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Answer:
The current drawn from the outlet is 0.2 A
The number of turns on the input side is 350 turns
Explanation:
Given;
number of turns of the secondary coil, Ns = 35 turns
the output current,
= 2 A
power supplied,
= 24 W
the standard wall outlet in most homes = 120 V = input voltage
For an ideal transformer; output power = input power
the current drawn from the outlet is calculated;

The number of turns on the input side is calculated as;

Answer:

Explanation:
Using the expression shown below as:

Where,
is the number of vacancies
N is the number of defective sites
k is Boltzmann's constant = 
is the activation energy
T is the temperature
Given that:

N = 10 moles
1 mole = 
So,
N = 
Temperature = 425°C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T = (425 + 273.15) K = 698.15 K
T = 698.15 K
Applying the values as:

![ln[\frac {2.3}{6.023}\times 10^{-11}]=-\frac {Q_v}{1.38\times 10^{-23}\times 698.15}](https://tex.z-dn.net/?f=ln%5B%5Cfrac%20%7B2.3%7D%7B6.023%7D%5Ctimes%2010%5E%7B-11%7D%5D%3D-%5Cfrac%20%7BQ_v%7D%7B1.38%5Ctimes%2010%5E%7B-23%7D%5Ctimes%20698.15%7D)

Question:
The question is not complete. See the complete question and the answer below.
A well that pumps at a constant rate of 0.5m3/s fully penetrates a confined aquifer of 34 m thickness. After a long period of pumping, near steady state conditions, the measured drawdowns at two observation wells 50m and 100m from the pumping well are 0.9 and 0.4 m respectively. (a) Calculate the hydraulic conductivity and transmissivity of the aquifer (b) estimate the radius of influence of the pumping well, and (c) calculate the expected drawdown in the pumping well if the radius of the well is 0.4m.
Answer:
T = 0.11029m²/sec
Radius of influence = 93.304m
expected drawdown = 3.9336m
Explanation:
See the attached file for the explanation.
Answer:
If you mean two sides are 7 and two sides are 14 then you'd have 42
and for the second you'd have 14
Explanation:
7 + 7 = 14, 14 + 14 = 28, 14 + 28 = 42
3 + 3 = 6, 4 + 4 = 8, 8 + 6 = 14