Answer:
how are supposed to help when you can't do anything?
Answer:
<em>a) 4.51 lbf-s^2/ft</em>
<em>b) 65.8 kg</em>
<em>c) 645 N</em>
<em>d) 23.8 lb</em>
<em>e) 65.8 kg</em>
<em></em>
Explanation:
Weight of the man on Earth = 145 lb
a) Mass in slug is...
32.174 pound = 1 slug
145 pound =
slug
= 145/32.174 = <em>4.51 lbf-s^2/ft</em>
b) Mass in kg is...
2.205 pounds = 1 kg
145 pounds =
kg
= 145/2.205 = <em>65.8 kg</em>
c) Weight in Newton = mg
where
m is mass in kg
g is acceleration due to gravity on Earth = 9.81 m/s^2
Weight in Newton = 65.8 x 9.81 = <em>645 N</em>
d) If on the moon with acceleration due to gravity of 5.30 ft/s^2,
1 m/s^2 = 3.2808 ft/s^2
m/s^2 = 5.30 ft/s^2
= 5.30/3.2808 = 1.6155 m/s^2
weight in Newton = mg = 65.8 x 1.6155 = 106
weight in pounds = 106/4.448 = <em>23.8 lb</em>
e) The mass of the man does not change on the moon. It will therefore have the same value as his mass here on Earth
mass on the moon = <em>65.8 kg</em>
Answer:
Coordination number:
Coordination number can be defined as the number of nearest atoms which touch the central atom.In simple word the number of neighbours atoms which touches a an atoms.
Atomic packing factor :
It is also known as packing efficiency.It is the fraction of volume in a Crystal which is filled by constituent particles.It is always less than 1.Generally it is represented in the fraction.
Ex:
Lets take FCC cubic cell
Coordination number = 12 .
Atomic packing factor = 74%
These parts are commonly called carburetor emulsion tubes. These tubes maintain the air-fuel ratio at different speeds.
The carburetor is a device of the combustion engine power supply system that mixes fuel and air in order to facilitate internal combustion.
The carburetor emulsion tubes are tubes that maintain the air-fuel ratio at different velocities.
These tubes (carburetor emulsion tubes) are small brass cylinders where the metering needle slides into them.
Learn more about carburetors here:
brainly.com/question/4237015
Answer:


Explanation:
= Area of section 1 = 
= Velocity of water at section 1 = 100 ft/min
= Specific volume at section 1 = 
= Density of fluid = 
= Area of section 2 = 
Mass flow rate is given by

The mass flow rate through the pipe is 
As the mass flowing through the pipe is conserved we know that the mass flow rate at section 2 will be the same as section 1

The speed at section 2 is
.