The available options are:
a. Want for a television set
b. Want for medical attention
c. Want for friendship
d. Want for new clothing
Answer:
Want for friendship
Explanation:
Given that economic want is what humans desire to have or possess such that they pay money to acquire them.
Hence, from the available options "want for friendship " is not economic want because it can't be bought with money, while other options can be bought with money or monetary transaction.
Answer:
m' = 4948.38 kg/s
Explanation:
For a case of 100% efficiency, the power produced must be equal to the rate of potential energy conversion
GIVEN THAT
Power = 100 MW
rate of Potential energy = (m')*g*h
100*10^6 = (m')*9.81*206
m' = 4948.38 kg/s
Answer:
Answer: ±0.02 units or 20±0.02 units or 19.98-20.02 units depending on how they prefer its written (typically the first or second one)
Explanation:
says on the sheet. Unless otherwise stated 0.XX = ±0.02 tolerance
(based on image sent in other post)
Answer:
26 lbf
Explanation:
The mass of the satellite is the same regardless of where it is.
The weight however, depends on the acceleration of gravity.
The universal gravitation equation:
g = G * M / d^2
Where
G: universal gravitation constant (6.67*10^-11 m^3/(kg*s))
M: mass of the body causing the gravitational field (mass of Earth = 6*10^24 kg)
d: distance to that body
15000 miles = 24140 km
The distance is to the center of Earth.
Earth radius = 6371 km
Then:
d = 24140 + 6371 = 30511 km
g = 6.67*10^-11 * 6*10^24 / 30511000^2 = 0.43 m/s^2
Then we calculate the weight:
w = m * a
w = 270 * 0.43 = 116 N
116 N is 26 lbf
Answer:
a) 0.684
b) 0.90
Explanation:
Catalyst
EO + W → EG
<u>a) calculate the conversion exiting the first reactor </u>
CAo = 16.1 / 2 mol/dm^3
Given that there are two stream one contains 16.1 mol/dm^3 while the other contains 0.9 wt% catalyst
Vo = 7.24 dm^3/s
Vm = 800 gal = 3028 dm^3
hence Im = Vin/ Vo = (3028 dm^3) / (7.24dm^3/s) = 418.232 secs = 6.97 mins
next determine the value of conversion exiting the reactor ( Xai ) using the relation below
KIm = 
------ ( 1 )
make Xai subject of the relation
Xai = KIm / 1 + KIm --- ( 2 )
<em>where : K = 0.311 , Im = 6.97 ( input values into equation 2 )</em>
Xai = 0.684
<u>B) calculate the conversion exiting the second reactor</u>
CA1 = CA0 ( 1 - Xai )
therefore CA1 = 2.5438 mol/dm^3
Vo = 7.24 dm^3/s
To determine the value of the conversion exiting the second reactor ( Xa2 ) we will use the relation below
XA2 = ( Xai + Im K ) / ( Im K + 1 ) ----- ( 3 )
<em> where : Xai = 0.684 , Im = 6.97, and K = 0.311 ( input values into equation 3 )</em>
XA2 = 0.90
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