Hydraulic fracturing is: Select one: a. a first step in storage of high-level radioactive waste b. a way of extracting natural gas shale formations c. a step in oil refining where oil is separated into different viscosities d. none of the above
Shale and other types of "tight" rock, or impermeable rock formations that lock in oil and gas and complicate the production of fossil fuels, can be extracted of their natural gas or oil using the modern high-volume hydraulic fracturing process. Thus, option B is correct.
What are the factor involving in Hydraulic fracturing?Shale gas is extracted through a procedure called hydraulic fracturing, also referred to as “fracking.” Due to the fact that shale reserves are often dispersed horizontally rather than vertically, deep holes are first drilled into the shale rock, then horizontal drilling is used to reach more of the gas.
A mixture of sand, chemicals, and water is pushed at high pressure down a borehole during hydraulic fracturing. The released gas can flow out of the rocks and back up the borehole because the water pressure causes cracks in the rock to widen, and the sand particles settle into the openings to keep them open.
Therefore, a way of extracting natural gas shale formations.
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an atom that gained an electron is called
Answer:
Hey mate......
Explanation:
This is ur answer.....
When an atom gains/loses an electron, the atom becomes charged, and is called an ion.Hope it helps!
Brainliest pls!
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hỗ trợ mình với được không các bạn
Answer:
Explanation:
Be bop
If OSHA determines that an employer's response to a non-formal complaint is adequate, what options does the employee filing the non-formal complaint have?
Solve using Matlab the problems:
One using the permutation of n objects formula
One using the permutation of r objects out of n objects
You can pick these problems from the textbook or you can make up your own questions.
Help me pleaseeeee
Answer:
Explanation:
% Clears variables and screen
clear; clc
% Asks user for input
n = input('Total number of objects: ');
r = input('Size of subgroup: ');
% Computes and displays permutation according to basic formulas
p = 1;
for i = n - r + 1 : n
p = p*i;
end
str1 = [num2str(p) ' permutations'];
disp(str1)
% Computes and displays combinations according to basic formulas
str2 = [num2str(p/factorial(r)) ' combinations'];
disp(str2)
=================================================================================
Example: check
How many permutations and combinations can be made of the 15 alphabets, taking four at a time?
The answer is:
32760 permutations
1365 combinations
==================================================================================
Use a truth table to verify the first De Morgan law ¬(p ∧ q) ≡ ¬p ∨ ¬q.
Answer:
p q output ¬(p ∧ q)
0 0 1
0 1 1
1 0 1
0 0 0
p q output ¬p ∨ ¬q
0 0 1
0 1 1
1 0 1
0 0 0
Explanation:
We'll create two separate truth tables for both sides of the equation, and see if they match.
The expressions in the question use AND, OR and NOT operators.
The AND operation needs both inputs to be 1 to return a 1.The OR operation needs at least 1 of the inputs to be 1 to return a 1. The NOT operation takes a 1 and turns it into a 0, or takes a 0 and turns it into a 1.Let's start with ¬(p ∧ q)
NOT (0 AND 0) = NOT (0) = 1NOT (0 AND 1) = NOT (0) = 1NOT (1 AND 0) = NOT (0) = 1NOT (1 AND 1) = NOT (1) = 0Now let's move on to the second expression ¬p ∨ ¬q
NOT(0) OR NOT(0) = 1 OR 1 = 1NOT(0) OR NOT(1) = 1 OR 0 = 1NOT(1) OR NOT(0) = 0 OR 1 = 1NOT(0) OR NOT(0) = 0 OR 0 = 0Therefore we can say the two expressions are equivalent.
Attached the truth table to verify the first De Morgan's law ¬(p ∧ q) ≡ ¬p ∨ ¬q:
What is the explanation of the truth table?As you can see from the attached truth table, the truth values for ¬(p ∧ q) and ¬p ∨ ¬q are the same for all combinations of p and q, confirming the validity of the first De Morgan's law.
De Morgan's law is a fundamental principle in propositional logic.
It states that the negation of a conjunction (AND) is equivalent to the disjunction (OR) of the negations of the individual propositions.
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Race cars at the Indianapolis Speedway average speeds of 185 mi/h. After determining the altitude of Indianapolis, find the Mach number of these cars and estimate whether compressibility might affect their aerodynamics.
Answer:
- the Mach number is 0.24.
- Compressibility becomes effective when Mach number is greater than 0.3, the Mach number of the race cars is less than 0.3, hence, compressibility will not affect their aerodynamics.
Explanation:
Given the data in the question;
Average speed V = 185 miles per hour = ( 185 /2.237 ) m/s = 82.7 m/s
From Almanac, we can find that Indianapolis is at 220 m altitude.
So from table, at that altitude, the standard speed of sound will be 339.4 m/s .
Mach number of the race car will be;
Mach Number = Velocity / sound speed
we substitute
Mach Number = ( 82.7 m/s ) / ( 339.4 m/s )
Mach Number = 0.24
Therefore the Mach number is 0.24.
We know that, compressibility becomes effective when the Mach number is greater than 0.3.
Since the Mach number of the race cars is less than 0.3, compressibility will not affect their aerodynamics.
