Answer: 71.72 days
Explanation:
This problem can be solved using the <u>Radioactive Half Life Formula: </u>
(1)
Where:
is the final amount of Iodine-131
is the initial amount of Iodine-131
is the time elapsed
is the half life of Iodine-131
Knowing this, let's substitute the values and find
from (1):
(2)
(3)
Applying natural logarithm in both sides:
(4)
(5)
Finding
:
(A) A device that converts heat into work with 100% efficiency
It clearly violates the second law of thermodynamics because it warns that while all work can be turned into heat, not all heat can be turned into work. Therefore, despite the innumerable efforts, the efficiencies of the bodies have only been able to reach 60% at present.
B. At the equator
Explanation:
The energy coming from the Sun hits the Earth's surface at different angles, depending on the latitude of the place. The more perpendicular the ray of lights hit the surface, the more the energy transmitted to the Earth's surface, the warmer the location.
The angle at which the ray of lights hit the Earth is related to the latitude: in particular, the ray of lights arrive perpendicular at the equator (
), they arrive at larger angle in the United States (which is located at intermediate latitudes) and they arrive at the largest angles at the poles. For this reason, the sun's most energy is concentrated at the equator.
A most probably because conduction transfer heat by movement of collisions of particles and movement of electrons within a body
Answer:
376966.991 Joules
Explanation:
Given that :
the height = 12 m
Let assume the tank have a thickness = dh
The radius of the tank by using the concept of similar triangle is :


The area of the tank =
The area of the tank = 
The area of the tank = 
The volume of the tank is = area × thickness
= 
Weight of the element = 
where;
= density of water ; which is given as 10000 N/m³
So;
Weight of the element = 
Weight of the element = 
However; the work required to pump this water = weight × height rise
where the height rise = 12 - h
the work required to pump this water =
(12 - h)
the work required to pump this water = 
We can determine the total workdone by integrating the work required to pump this water
SO;
Workdone = 
= 
= ![\mathbf{ 69.44 \pi[ \frac{12h^3}{3}- \frac{h^4}{4}]^{12}}_0} }](https://tex.z-dn.net/?f=%5Cmathbf%7B%2069.44%20%5Cpi%5B%20%5Cfrac%7B12h%5E3%7D%7B3%7D-%20%20%5Cfrac%7Bh%5E4%7D%7B4%7D%5D%5E%7B12%7D%7D_0%7D%20%7D)
= ![\mathbf{69.44 \pi [ \frac{12^4}{3}-\frac{12^4}{4}]}](https://tex.z-dn.net/?f=%5Cmathbf%7B69.44%20%5Cpi%20%5B%20%5Cfrac%7B12%5E4%7D%7B3%7D-%5Cfrac%7B12%5E4%7D%7B4%7D%5D%7D)
= ![\mathbf{69.44 \pi*12^4 [ \frac{4-3}{12}]}](https://tex.z-dn.net/?f=%5Cmathbf%7B69.44%20%5Cpi%2A12%5E4%20%5B%20%5Cfrac%7B4-3%7D%7B12%7D%5D%7D)
= 
= 376966.991 Joules