Answer:
The rate at which the container is losing water is 0.0006418 g/s.
Explanation:
- Under the assumption that the can is a closed system, the conservation law applied to the system would be: , where is all energy entering the system, is the total energy leaving the system and, is the change of energy of the system.
- As the purpose is to kept the beverage can at constant temperature, the change of energy () would be 0.
- The energy that goes into the system, is the heat transfer by radiation from the environment to the top and side surfaces of the can. This kind of transfer is described by: where is the emissivity of the surface, known as the Stefan–Boltzmann constant, is the total area of the exposed surface, is the temperature of the surface in Kelvin, is the environment temperature in Kelvin.
- For the can the surface area would be ta sum of the top and the sides. The area of the top would be , the area of the sides would be . Then the total area would be
- Then the radiation heat transferred to the can would be .
- The can would lost heat evaporating water, in this case would be , where is the rate of mass of water evaporated and, is the heat of vaporization of the water ().
- Then in the conservation balance: , it would be.
- Recall that , then solving for :
Nonmetals often share or gain
electrons. The nonmetals in the periodic table increases as you move to the
right and decreases as you go down. This is because, the smaller the atom, the
reactive it gets due to less electron attached to the orbits of the atom. The
reactivity of nonmetals is arranged in decreasing order.
<span>
Carbon
</span>
Nitrogen
Oxygen
Fluorine
Phosphorus
<span>
Sulfur</span>
Chlorine
<span>
Selenium</span>
<span>
Bromine</span>
<span>
Iodine</span>
Answer:
They are 7.4m apart.
Explanation:
Here we have a parabolic motion problem. we need the time taken to land so:
considerating only the movement on Y axis:
Because we have a contant velocity motion on X axis:
and
the distance between them is given by:
The sample appears to have gone through 3 half-lives
1st half life: 1000 to 500 g
2nd half life: 500 to 250 g
3rd half life: 250 to 125 g
The duration of a half-life, therefore, can be inferred to be 66 ÷ (3) = 22 days.
After a 4th half life, there will be 125÷2= 62.5 g.
At this point, an additional 22 days will have passed, for a total of 88 days.
Answer is C.