Answer:
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Answer:
The mass of the Al-duckie should be 30 kg.
Explanation:
We will use the first law of thermodynamics:
ΔU = m·Cv·ΔT
Since the specific heat of water is 4.185 J(gºC), the change in the water's internal energy would be:
ΔU = 100 kg · 4.185 J(gºC) · (42ºC - 38ºC) = 1674 KJ
Given that no heat is lost, all the internal energy that the water loses while cooling down will transfer to the duckie. So, if the duckie has ΔU = 1674 KJ and its final temperature is the desired 38 ºC, we can calculate its mass using the first law again:
![m=\frac{\Delta{U}}{Cv{\Delta{T}}}=\frac{1674}{0.9*[38-(-24)]}=30Kg](https://tex.z-dn.net/?f=m%3D%5Cfrac%7B%5CDelta%7BU%7D%7D%7BCv%7B%5CDelta%7BT%7D%7D%7D%3D%5Cfrac%7B1674%7D%7B0.9%2A%5B38-%28-24%29%5D%7D%3D30Kg)
Answer:
Explanation:
According to the law of conservation of energy, the energy of the absorbed photon must be equal to the binding energy of the electron plus the energy of the released electron:
1 eV is equal to 
, so:
Solving for 
and replacing the given values:
Explanation:
The theory of speciation confirms what happens with fireflies, only fireflies that are part of the same species can reproduce among themselves, which means that fireflies that use pheromones as mating signals will attract fireflies that use that same form or mechanism of reproduction.
Some mechanisms that allow this type of differentiation or speciation to occur are: seasonal or geographic isolation and sexual isolation due to behavior or conduct.
Speciation allows the formation of new populations of organisms that share the same physiological and genetic characteristics. Therefore, the adult fireflies that shine as a mating signal are possibly found in the same geographical position and their physiological and genetic characteristics are compatible with those of his own species.
(a) Differentiate the position vector to get the velocity vector:
<em>r</em><em>(t)</em> = (3.00 m/s) <em>t</em> <em>i</em> - (4.00 m/s²) <em>t</em>² <em>j</em> + (2.00 m) <em>k</em>
<em>v</em><em>(t)</em> = d<em>r</em>/d<em>t</em> = (3.00 m/s) <em>i</em> - (8.00 m/s²) <em>t</em> <em>j</em>
<em></em>
(b) The velocity at <em>t</em> = 2.00 s is
<em>v</em> (2.00 s) = (3.00 m/s) <em>i</em> - (16.0 m/s) <em>j</em>
<em></em>
(c) Compute the electron's position at <em>t</em> = 2.00 s:
<em>r</em> (2.00 s) = (6.00 m) <em>i</em> - (16.0 m) <em>j</em> + (2.00 m) <em>k</em>
The electron's distance from the origin at <em>t</em> = 2.00 is the magnitude of this vector:
||<em>r</em> (2.00 s)|| = √((6.00 m)² + (-16.0 m)² + (2.00 m)²) = 2 √74 m ≈ 17.2 m
(d) In the <em>x</em>-<em>y</em> plane, the velocity vector at <em>t</em> = 2.00 s makes an angle <em>θ</em> with the positive <em>x</em>-axis such that
tan(<em>θ</em>) = (-16.0 m/s) / (3.00 m/s) ==> <em>θ</em> ≈ -79.4º
or an angle of about 360º + <em>θ</em> ≈ 281º in the counter-clockwise direction.
