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Svetach [21]
2 years ago
15

Two equal quantities of water, of mass m and at temperatures T1 and T2, (T1 > T2) are mixed together with the pressure kept c

onstant. Assume heat exchange occurs only between the two water containers. (a) Show that the entropy change of the universe is delta S = 2mc_P ln(T_1 + T_2/2 squareroot T_1T_2), where c_P is the specific heat capacity of water at constant pressure. Show that delta S > 0 for any finite temperatures T_1 and T_2.

Physics
1 answer:
Ray Of Light [21]2 years ago
4 0

Answer:

ΔS=2*m*Cp*ln((T1+T2)/(2*(T1*T2)^1/2))

Explanation:

The concepts and formulas that I will use to solve this exercise are the integration and the change in the entropy of the universe. To calculate the final temperature of the water the expression for the equilibrium temperature will be used. Similarly, to find the change in entropy from cold to hot water, the equation of the change of entropy will be used. In the attached image is detailed the step by step of the resolution.

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two charges having the same charge magnitude experiencing an attracting force of 3.60N when the charges are 30cm apart.what is t
Tomtit [17]

The charges have opposite sign and magnitude 6 \mu C

Explanation:

The magnitude of the electrostatic force between two electric charges is given by Coulomb's law:

F=k\frac{q_1 q_2}{r^2}

where:

k=8.99\cdot 10^9 Nm^{-2}C^{-2} is the Coulomb's constant

q_1, q_2 are the two charges

r is the separation between the two charges

In this problem, we have:

F = 3.60 N is the force between the two charges

r = 30 cm = 0.30 m is their separation

The two charges have same magnitude, so

q_1 = q_2 = q

So we can rewrite the equation as

F=\frac{kq^2}{r^2}

And solving for q:

q=\sqrt{\frac{Fr^2}{k}}=\sqrt{\frac{(3.60)(0.30)^2}{8.99\cdot 10^9}}=6\cdot 10^{-6} C = 6\mu C

Moreover, the force between the charges is attractive: we know that charges of same sign repel each other while charges of opposite sign attract each other, therefore the charges in this problem have opposite sign, so

q_1 = 6 \mu C\\q_2 = -6 \mu C

Learn more about electric force:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

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3 years ago
The reflection of a sound wave from a wal is called
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I think its an echo?
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3 years ago
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A light, flexible cable is wrapped around a solid cylinder with mass 3.3 kg and a radius of 0.8 meters. The cylinder rotates on
kari74 [83]

Answer:

9.16rad/s^2

Explanation:

We are given that

Mass,m_1=3.3 kg

Radius,r=0.8 m

m_2=4.9 kg

Height,h=2.9 m

We have to find the angular acceleration of the cylinder.

According to question

4.9g-T=4.9a

Tr=I\alpha

Where

\alpha=\frac{a}{r}

Tr=\frac{1}{2}m_1ra

T=\frac{1}{2}m_1a=\frac{1}{2}(3.3)a

Substitute the value

4.9g-\frac{1}{2}(3.3a)=4.9a

4.9\times 9.8=4.9a+\frac{3.3a}{2}

Where g=9.8 m/s^2

48.02=a(4.9+1.65)=6.55a

a=\frac{48.02}{6.55}=7.33m/s^2

Angular acceleration,\alpha=\frac{a}{r}=\frac{7.33}{0.8}=9.16rad/s^2

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3 years ago
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