TIMED TEST!!!!!!!!!!!!What process that occurred in the Midwestern United States in the 1930s is now occurring in the Sahel regi
1 answer:
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Answer:
A) 11.1 ms
B) 5.62 Ω
Explanation:
L ( inductance ) = 10 mH
Vcc = 14V
<u>A) determine the required on time of the switch such that the peak energy stored in the inductor is 1.2J </u>
first calculate for the current ( i ) using the equation for energy stored in an inductor hence
i = ----- ( 1 )
where : W = 1.2j , L = 10 mH
Input values into equation 1
i = 15.49 A
Now determine the time required with expression below
i( t ) = 15.49 A
L = 10 mH, Vcc = 14
hence the time required ( T-on ) = 11.1 ms
attached below is detailed solution
B) <u>select the value of R such that switching cycle can be repeated every 20 ms </u>
using the expression below
τ = ---- ( 2 )
but first we will determine the value of τ
τ = t-off / 5 time constants
= (20 - 11.1 ) / 5 = 1.78 ms
Back to equation 2
R = L / τ
= (10 * 10^-3) / (1.78 * 10^-3)
= 5.62 Ω
Answer:
48.61
Explanation:
See attached diagram.
The level rise in the tube is l sin α.
The level drop in the cylinder (let's call it y) is:
π/4 D² y = π/4 d² l
D² y = d² l
y = l (d/D)²
The elevation difference is the sum:
l sin α + l (d/D)²
l (sin α + (d/D)²)
From Bernoulli's principle:
P = ρgl (sin α + (d/D)²)
Divide both sides by density of water (ρw) and gravity:
P/(ρw g) = (ρ/ρw) l (sin α + (d/D)²)
h = S l (sin α + (d/D)²)
If we disregard the level change in the cylinder:
h = S l (sin α)
We want the percent error between these two expressions for h to be 0.1% when α = 25°.
[ S l (sin α + (d/D)²) − S l (sin α) ] / [ S l (sin α + (d/D)²) ] = 0.001
[ S l sin α + S l (d/D)² − S l sin α ] / [ S l (sin α + (d/D)²) ] = 0.001
[ S l (d/D)² ] / [ S l (sin α + (d/D)²)] = 0.001
(d/D)² / (sin α + (d/D)²) = 0.001
(d/D)² = 0.001 (sin α + (d/D)²)
(d/D)² = 0.001 sin α + 0.001 (d/D)²
0.999 (d/D)² = 0.001 sin α
d/D = √(0.001 sin α / 0.999)
When α = 25°:
d/D ≈ 0.02057
D/d ≈ 48.61
Answer:
367.43 mm²
Explanation:
Given:
Flow rate, Q = 0.7 L/s
1000 L = 1 m³ = 10⁹ mm³
thus,
1 L = 10⁶ mm³
Therefore,
Q = 0.7 × 10⁶ mm³/s
Cross-sectional area at the top of the sprue = 750 mm²
Length of the sprue = 185 mm
Now,
Velocity =
where, g is the acceleration due to gravity = 9.81 m/s²
h is the height through which flow is taking place = 185 mm = 0.185
thus,
Velocity =
or
velocity = 1.9051 m/s = 1905.1 mm/s
Also,
Q = Area × Velocity
thus,
0.7 × 10⁶ = Area × 1905.1
or
Area = 367.43 mm²