Answer:
1387908 lbm/h
Explanation:
Air flowing into jet engine = 70 lbm/s
ρ = Exhaust gas density = 0.1 lbm/ft³
r = Radius of exit with a circular cross section = 1 ft
v = Exhaust gas velocity = 1450 ft/s
Exhaust gas mass (flow rate)= Air flowing into jet engine + Fuel
Q = (70+x) lbm/s
Area of exit with a circular cross section = π×r² = π×1²= π m²
Now from energy balance
Q = ρ×A×v
⇒70+x = 0.1×π×1450
⇒70+x = 455.53
⇒ x = 455.53-70
⇒ x = 385.53 lbm/s
∴ Mass of fuel which is supplied to the engine each minute is 1387908 lbm/h
You are Thomas Alva Edison. You also invented the phonograph
and the first practical movie camera. Sadly, you died almost exactly
9 years before I was born.
Answer:
76.73 ft/s
Explanation:
Let the final velocity is v.
initial velocity, u = 96 ft/s
g = 32 ft/s²
height, h = 52 feet
use third equation of motion
v² = u² - 2 gh
v² = 96 x 96 - 2 x 32 x 52
v = 76.73 ft/s
Thus, the speed of the ball as it reaches the ground is 76.73 ft/s.
Answer:
radius r is 0.414 R
Explanation:
Given data
FCC octahedral site
atomic radius R
to find out
radius r
solution
we know that at FCC octahedral
length of side = 2 R + 2r
and by pythagorean theorem
a = 2√2R
here a = 2R + 2r
so 2R + 2r = 2√2R
so r = ( √2R )- R
r = 0.414 R
so radius r is 0.414 R
Answer:
left is the left side of your hand when looking in a way
Explanation:
