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dybincka [34]
3 years ago
9

Wn sóiEoEfind the transfer function​

Engineering
1 answer:
zaharov [31]3 years ago
8 0
Kamwiisyveu ajoae iosiwnjije j’dis ke si
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In a steady flow device, the properties of the system remains constant with time. a)True b) False
Leviafan [203]

Answer:

True

Explanation:

By definition of steady flow we have

\frac{\partial f(x,y,z,t) }{\partial t}=0

where f(x,y,z,t) is any property of the system under consideration

=> f(x,y,z,t) = constant

7 0
3 years ago
A constant-head permeability test gives the following information: - Water flows horizontally through the soil sample. - The hei
Marysya12 [62]

Answer:

Complete answer to the question is explained in the attached files.please have a look on it.

Explanation:

5 0
3 years ago
State the number of terms for each of the following algebraic expression 2x+1
harina [27]

Answer:

Expressions are made up of terms.

A term is a product of factors.

Coefficient is the numerical factor in the term

Before moving to terms like monomials, binomials, and polynomials, like and unlike terms are discussed.

When terms have the same algebraic factors, they are like terms.

When terms have different algebraic factors, they are unlike terms.

Explanation:

Hi please follow me also if you can and thanks.

6 0
3 years ago
2. The moist weight of 0.1 ft3 of soil is 12.2 lb. If the moisture content is 12% and the specific gravity of soil solids is 2.7
adell [148]

The answers to dry unit weight, void ratio, porosity, degree of saturation, volume occupied by water are respectively;

γ_d = 108.93 lb/ft³; e = 0.56; n = 0.36; S = 0.58; V_w = 0.021 ft³

<h3>Calculation of Volume and Weight of soil</h3>

We are given;

Moist weight; W = 12.2 lb

Volume of moist soil; V = 0.1 ft³

moisture content; w = 12% = 0.12

Specific gravity of soil solids; G_s = 2.72

A) Formula for dry unit weight is;

γ_d = γ/(1 + w)

where γ_w is moist unit weight as;

γ_w = W/V

γ_w = 122/0.1 = 122 lb/ft³

Thus;

γ_d = 122/(1 + 0.12)

γ_d = 108.93 lb/ft³

B) Formula for void ratio is;

e = [(G_s * γ_w)/γ_d] - 1

e = [(2.72 * 122)/108.93] - 1

e = 0.56

C) Formula for porosity is;

n = e/(1 + e)

n = 0.56/(1 + 0.56)

n = 0.36

D) Formula for degree of saturation is;

S = (w * G_s)/e

S = (0.12 * 2.72)/0.56

S = 0.58

E) Volume occupied by water is gotten from;

V_w = S*V_v

where;

V_v is volume of voids = nV

V_v = 0.36*0.1

V_v = 0.036 ft³

Thus;

V_w = 0.58 * 0.036

V_w = 0.021 ft³

Read more about Specific Gravity of Soil at; brainly.com/question/14932758

4 0
3 years ago
When a variable is stored in memory, it is associated with an address. To obtain the address of a variable, the &amp; operator c
liubo4ka [24]

Answer:

Explanation:

1) C program file addressOfScalar.c

#include <stdio.h>

int main()

{

//intialize a char variable, print its address and the next address

char charvar = 'a';

printf("address of charvar = %p\n", (void *)(&charvar));

printf("address of charvar - 1 = %p\n", (void *)(&charvar - 1));

printf("address of charvar + 1 = %p\n", (void *)(&charvar + 1));

//intialize a int variable, print its address and the next address

int intvar = 1;

printf("address of intvar = %p\n", (void *)(&intvar));

printf("address of intvar - 1 = %p\n", (void *)(&intvar - 1));

printf("address of intvar + 1 = %p\n", (void *)(&intvar + 1));

}

In C programming language, an int variable takes 4 bytes of memory. So any arithmetic on integer address, always considers it as 4 bytes of data. So intvar-1 refers to a location 4 bytes before intvar's address and intvar+1 refers to 4 bytes after intvar's address.

3 0
3 years ago
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