Answer:
the surface heat-transfer coefficient due to natural convection during the initial cooling period. = 4.93 w/m²k
Explanation:
check attachement for answer explanation
Answer:
y ≈ 2.5
Explanation:
Given data:
bottom width is 3 m
side slope is 1:2
discharge is 10 m^3/s
slope is 0.004
manning roughness coefficient is 0.015
manning equation is written as

where R is hydraulic radius
S = bed slope



P is perimeter 

![Q = (2+2y) y) \times 1/0.015 [\frac{(3+2y) y}{(3+2\sqrt{5} y)}]^{2/3} 0.004^{1/2}](https://tex.z-dn.net/?f=Q%20%3D%20%282%2B2y%29%20y%29%20%5Ctimes%201%2F0.015%20%5B%5Cfrac%7B%283%2B2y%29%20y%7D%7B%283%2B2%5Csqrt%7B5%7D%20y%29%7D%5D%5E%7B2%2F3%7D%200.004%5E%7B1%2F2%7D)
solving for y![100 =(2+2y) y) \times (1/0.015) [\frac{(3+2y) y}{(3+2\sqrt{5} y)}]^{2/3} \times 0.004^{1/2}](https://tex.z-dn.net/?f=100%20%3D%282%2B2y%29%20y%29%20%5Ctimes%20%281%2F0.015%29%20%5B%5Cfrac%7B%283%2B2y%29%20y%7D%7B%283%2B2%5Csqrt%7B5%7D%20y%29%7D%5D%5E%7B2%2F3%7D%20%5Ctimes%200.004%5E%7B1%2F2%7D)
solving for y value by using iteration method ,we get
y ≈ 2.5
Answer: 24 pA
Explanation:
As pure silicon is a semiconductor, the resistivity value is strongly dependent of temperature, as the main responsible for conductivity, the number of charge carriers (both electrons and holes) does.
Based on these considerations, we found that at room temperature, pure silicon resistivity can be approximated as 2.1. 10⁵ Ω cm.
The resistance R of a given resistor, is expressed by the following formula:
R = ρ L / A
Replacing by the values for resistivity, L and A, we have
R = 2.1. 10⁵ Ω cm. (10⁴ μm/cm). 50 μm/ 0.5 μm2
R = 2.1. 10¹¹ Ω
Assuming that we can apply Ohm´s Law, the current that would pass through this resistor for an applied voltage of 5 V, is as follows:
I = V/R = 5 V / 2.1.10¹¹ Ω = 2.38. 10⁻¹¹ A= 24 pA
Answer:
a) 23.89 < -25.84 Ω
b) 31.38 < 25.84 A
c) 0.9323 leading
Explanation:
A) Calculate the load Impedance
current on load side = 0.75 p.u
power factor angle = 25.84
= 0.75 < 25.84°
attached below is the remaining part of the solution
<u>B) Find the input current on the primary side in real units </u>
load current in primary = 31.38 < 25.84 A
<u>C) find the input power factor </u>
power factor = 0.9323 leading
<em></em>
<em>attached below is the detailed solution </em>