The equilibrium temperature is T13=3.12 ◦C
<u>Explanation:</u>
<u>Given </u>
The temperature of liquids: T1=6◦C, T2=23◦C, T3=38◦C
The temperature of 1+2 liquids mix: T12= 13◦C.
The temperature of 2+3 liquids mix: T23=26.8 ◦C.
The temperature of 1+3 liquids mix: T13= ??
<u>1.When the first two liquids are mixed:</u>
- mC1(T1-T12)+mC2(T2-T12)=0
- C1(6-13)=C2(23-13)=0
- 7C1=10C2
- C1=1.42C2
<u>2.When the second and third liquids are mixed</u><u>:</u>
- mC2(T2-T23)+mC3(T3-T23)=0
- C2(23-26.8)=C3(38-26.8)=0
- 3.8C2=12.8C3
- C2=3.36C3
<u>3.When the first and third liquids are mixed:</u>
- mC1(T1-T13)+mC3(T3-T13)=0
- C1(6-T13)+C3(38-T13)=0
- C1=1.42C2 C2=3.36C3
- C1=1.42C2(3.36C3)
- C1=4.77C3
- C1(6-T13)+C3(38-T13)=0
- 4.77C3(6-T13)+C3(38-T13)=0
- By solving the equation we get,
- T13=3.12 ◦C
- The equilibrium temperature is T13=3.12 ◦C
<u></u>
Answer: option 4: A wire that is 2-mm thick and coiled.
Explanation:
The current in each wire is same. The magnetic field due to a current carrying wire increases if the wire is coiled with the more number of turns. A thick wire would cause low resistance to the current. Hence, a 2-mm thick wire which is coiled would produce the strongest magnetic field.
The electric current amount will vary according to the resistance value when connected to a voltage of a power supply. If resistance is reduced, current will increase. So, if the resistance is infinite (open circuit), the current could be considered zero, or infinitesimal small, if the resistance is zero (of infinitesimal small), the current will tend to be the maximum the power supply can provide. Here is the case of the power supply capability, input resistance, capacity to hold and regulate the output voltage to a certain maximum current, etc. Normally when you short circuit (zero ohms resi
The magnitude of the electric field for 60 cm is 6.49 × 10^5 N/C
R(radius of the solid sphere)=(60cm)( 1m /100cm)=0.6m
Since the Gaussian sphere of radius r>R encloses all the charge of the sphere similar to the situation in part (c), we can use Equation (6) to find the magnitude of the electric field:
Substitute numerical values:
The spherical Gaussian surface is chosen so that it is concentric with the charge distribution.
As an example, consider a charged spherical shell S of negligible thickness, with a uniformly distributed charge Q and radius R. We can use Gauss's law to find the magnitude of the resultant electric field E at a distance r from the center of the charged shell. It is immediately apparent that for a spherical Gaussian surface of radius r < R the enclosed charge is zero: hence the net flux is zero and the magnitude of the electric field on the Gaussian surface is also 0 (by letting QA = 0 in Gauss's law, where QA is the charge enclosed by the Gaussian surface).
Learn more about Gaussian sphere here:
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