To find the temperature it is necessary to use the expression and concepts related to the ideal gas law.
Mathematically it can be defined as

Where
P = Pressure
V = Volume
n = Number of moles
R = Gas constant
T = Temperature
When the number of moles and volume is constant then the expression can be written as

Or in practical terms for this exercise depending on the final temperature:

Our values are given as

Replacing

Therefore the final temperature of the gas is 800K
that would be the energy from the core. because of the fact that ht eearths core is producing 45mill killawhats and giggawats
I believe that if you touch a metal sphere with a plastic straw, the straw would not have enough strength to push it. So in that case, the metal sphere will not move and will stay in one place.
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First overtone of open organ pipe is given as

first overtone of closed organ pipe is given as

now they are in unison so we will have




so end correction of both pipes is e = 1 cm
Answer: Option (b) is the correct answer.
Explanation:
Since, there is a negative charge present on the ball and a positive charge present on the rod. So, when the negatively charged metal ball will come in contact with the rod then positive charges from rod get conducted towards the metal ball.
Hence, the rod gets neutralized. But towards the metal ball there is a continuous supply of negative charges. Therefore, after the neutralization of positive charge from the rod there will be flow of negative charges from the metal ball towards the rod.
Thus, we can conclude that negative charge spread evenly on both ends.