Answer:
Temperature of the air in the balloon = 272°C
Explanation:
Given:
Volume of balloon = 500 m³
Air temperature = 15° C = 273 + 15 = 288 K
Total weight = 290 kg
Density of air = 1.23 kg/m³
Find:
Temperature of the air in the balloon
Computation:
Density of hot air = Density of air - [Total weight / Volume of balloon]
Density of hot air = 1.23 - [290 - 500]
Density of hot air = 0.65 kg/m³
[Density of hot air][Temperature of the air in the balloon] = [Density of air][Air temperature ]
Temperature of the air in the balloon = [(1.23)(288)]/(0.65)
Temperature of the air in the balloon = 544.98
Temperature of the air in the balloon = 545 K
Temperature of the air in the balloon = 545 - 273 = 272°C
Yes. According to the coach's mathematical criteria, Tommy had a great season.
Sadly, Tommy doesn't even know it. His tone-deaf coach decided to describe success in terms that are absurd for 7-yr-olds, as well as for most of their parents.
Answer:
The maximum no. of electrons- 
Solution:
As per the question:
Maximum rate of transfer of charge, I = 1.0 C/s
Time, t = 1.0 h = 3600 s
Rate of transfer of charge is current, I
Also,
Q = ne
where
n = no. of electrons
Q = charge in coulomb
I = current
Thus
Q = It
Thus the charge flow in 1. 0 h:
Maximum number of electrons, n is given by:
where
e = charge on an electron = 
Thus
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