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mylen [45]
3 years ago
8

A steel block has a volume of 0.08 m³ and a density of 7,840 kg/m³. What is the force of gravity acting on the block (the weight

) in water?
A. 6,150.64 N
B. 5,362.56 N
C. 7,600.18 N
D. 6,700.56 N
Chemistry
1 answer:
melomori [17]3 years ago
7 0

Given:

volume of 0.08 m³

density of 7,840 kg/m³

Required:

force of gravity

Solution:

Find the mass using density equation.

D = M/V

M = DV

M = (7,840 kg/m³)(0.08 m³)

M = 627.2kg

 

F = Mg

F = (627.2kg)(9.8m/s2)

F = 6147N

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write chemical equation for the following decomposition reactions. Aluminum oxide(s) decomposes when electricity is passed throu
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Answer:

2Al2O3 (l) ---> 4Al (l) + 3O2 (g)

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Al2O3 (l) ---> Al (l) + O2 (g)

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Reynolds number, for an aorta of 0.9-centimeter diameter, calculate the blood flow in liters per minute when the flow regime in
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Answer:

Flow in liters per minute is 4 L/min

Explanation:

For this case, we have an aorta of 0.9 cm of diameter (D). Let's suppose an uniform and constant diameter for calculation purposes.  

D = (0.9 cm)(1m/100cm)

D = 0.009 m

It is required to calculate the blood flow in liters per minute when the flow regime changes from laminar to turbulent. Laminar flow is usually less than 2500 for Reynolds value, and Turbulent flow when Re is higher than 2500. So, we need to study the phenomenon for  

Re = 2500.

Using the definition of Reynolds we can find the velocity average of the blood, and use it to find flow. Where blood density is \rho, aorta diameter is D, average velocity is v and blood viscosity is \mu

Re = \frac{\rho v D}{\mu } \\v = \frac{Re*\mu}{\rho D}

From data problem, we have Re, D values. As we need blood density and blood viscosity we can find them in medical studies. For example: in this online document: Blood flow analysis of the aortic arch using computational fluid dynamics †  

Satoshi Numata, Keiichi Itatani, Keiichi Kanda, Kiyoshi Doi, Sachiko Yamazaki, Kazuki Morimoto, Kaichiro Manabe, Koki Ikemoto, Hitoshi Yaku Author Notes  

European Journal of Cardio-Thoracic Surgery, Volume 49, Issue 6, June 2016, Pages 1578–1585, https://doi.org/10.1093/ejcts/ezv459  

Published: 20 January 2016

\rho = 1060 kg/m^3\\\mu = 0.0004 kg/(ms)\\

v = \frac{Re*\mu}{\rho D}\\v = \frac{2500 * 0.004 kg/(ms)}{(1060 kg/m3)(0.009 m)}\\v = 1.04 m/s\\

Average blood velocity is 1.04 m/s. After that, we can calculate the flow (Q) using the flow are (Ao) of aorta.

Q = v*Ao

Q = (1.4m/s)*(\pi*(0.009m/2)^2)\\Q = 6.67 * 10^-5 \frac{m^3}{s}\\Q = (6.67 * 10^-5 \frac{m^3}{s})(\frac{1000 L}{1 m^3} ) (\frac{60 s}{1 min} )\\Q = 4 L/min

Finally, the blood flow in liters per minute is 4 L/min when the flow regime in the aorta changes from laminar to turbulent.

7 0
3 years ago
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