 more answers

A bathtub contains 65 gallons of water and the total weight of the tub and water is approximately 931.925 pounds. You pull the p

levacccp [35]

Answer:

$Q = 8,345 * v$

Explanation:

So, we are looking for an expression of the amount of water that has been drained from the tub. The expression is in terms of v that represent the number of gallons of water drained since the plug was pulled. Since we are interested in the pounds of water that has been drained from the tub we need to take into account that for every gallon of water drained, 8.345 pounds have left the tub. Therefore, the expression for the weight of water Q that has been drained from the tub in terms of v is simply :

$Q = 8,345 * v$

Where v is the amount of gallons that has been drained from the tub.

Have a nice day. let me know if I can help with anything else

8 0

3 years ago