The difference between the electrical gradient is a electrical issue in power
Answer:
The answer is "conditionally unstable"
Explanation:
The conditional volatility is really a condition of uncertainty, which reflects on whether increasing air is polluted or not. It determines the rate of ambient delay, which has been between humid and dry adiabatic rates. In general, the environment is in an unilaterally unhealthy region.
Classification dependent on ELR:
Larger than 10 
m Around 10 and 6 m or less 6 m volatile implicitly unreliable Therefore ELR is implicitly unreliable 9 m, that's why it is "conditionally unstable".
The design speed was used for the freeway exit ramp is 11 mph.
<h3>Design speed used in the exit ramp</h3>
The design speed used in the exit ramp is calculated as follows;
f = v²/15R - 0.01e
where;
v = ωr
v = (θ/t) r
θ = 90⁰ = 1.57 rad
v = (1.57 x 19.4)/2.5 s
v = 12.18 ft/s = 8.3 mph
<h3>Design speed</h3>
f = v²/15R - 0.01e
let the maximum superelevation, e = 1%
f = (8.3)²/(15 x 19.4) - 0.01
f = 0.22
0.22 is less than value of f which is 0.4
<h3>next iteration, try 10 mph</h3>
f = (10)²/(15 x 19.4) - 0.01
f = 0.33
0.33 is less than 0.4
<h3>next iteration, try 11 mph</h3>
f = (11)²/(15 x 19.4) - 0.01
f = 0.4
Thus, the design speed was used for the freeway exit ramp is 11 mph.
Learn more about design speed here: brainly.com/question/22279858
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Answer:
\epsilon = 0.028*0.3 = 0.0084
Explanation:
\frac{P_1}{\rho} + \frac{v_1^2}{2g} +z_1 +h_p - h_l =\frac{P_2}{\rho} + \frac{v_2^2}{2g} +z_2
where P_1 = P_2 = 0
V1 AND V2 =0
Z1 =0
h_P = \frac{w_p}{\rho Q}
=\frac{40}{9.8*10^3*0.2} = 20.4 m
20.4 - (f [\frac{l}{d}] +kl) \frac{v_1^2}{2g} = 10
we know thaTV =\frac{Q}{A}
V = \frac{0.2}{\pi \frac{0.3^2}{4}} =2.82 m/sec
20.4 - (f \frac{60}{0.3} +14.5) \frac{2.82^2}{2*9.81} = 10
f = 0.0560
Re =\frac{\rho v D}{\mu}
Re =\frac{10^2*2.82*0.3}{1.12*10^{-3}} =7.53*10^5
fro Re = 7.53*10^5 and f = 0.0560
\frac{\epsilon}{D] = 0.028
\epsilon = 0.028*0.3 = 0.0084
