Answer:
The temperature of silver at this given resistivity is 2971.1 ⁰C
Explanation:
The resistivity of silver is calculated as follows;
![R_t = R_o[1 + \alpha(T-T_o)]\\\\](https://tex.z-dn.net/?f=R_t%20%3D%20R_o%5B1%20%2B%20%5Calpha%28T-T_o%29%5D%5C%5C%5C%5C)
where;
Rt is the resistivity of silver at the given temperature
Ro is the resistivity of silver at room temperature
α is the temperature coefficient of resistance
To is the room temperature
T is the temperature at which the resistivity of silver will be two times the resistivity of iron at room temperature
![R_t = R_o[1 + \alpha(T-T_o)]\\\\\R_t = 1.59*10^{-8}[1 + 0.0038(T-20)]](https://tex.z-dn.net/?f=R_t%20%3D%20R_o%5B1%20%2B%20%5Calpha%28T-T_o%29%5D%5C%5C%5C%5C%5CR_t%20%3D%201.59%2A10%5E%7B-8%7D%5B1%20%2B%200.0038%28T-20%29%5D)
Resistivity of iron at room temperature = 9.71 x 10⁻⁸ ohm.m
When silver's resistivity becomes 2 times the resistivity of iron, we will have the following equations;
![R_t,_{silver} = 2R_o,_{iron}\\\\1.59*10^{-8}[1 + 0.0038(T-20)] =(2 *9.71*10^{-8})\\\\\ \ (divide \ through \ by \ 1.59*10^{-8})\\\\1 + 0.0038(T-20) = 12.214\\\\1 + 0.0038T - 0.076 = 12.214\\\\0.0038T +0.924 = 12.214\\\\0.0038T = 12.214 - 0.924\\\\0.0038T = 11.29\\\\T = \frac{11.29}{0.0038} \\\\T = 2971.1 \ ^0C](https://tex.z-dn.net/?f=R_t%2C_%7Bsilver%7D%20%3D%202R_o%2C_%7Biron%7D%5C%5C%5C%5C1.59%2A10%5E%7B-8%7D%5B1%20%2B%200.0038%28T-20%29%5D%20%3D%282%20%2A9.71%2A10%5E%7B-8%7D%29%5C%5C%5C%5C%5C%20%5C%20%28divide%20%5C%20through%20%5C%20by%20%5C%201.59%2A10%5E%7B-8%7D%29%5C%5C%5C%5C1%20%2B%200.0038%28T-20%29%20%3D%2012.214%5C%5C%5C%5C1%20%2B%200.0038T%20-%200.076%20%3D%2012.214%5C%5C%5C%5C0.0038T%20%2B0.924%20%3D%2012.214%5C%5C%5C%5C0.0038T%20%20%3D%2012.214%20-%200.924%5C%5C%5C%5C0.0038T%20%3D%2011.29%5C%5C%5C%5CT%20%3D%20%5Cfrac%7B11.29%7D%7B0.0038%7D%20%5C%5C%5C%5CT%20%3D%202971.1%20%5C%20%5E0C)
Therefore, the temperature of silver at this given resistivity is 2971.1 ⁰C
Answer:(c)
Explanation:
Charge particle posses a region of influence called a filed i.e. Electric field.
It is a region around the charged particle such that other charge particle experiences force due to this field. The nature of force is decided by the charge on the particle creating an electric field.
For a positive charge, electric field lines emerge out of it, and for a negative charge, it acts as radially inwards.
Answer: 0.48W
useful power output=total power output*efficency
useful power output=2.4W*0.20=0.48W
If the echo (the reflected sound) reaches your ear less than about
0.1 second after the original sound, your brain doesn't separate them,
and you're not aware of the echo even though it's there.
If the echo comes from, say, a wall, 0.1 second means you'd have to be
about 17 meters away from the wall. If you're closer than that, then the
echo reaches you in less than 0.1 second and you're not aware of it.
A. 30 meters . . .
No. You hear that echo easily
B. you're standing within range of both sounds . . .
No. You hear that echo easily, if you're at least 17 meters from the wall.
C. less than 0.1 second later . . .
That's it. The echo is there but your brain doesn't know it.
D. 21.5 meters
No. You hear that echo easily.
Answer:
2274 J/kg ∙ K
Explanation:
The complete statement of the question is :
A lab assistant drops a 400.0-g piece of metal at 100.0°C into a 100.0-g aluminum cup containing 500.0 g of water at 15 °C. In a few minutes, she measures the final temperature of the system to be 40.0°C. What is the specific heat of the 400.0-g piece of metal, assuming that no significant heat is exchanged with the surroundings? The specific heat of this aluminum is 900.0 J/kg ∙ K and that of water is 4186 J/kg ∙ K.
= mass of metal = 400 g
= specific heat of metal = ?
= initial temperature of metal = 100 °C
= mass of aluminum cup = 100 g
= specific heat of aluminum cup = 900.0 J/kg ∙ K
= initial temperature of aluminum cup = 15 °C
= mass of water = 500 g
= specific heat of water = 4186 J/kg ∙ K
= initial temperature of water = 15 °C
= Final equilibrium temperature = 40 °C
Using conservation of energy
heat lost by metal = heat gained by aluminum cup + heat gained by water
