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sineoko [7]
3 years ago
14

Question 1: Final Results = What are the values of the resistances such that the gain = -100, Rin = 1 MI2. Don't use resistances

greater than 1 M2.
is this question wrong??? me and mh freinds are trying to figure oyt the answer for 3 days but we can't ​
Engineering
1 answer:
lidiya [134]3 years ago
4 0

Answer:

Explanation:

In a study of algebra, you will encounter many families of equations, or groups of

equations that share common characteristics. Of interest to us here is the family of

linear equations in one variable, a study that lays the foundation for understanding

more advanced families. In addition to solving linear equations, we’ll use the skills we

develop to solve for a specified variable in a formula, a practice widely used in science,

business, industry, and research.

A. Solving Linear Equations Using Properties of Equality

An equation is a statement that two expressions are

equal. From the expressions and

we can form the equation

which is a linear equation in one variable. To solve

an equation, we attempt to find a specific input or xvalue that will make the equation true, meaning the

left-hand expression will be equal to the right. Using

Table 1.1, we find that is a

true equation when x is replaced by 2, and is a false

equation otherwise. Replacement values that make

the equation true are called solutions or roots of the equation.

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pychu [463]

Answer:

c

Explanation:

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7 0
3 years ago
What is the weight of a glider with a mass of 5.3 grams? (Hint: watch your units!)
Vinvika [58]

Answer: 0.053

Explanation:

So, we convert 5.3 grams into kilograms.

5.3g = 0.0053 kg (Since 1kg equals 1000g)

On Earth, gravity is 10 N/kg.

Weight = mass x gravity

Weight = 0.0053kg x 10 N/kg

Weight = 0.053 Newtons (On Earth)

6 0
3 years ago
Determine whether or not each of the following signals is periodic.
Sloan [31]

Answer:

a) periodic (N = 1)

b) not periodic

c) not periodic

d) periodic (N = 8)

e) periodic (N = 16)

Explanation:

For function to be a periodic: f(n) = f(n+N)

a) x[n]=sin(\frac{8\pi}{2}n+1)\\\\sin(\frac{8\pi}{2}n+1)=sin(4\pi n+1)

It is periodic with fundamental period N = 1

b) x[n]=cos(\frac{n}{8} -\pi)\\\\\frac{1}{8} N=2\pi k

N must be integer. So it is nor periodic

c) x[n]=cos(\frac{\pi}{8} n^2)\\\\cos(\frac{\pi}{8} (n+N)^2)=cos(\frac{\pi}{8} (n^2+N^2+2nN)\\\\N^2 = 16 \:\:or\:\:2nN=16

Since N is dependent to n. So it is not periodic.

d) x[n]=cos(\frac{\pi }{2}  n) cos(\frac{\pi }{4}  n)\\\\x[n] = \frac{1}{2} cos(\frac{3\pi }{4} n) + \frac{1}{2} cos(\frac{\pi }{4} n)\\\\N_1=8\:\:and\:\:N_2=8\\

So it is periodic with fundamental period N = 8.

e) x[n]=2cos(\frac{\pi }{4}  n)+sin(\frac{\pi }{8} n)-2cos(\frac{\pi }{2} n+\frac{\pi }{6} )\\\\N_1=8\:\:and\:\:N_2=16\:\:and\:\:N_3=4

So it is periodic with N = 16.

3 0
3 years ago
An ideal Diesel cycle has a maximum cycle temperature of 2000°C. The state of the air at the beginning of the compression is P1
IrinaVladis [17]

Answer:

Power produced = 90.47 KW

Explanation:

We are given;

R = 0.287 kJ/kg·K

T1 = 15°C = 15 + 273 = 288 K

P1 = 95 KPa

Number of cylinders;N_cyl = 8

Bore;B = 10cm = 0.1m

Stroke;S = 12cm = 0.12m

cp = 1.005 kJ/kg·K

cv = 0.718 kJ/kg·K

k = 1.4

First of all let's find the initial specific volume;

α1 = RT1/P1

α1 = 0.287 * 288/95

α1 = 0.87 m³/kg

Now, let's find the total mass of air from the formula;

m =(V•N_cyl)/α1 =(B²•N_cyl•S•π)/4α1

So, m = (B²•N_cyl•S•π)/4α1

m = (0.1²•8•0.12•π)/(4*0.87)

m = 0.00867 Kg

Now, let's calculate the total mass flow rate;

m' = (m*N_rev)/n'

Where;

N_rev is number of revolutions given as 1500 rpm = 1500/60 rev/s = 25 rev/s

n' is the repetitions per circle = 2.

Thus;

m' = (0.00867*25)/2

m' = 0.108375 kg/s

The temperature at state 2 is gotten from the formula;

T2 = T1*r^(k - 1)

Where r is compression ratio.

We know that formula for compression ratio is;

the ratio of the maximum to minimum volume in the cylinder of an internal combustion engine.

In the question, we are told that minimum volume enclosed in the cylinder is 5 percent of the maximum cylinder volume.

Thus,

r = 100/5 = 20

So, T2 = 288*20^(1.4 - 1)

T2 = 954.563 K

The cut off ratio is gotten from the formula;

r_c = α3/α2 = T3/T2

T3 = 2000°C = 2000 + 273K = 2273K

Thus; r_c = 2273/954.563

r_c = 2.38

The heat input is gotten from the formula;

q_in = cp(T3 - T2)

q_in = 1.005(2273 - 954.563)

q_in = 1325.03 KJ/Kg

The efficiency is gotten from;

η = 1 - [1/(r^(k - 1)]*[((r_c)^(k) - 1)/(k(r_c - 1))]

Thus;

η = 1 - [1/(20^(1.4 - 1)]*[((2.38)^(1.4) - 1)/(1.4(2.38 - 1))]

η = 0.63

Now, the power output is gotten from the equation;

W' = m'•η•q_in

W' = 0.108375*0.63*1325.03

W' = 90.47 KW

7 0
3 years ago
What's the function of the rubber, square-cut seal positioned around the caliper piston?
lozanna [386]

Explanation:

The square cut seal is what allows the caliper piston to retract back into its housing. Just because you can manually push the piston in, doesn't mean the square cut seal is working. Rapid brake pad wear, brake drag or brake pull are key indicators that the square cut seal is comprised

5 0
3 years ago
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