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:
a) the inductance of the coil is 6 mH
b) the emf generated in the coil is 18 mV
Explanation:
Given the data in the question;
N = 570 turns
diameter of tube d = 8.10 cm = 0.081 m
length of the wire-wrapped portion l = 35.0 cm = 0.35 m
a) the inductance of the coil (in mH)
inductance of solenoid
L = N²μA / l
A = πd²/4
so
L = N²μ(πd²/4) / l
L = N²μ(πd²) / 4l
we know that μ = 4π × 10⁻⁷ TmA⁻¹
we substitute
L = [(570)² × 4π × 10⁻⁷× ( π × (0.081)² )] / 4(0.35)
L = 0.00841549 / 1.4
L = 6 × 10⁻³ H
L = 6 × 10⁻³ × 1000 mH
L = 6 mH
Therefore, the inductance of the coil is 6 mH
b)
Emf ( ∈ ) = L di/dt
given that; di/dt = 3.00 A/sec
{∴ di = 3 - 0 = 3 and dt = 1 sec}
Emf ( ∈ ) = L di/dt
we substitute
⇒ 6 × 10⁻³ ( 3/1 )
= 18 × 10⁻³ V
= 18 × 10⁻³ × 1000
= 18 mV
Therefore, the emf generated in the coil is 18 mV
Answer:Counter,
0.799,
1.921
Explanation:
Given data




Since outlet temperature of cold liquid is greater than hot fluid outlet temperature therefore it is counter flow heat exchanger
Equating Heat exchange
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=
we can see that heat capacity of hot fluid is minimum
Also from energy balance

=


NTU=1.921





Advantages include low costs and minimal labor.Water stays in the root zone, and foliage stays dry. Drawbacks to surface irrigation include potential overwatering and wasteful runoff.
Answer:
the rate of increase of radius is dR/dt = 0.804 m/hour = 80.4 cm/hour
Explanation:
the slick of oil can be modelled as a cylinder of radius R and thickness h, therefore the volume V is
V = πR² * h
thus
h = V / (πR²)
Considering that the volume of the slick remains constant, the rate of change of radius will be
dh/dt = V d[1/(πR²)]/dt
dh/dt = (V/π) (-2)/R³ *dR/dt
therefore
dR/dt = (-dh/dt)* (R³/2) * (π/V)
where dR/dt = rate of increase of the radius , (-dh/dt)= rate of decrease of thickness
when the radius is R=8 m , dR/dt is
dR/dt = (-dh/dt)* (R³/2) * (π/V) = 0.1 cm/hour *(8m)³/2 * π/1m³ *(1m/100 cm)= 0.804 m/hour = 80.4 cm/hour