Answer:
a=2 ok do it and ........
Answer:
Turn the heater on
Explanation:
There are two main forces involved in a balloon flight
The downward force is the total weight of the balloon: the air it contains, the gas bag, the basket, the passengers, etc.
The upward force is the weight of the of the air the balloon displaces.
During level flight
,
buoyant force = weight of displaced air - total weight of balloon
If you increase the temperature of the air in the bag, the air molecules spread out and leave through the bottom of the bag.
The balloon still has the same volume, so the weight of displaced outside air stays the same.
However, the balloon has lost some hot inside air, so its total weight decreases.
The upward force is greater than the downward force, so the balloon rises.
Answer:
a)η = 69.18 %
b)W= 1210 J
c)P=3967.21 W
Explanation:
Given that
Q₁ = 1749 J
Q₂ = 539 J
From first law of thermodynamics
Q₁ = Q₂ +W
W=Work out put
Q₂=Heat rejected to the cold reservoir
Q₁ =heat absorb by hot reservoir
W= Q₁- Q₂
W= 1210 J
The efficiency given as
![\eta=\dfrac{W}{Q_1}](https://tex.z-dn.net/?f=%5Ceta%3D%5Cdfrac%7BW%7D%7BQ_1%7D)
![\eta=\dfrac{1210}{1749}](https://tex.z-dn.net/?f=%5Ceta%3D%5Cdfrac%7B1210%7D%7B1749%7D)
![\eta=0.6918](https://tex.z-dn.net/?f=%5Ceta%3D0.6918)
η = 69.18 %
We know that rate of work done is known as power
![P=\dfrac{W}{t}](https://tex.z-dn.net/?f=P%3D%5Cdfrac%7BW%7D%7Bt%7D)
![P=\dfrac{1210}{0.305}\ W](https://tex.z-dn.net/?f=P%3D%5Cdfrac%7B1210%7D%7B0.305%7D%5C%20W)
P=3967.21 W
To solve this problem we will use the basic concept given by the Volume of a sphere with which the atom approaches. The fraction in percentage terms would be given by the division of the total volume of the nucleus by that of the volume of the atom, that is,
![\% Percent = \frac{V_{nucleus}}{V_{atom}}*100](https://tex.z-dn.net/?f=%5C%25%20Percent%20%3D%20%5Cfrac%7BV_%7Bnucleus%7D%7D%7BV_%7Batom%7D%7D%2A100)
![\% Percent = \frac{4/3 \pi (d_{nucleus}/2)^3}{4/3 \pi (d_{atom}/2)^3}*100](https://tex.z-dn.net/?f=%5C%25%20Percent%20%3D%20%5Cfrac%7B4%2F3%20%5Cpi%20%28d_%7Bnucleus%7D%2F2%29%5E3%7D%7B4%2F3%20%5Cpi%20%28d_%7Batom%7D%2F2%29%5E3%7D%2A100)
![\% Percent = \frac{(d_{nucleus}/2)^3}{ (d_{atom}/2)^3}*100](https://tex.z-dn.net/?f=%5C%25%20Percent%20%3D%20%5Cfrac%7B%28d_%7Bnucleus%7D%2F2%29%5E3%7D%7B%20%28d_%7Batom%7D%2F2%29%5E3%7D%2A100)
![\% Percent =\frac{(1.0*10^{-14}/2)^3}{ (1.1*10^{-10}/2)^3}*100](https://tex.z-dn.net/?f=%5C%25%20Percent%20%3D%5Cfrac%7B%281.0%2A10%5E%7B-14%7D%2F2%29%5E3%7D%7B%20%281.1%2A10%5E%7B-10%7D%2F2%29%5E3%7D%2A100)
![\% Percent = 7.51*10^{-13}*100](https://tex.z-dn.net/?f=%5C%25%20Percent%20%3D%207.51%2A10%5E%7B-13%7D%2A100)
![\% Percent = 7.51*10^{-11}\%](https://tex.z-dn.net/?f=%5C%25%20Percent%20%3D%207.51%2A10%5E%7B-11%7D%5C%25)
Therefore the percent of the atom's volume is occupied by mass is ![7.51*10^{-11}\%](https://tex.z-dn.net/?f=7.51%2A10%5E%7B-11%7D%5C%25)