Answer:
Thus the time taken is calculated as 387.69 years
Solution:
As per the question:
Half life of
= 28.5 yrs
Now,
To calculate the time, t in which the 99.99% of the release in the reactor:
By using the formula:

where
N = No. of nuclei left after time t
= No. of nuclei initially started with

(Since, 100% - 99.99% = 0.01%)
Thus

Taking log on both the sides:


t = 387.69 yrs
Answer:
k = 104.46 N/m
Explanation:
Here we can use energy conservation
so we will have
initial gravitational potential energy = final total spring potential energy
as we know that she falls a total distance of 31 m
while the unstretched length of the string is 12 m
so the extension in the string is given as


so we have



Answer:
The rate of heat removed from inside the refrigerator is 300 watts.
Explanation:
By the First Law of Thermodynamics and the definition of a Refrigeration Cycle, we have the following formula to determine the rate of heat removed from inside the refrigerator (
), in watts:
(1)
Where:
- Rate of heat released to the room, in watts.
- Rate of electric energy needed by the refrigerator, in watts.
If we know that
and
, then the rate of heat removed from inside the refrigerator is:


The rate of heat removed from inside the refrigerator is 300 watts.
The magnitudes of his q and ∆H for the copper trial would be lower than the aluminum trial.
The given parameters;
- <em>initial temperature of metals, = </em>
<em /> - <em>initial temperature of water, = </em>
<em> </em> - <em>specific heat capacity of copper, </em>
<em> = 0.385 J/g.K</em> - <em>specific heat capacity of aluminum, </em>
= 0.9 J/g.K - <em>both metals have equal mass = m</em>
The quantity of heat transferred by each metal is calculated as follows;
Q = mcΔt
<em>For</em><em> copper metal</em><em>, the quantity of heat transferred is calculated as</em>;

<em>The </em><em>change</em><em> in </em><em>heat </em><em>energy for </em><em>copper metal</em>;

<em>For </em><em>aluminum metal</em><em>, the quantity of heat transferred is calculated as</em>;

<em>The </em><em>change</em><em> in </em><em>heat </em><em>energy for </em><em>aluminum metal </em><em>;</em>

Thus, we can conclude that the magnitudes of his q and ∆H for the copper trial would be lower than the aluminum trial.
Learn more here:brainly.com/question/15345295
We can answer the problem by Snell's Law:
Snell's law<span> (also known as </span>Snell<span>–Descartes </span>law<span> and the </span>law<span> of refraction) is a formula used to describe the relationship between the angles of incidence and refraction, when referring to light or other waves passing through a boundary between two different isotropic media, such as water, glass, or air.</span>