Answer:
L= 50000 lb
D = 5000 lb
Explanation:
To maintain a level flight the lift must equal the weight in magnitude.
We know the weight is of 50000 lb, so the lift must be the same.
L = W = 50000 lb
The L/D ratio is 10 so
10 = L/D
D = L/10
D = 50000/10 = 5000 lb
To maintain steady speed the thrust must equal the drag, so
T = D = 5000 lb
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
Answer:
a) V =10¹¹*(1.5q₁ + 3q₂)
b) U = 1.34*10¹¹q₁q₂
Explanation:
Given
x₁ = 6 cm
y₁ = 0 cm
x₂ = 0 cm
y₂ = 3 cm
q₁ = unknown value in Coulomb
q₂ = unknown value in Coulomb
A) V₁ = Kq₁/r₁
where r₁ = √((6-0)²+(0-0)²)cm = 6 cm = 0.06 m
V₁ = 9*10⁹q₁/(0.06) = 1.5*10¹¹q₁
V₂ = Kq₂/r₂
where r₂ = √((0-0)²+(3-0)²)cm = 3 cm = 0.03 m
V₂ = 9*10⁹q₂/(0.03) = 3*10¹¹q₂
The electric potential due to the two charges at the origin is
V = ∑Vi = V₁ + V₂ = 1.5*10¹¹q₁ + 3*10¹¹q₂ = 10¹¹*(1.5q₁ + 3q₂)
B) The electric potential energy associated with the system, relative to their infinite initial positions, can be obtained as follows
U = Kq₁q₂/r₁₂
where
r₁₂ = √((0-6)²+(3-0)²)cm = √45 cm = 3√5 cm = (3√5/100) m
then
U = 9*10⁹q₁q₂/(3√5/100)
⇒ U = 1.34*10¹¹q₁q₂
Answer:
11/8
Explanation:
7/8+4/8=11/8 if you want it on a ruler you need to go online and type fraction to decimal converter and put in 11/8 into a decimal