The answers are:
a) 
b) 
Why?
It seems that you forgot to write the questions of the problem, however, in order to help you, I will try to complete it.
The questions are:
a) How much work does the heart do in a day?
b) What is its power output in watts?
So, solving we have:
We need to convert from liter to cubic meters in order to use the given information, so:
Also, we need to find the mass given the density of the blood.
Now, calculating how much work does the heart do in a day, we have:
Then, calculating what is the power output and its horsepower, we have:
Have a nice day!
Answer:
Explanation:
40 divided by 10 then which would equal 4. Add the 1.0 , 2 ,and 15 together. Then multply the 60 by 18.0 after you are done dividing the answer is 3 with a remainder of 6.
Answer:
F = Gm1m2/r^2 where G = 6.67x10^-11, m1 =1300, m2 = 7800, r = 0.23m
F = 6.67x10^-11 *1300*7800/(0.23)^2 = 0.0127852N
Explanation:
Answer:
In the case of a solution transition metal complex that has an absorption peak at 450 nm in the blue region of the visible spectrum, the (complementary) color of this solution is orange (option B).
Explanation:
The portion of UV-visible radiation that is absorbed implies that a portion of electromagnetic radiation is not absorbed by the sample and is therefore transmitted through it and can be captured by the human eye. That is, in the visible region of a complex, the visible color of a solution can be seen and that corresponds to the wavelengths of light it transmits, not absorbs. The absorbing color is complementary to the color it transmits.
So, in the attached image you can see the approximate wavelengths with the colors, where they locate the wavelength with the absorbed color, you will be able to observe the complementary color that is seen or reflected.
<u><em>
In the case of a solution transition metal complex that has an absorption peak at 450 nm in the blue region of the visible spectrum, the (complementary) color of this solution is orange (option B).</em></u>
Answer:
75.6J
Explanation:
Hi!
To solve this problem we must use the first law of thermodynamics that states that the heat required to heat the air is the difference between the energy levels of the air when it enters and when it leaves the body,
Given the above we have the following equation.
Q=(m)(h2)-(m)(h1)
where
m=mass=1.3×10−3kg.
h2= entalpy at 37C
h1= entalpy at -20C
Q=m(h2-h1)
remember that the enthalpy differences for the air can approximate the specific heat multiplied by the temperature difference
Q=mCp(T2-T1)
Cp= specific heat of air = 1020 J/kg⋅K
Q=(1.3×10−3)(1020)(37-(-20))=75.6J
