Answer:
1 / f = 1 / o + 1 / i = (i + o) / o * 1
f = o * i / (o + i) = 60 * 30 / (60 + 30) = 1800 / 90 = 20 cm
Both the object and image are in positive space for a mirror
Answer:
The answer is Producer. Hope this helps!
Answer: Your answer is<u> 1.36.</u>
Hope this helps!
Answer : The temperature when the water and pan reach thermal equilibrium short time later is, ![59.10^oC](https://tex.z-dn.net/?f=59.10%5EoC)
Explanation :
In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.
![q_1=-q_2](https://tex.z-dn.net/?f=q_1%3D-q_2)
![m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)](https://tex.z-dn.net/?f=m_1%5Ctimes%20c_1%5Ctimes%20%28T_f-T_1%29%3D-m_2%5Ctimes%20c_2%5Ctimes%20%28T_f-T_2%29)
where,
= specific heat of aluminium = ![0.90J/g^oC](https://tex.z-dn.net/?f=0.90J%2Fg%5EoC)
= specific heat of water = ![4.184J/g^oC](https://tex.z-dn.net/?f=4.184J%2Fg%5EoC)
= mass of aluminum = 0.500 kg = 500 g
= mass of water = 0.250 kg = 250 g
= final temperature of mixture = ?
= initial temperature of aluminum = ![150^oC](https://tex.z-dn.net/?f=150%5EoC)
= initial temperature of water = ![20^oC](https://tex.z-dn.net/?f=20%5EoC)
Now put all the given values in the above formula, we get:
![500g\times 0.90J/g^oC\times (T_f-150)^oC=-250g\times 4.184J/g^oC\times (T_f-20)^oC](https://tex.z-dn.net/?f=500g%5Ctimes%200.90J%2Fg%5EoC%5Ctimes%20%28T_f-150%29%5EoC%3D-250g%5Ctimes%204.184J%2Fg%5EoC%5Ctimes%20%28T_f-20%29%5EoC)
![T_f=59.10^oC](https://tex.z-dn.net/?f=T_f%3D59.10%5EoC)
Therefore, the temperature when the water and pan reach thermal equilibrium short time later is, ![59.10^oC](https://tex.z-dn.net/?f=59.10%5EoC)