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Answer:
minimum electric power consumption of the fan motor is 0.1437 Btu/s
Explanation:
given data
area = 3 ft by 3 ft
air density = 0.075 lbm/ft³
to find out
minimum electric power consumption of the fan motor
solution
we know that energy balance equation that is express as
E in - E out =
......................1
and at steady state
= 0
so we can say from equation 1
E in = E out
so
minimum power required is
E in = W = m
=
put here value
E in =
E in =
E in = 0.1437 Btu/s
minimum electric power consumption of the fan motor is 0.1437 Btu/s
The number of hectares of each crop he should plant are; 250 hectares of Corn, 500 hectares of Wheat and 450 hectares of soybeans
<h3>How to solve algebra word problem?</h3>
He grows corn, wheat and soya beans on the farm of 1200 hectares. Thus;
C + W + S = 12 ----(1)
It costs $45 per hectare to grow corn, $60 to grow wheat, and $50 to grow soybeans. Thus;
45C + 60W + 50S = 63750 -----(2)
He will grow twice as many hectares of wheat as corn. Thus;
W = 2C ------(3)
Put 2C for W in eq 1 and eq 2 to get;
C + 2C + S = 1200
3C + S = 1200 -----(4)
45C + 60(2C) + 50S = 63750
45C + 120C + 50S = 63750
165C + 50S = 63750 ------(5)
Solving eq 4 and 5 simultaneosly gives;
C = 250 and W = 500
Thus; S = 1200 - 3(250)
S = 450
Read more about algebra word problems at; brainly.com/question/13818690
Answer:
work done = 48.88 ×
J
Explanation:
given data
mass = 100 kN
velocity = 310 m/s
time = 30 min = 1800 s
drag force = 12 kN
descends = 2200 m
to find out
work done by the shuttle engine
solution
we know that work done here is
work done = accelerating work - drag work - descending work
put here all value
work done = ( mass ×velocity ×time - force ×velocity ×time - mass ×descends ) 10³ J
work done = ( 100 × 310 × 1800 - 12×310 ×1800 - 100 × 2200 ) 10³ J
work done = 48.88 ×
J