Please help on this physics question
1 answer:
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Answer:
The resistance that will provide this potential drop is 388.89 ohms.
Explanation:
Given;
Voltage source, E = 12 V
Voltage rating of the lamp, V = 5 V
Current through the lamp, I = 18 mA
Extra voltage or potential drop, IR = E- V
IR = 12 V - 5 V = 7 V
The resistance that will provide this potential drop (7 V) is calculated as follows:
IR = V
Therefore, the resistance that will provide this potential drop is 388.89 ohms.
The final temperature of the system is 32.5°
we know, H = mcT
where, H = Heat content of the body
m = Mass,
c = Specific heat
T = Change in temperature
According to to the Principle of Calorimetry
The net heat remains constant i.e.
⇒ the heat given by water = heat accepted by the aluminum container.
⇒ 330 x 1 x (45 - T) = 855 x
x (T - 10)
⇒ 14,850 - 330T = 183.21T - 1832
⇒ - 513.21 T = - 16682
or T = 32.5°
we know, H = mcT
where, H = Heat content of the body
m = Mass,
c = Specific heat
T = Change in temperature
According to to the Principle of Calorimetry
The net heat remains constant i.e.
⇒ the heat given by water = heat accepted by the aluminum container.
⇒ 330 x 1 x (45 - T) = 855 x
⇒ 14,850 - 330T = 183.21T - 1832
⇒ - 513.21 T = - 16682
or T = 32.5°