Answer:
165 mm
Explanation:
The mass on the piston will apply a pressure on the oil. This is:
p = f / A
The force is the weight of the mass
f = m * a
Where a in the acceleration of gravity
A is the area of the piston
A = π/4 * D1^2
Then:
p = m * a / (π/4 * D1^2)
The height the oil will raise is the heignt of a colum that would create that same pressure at its base:
p = f / A
The weight of the column is:
f = m * a
The mass of the column is its volume multiplied by its specific gravity
m = V * S
The volume is the base are by the height
V = A * h
Then:
p = A * h * S * a / A
We cancel the areas:
p = h * S * a
Now we equate the pressures form the piston and the pil column:
m * a / (π/4 * D1^2) = h * S * a
We simplify the acceleration of gravity
m / (π/4 * D1^2) = h * S
Rearranging:
h = m / (π/4 * D1^2 * S)
Now, h is the heigth above the interface between the piston and the oil, this is at h1 = 42 mm. The total height is
h2 = h + h1
h2 = h1 + m / (π/4 * D1^2 * S)
h2 = 0.042 + 10 / (π/4 * 0.14^2 * 0.8) = 0.165 m = 165 mm
Answer:
a) 199.04 ohms
b) attached in image
c) -0.696dB
Explanation:
We are given:
Fc = 8Khz = 8000hz
![C = 100nF = 100*10^-^9F](https://tex.z-dn.net/?f=%20C%20%3D%20100nF%20%3D%20100%2A10%5E-%5E9F%20)
a)Using the formula:
![F_c = \frac{1}{2pie*Rc}](https://tex.z-dn.net/?f=%20F_c%20%3D%20%5Cfrac%7B1%7D%7B2pie%2ARc%7D)
![8000= \frac{1}{2*3.14*R*100*10^-^9}](https://tex.z-dn.net/?f=8000%3D%20%5Cfrac%7B1%7D%7B2%2A3.14%2AR%2A100%2A10%5E-%5E9%7D)
![R =\frac{1}{2*3.14*100*10^-^9*8000}](https://tex.z-dn.net/?f=%20R%20%3D%5Cfrac%7B1%7D%7B2%2A3.14%2A100%2A10%5E-%5E9%2A8000%7D)
R = 199.04 ohms
b) diagram is attached
c) ![H(w) = \frac{V_out(w)}{Vin(w)} = \frac{1}{1-j\frac{wc}{w}}](https://tex.z-dn.net/?f=H%28w%29%20%3D%20%5Cfrac%7BV_out%28w%29%7D%7BVin%28w%29%7D%20%3D%20%5Cfrac%7B1%7D%7B1-j%5Cfrac%7Bwc%7D%7Bw%7D%7D)
![H(F) = \frac{1}{1-j\frac{fc}{f}}](https://tex.z-dn.net/?f=H%28F%29%20%3D%20%5Cfrac%7B1%7D%7B1-j%5Cfrac%7Bfc%7D%7Bf%7D%7D)
At F = 20KHz and Fc= 8KHz we have:
![H(F)= \frac{1}{1-j\frac{8}{20}} = \frac{1}{1-j(0.4)}](https://tex.z-dn.net/?f=%20H%28F%29%3D%20%5Cfrac%7B1%7D%7B1-j%5Cfrac%7B8%7D%7B20%7D%7D%20%3D%20%5Cfrac%7B1%7D%7B1-j%280.4%29%7D)
![|H(F)|= \frac{1}{\sqrt{1^2+(0.4)^2}}](https://tex.z-dn.net/?f=%20%7CH%28F%29%7C%3D%20%5Cfrac%7B1%7D%7B%5Csqrt%7B1%5E2%2B%280.4%29%5E2%7D%7D)
=0.923
|H(F)| in dB = 20log |H(F)|
=20log0.923
= -0.696dB
<em>In Star connection, the line voltage is equal to root three times of the phase voltage, whereas in delta connection line voltage is equal to the phase voltage. ... In star connection, phase voltage is low as 1/√3 times the line voltage, whereas in delta connection phase voltage is equal to the line voltage.</em>
The load is 17156 N.
<u>Explanation:</u>
First compute the flexural strength from:
σ = FL / π![R^{3}](https://tex.z-dn.net/?f=R%5E%7B3%7D)
= 3000
(40
10^-3) / π (5
10^-3)^3
σ = 305
10^6 N / m^2.
We can now determine the load using:
F = 2σd^3 / 3L
= 2(305
10^6) (15
10^-3)^3 / 3(40
10^-3)
F = 17156 N.
That due to the specific tasks that needs to be accomplished by each program to make an all encompassing program would be inefficient and full of bugs