One piece of evidence that supports the Theory of Plate Tectonics is the discovery of what in both South America and Africa? The ancient atmosphere in both places was identical. The rates of weathering of rock are similar. Fossil remains of the same land-dwelling animal. Plants on both continents have similar flowers.

Answers

Answer 1
Answer:

Answer: Fossil remains of the same land-dwelling animal.

Explanation: Fossil remains which were found to belong to same land dwelling animals, in South America and Africa was used as evidence to help support the theory of Tectonics plates, what this theory simply means is that the whole continents of earth were once fused together until a tectonic plate caused it’s division. Since same remains were found in Africa and South America this shows that both continents were once fused together.

Answer 2
Answer:

Answer:

Fossil remains of the same land-dwelling animal

Explanation:

Fossil remains tell us where the animals once lived and how by the movement of plate spearated their remaind that was burried thousands of years ago.


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An industrial load with an operating voltage of 480/0° V is connected to the power system. The load absorbs 120 kW with a lagging power factor of 0.77. Determine the size of the capacitor in vars that is necessary to raise the power factor to 0.9 lagging.
The following program includes fictional sets of the top 10 male and female baby names for the current year. Write a program that creates: A set all_names that contains all of the top 10 male and all of the top 10 female names. A set neutral_names that contains only names found in both male_names and female_names. A set specific_names that contains only gender specific names. Sample output for all_names: {'Michael', 'Henry', 'Jayden', 'Bailey', 'Lucas', 'Chuck', 'Aiden', 'Khloe', 'Elizabeth', 'Maria', 'Veronica', 'Meghan', 'John', 'Samuel', 'Britney', 'Charlie', 'Kim'}
1.19. A gas is confined in a 0.47 m diameter cylinder by a piston, on which rests a weight. The mass of the piston and weight together is 150 kg. The local acceleration of gravity is 9.813 m·s−2, and atmospheric pressure is 101.57 kPa. (a) What is the force in newtons exerted on the gas by the atmosphere, the piston, and the weight, assuming no friction between the piston and cylinder? (b) What is the pressure of the gas in kPa? (c) If the gas in the cylinder is heated, it expands, pushing the piston and weight upward. If the piston and weight are raised 0.83 m, what is the work done by the gas in kJ? What is the change in potential energy of the piston and weight?\
Why do we care about a material's ability to resist torsional deformation?(A) Because the angle of twist of a material is often used to predict its shear toughness(B) Because rotating shafts are used in engineering applications(C) We don't care - simply an academic exercise.(D) Because it can determine G and inform us of a materials ability to resist shear deformation
Calculate how large a mass would be necessary to obtain a mechanical noise limit of [Equation] = 1 nG, 1 µG, and 1 mG if the mechanical resonance frequency is [Equation] = 100 Hz. If the mass is to be made of cube of `silicon, what would its physical dimensions be?

Describe how a feeler gauge can be used to assist in the adjustment of a spark plug electrode gap​

Answers

Answer:

Explanation:

Adjusting the distance between the two electrodes is called gapping your spark plugs. You need a feeler gauge to gap your spark plugs properly If you're re-gapping a used plug, make sure that it's clean (gently scrub it with a wire brush)

hope this you

Q1. Basic calculation of the First law (2’) (a) Suppose that 150 kJ of work are used to compress a spring, and that 25 kJ of heat are given off by the spring during this compression. What is the change in internal energy of the spring during the process? (b) Suppose that 100 kJ of work is done by a motor, but it also gives off 10 kJ of heat while carrying out this work. What is the change in internal energy of the motor during the process?

Answers

Answer:

(a) ΔU = 125 kJ

(b) ΔU = -110 kJ

Explanation:

(a) Suppose that 150 kJ of work are used to compress a spring, and that 25 kJ of heat are given off by the spring during this compression. What is the change in internal energy of the spring during the process?

The work is done to the system so w = 150 kJ.

The heat is released by the system so q = -25 kJ.

The change in internal energy (ΔU) is:

ΔU = q + w

ΔU = -25 kJ + 150 kJ = 125 kJ

(b) Suppose that 100 kJ of work is done by a motor, but it also gives off 10 kJ of heat while carrying out this work. What is the change in internal energy of the motor during the process?

The work is done by the system so w = -100 kJ.

The heat is released by the system so q = -10 kJ.

The change in internal energy (ΔU) is:

ΔU = q + w

ΔU = -10 kJ - 100 kJ = -110 kJ

for a rankine cycle with one stage of reheat between turbines, there are how many relevant pressures?

Answers

The four relevant pressures in a Rankine cycle with one stage of reheat are P1, P2, P3, and P4.

For a Rankine cycle with one stage of reheat between turbines, there are typically four relevant pressures:

  1. Boiler pressure (P1): This is the pressure at which the water is heated in the boiler before entering the first turbine.
  2. High-pressure turbine outlet pressure (P2): This is the pressure at the outlet of the first turbine before the steam is sent to the reheater.
  3. Reheat pressure (P3): This is the pressure at which the steam is reheated before entering the second turbine.
  4. Low-pressure turbine outlet pressure (P4): This is the pressure at the outlet of the second turbine, which is also the condenser pressure.

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Write two scnr.nextInt statements to get input values into birthMonth and birthYear. Then write a statement to output the month, a slash, and the year. End with newline. The program will be tested with inputs 1 2000 and then with inputs 5 1950. Ex: If the input is 1 2000, the output is: 1/2000

Answers

Answer:

import java.util.Scanner;

public class InputExample {

   public static void main(String[] args) {

       Scanner scnr = new Scanner(System.in);

       int birthMonth;

       int birthYear;

       birthMonth = scnr.nextInt();

       birthYear = scnr.nextInt();

       System.out.println(birthMonth+"/"+birthYear);

   }

}

import java.util.Scanner;

public class InputExample {

   public static void main(String[] args) {

       Scanner scnr = new Scanner(System.in);

       int birthMonth;

       int birthYear;

       // Get input values for birthMonth and birthYear from the user

       birthMonth = scnr.nextInt();

       birthYear = scnr.nextInt();

       // Output the month, a slash, and the year

       System.out.println(birthMonth + "/" + birthYear);

   }

}

When the program is tested with inputs 1 2000, the output will be:

1/2000

And when tested with inputs 5 1950, the output will be:

5/1950

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For a steel alloy it has been determined that a carburizing heat treatment of 11-h duration will raise the carbon concentration to 0.38 wt% at a point 2.3 mm from the surface. Estimate the time (in h) necessary to achieve the same concentration at a 5.3 mm position for an identical steel and at the same carburizing temperature.

Answers

Answer:

Time =t2=58.4 h

Explanation:

Since temperature is the same hence using condition

x^2/Dt=constant

where t is the time as temperature so D also remains constant

hence

x^2/t=constant

2.3^2/11=5.3^2/t2

time=t^2=58.4 h

Given asphalt content test data: a. Calculate the overall mean and standard deviation for the entire test period.
b. The contract specifications require an average asphalt content of 5.5% +/- 0.5% every day. Plot the daily average asphalt content. Show upper and lower control limits.
c. Do all of these samples meet the contract specifications? Explain your answer.
d. What trend do you observe based on the data? What could cause this trend?"

Answers

Answer:

hello your question is incomplete attached below is the complete question

A) overall mean = 5.535,  standard deviation ≈ 0.3239

B ) upper limit = 5.85, lower limit = 5.0

C) Not all the samples meet the contract specifications

D) fluctuation ( unstable Asphalt content )

Explanation:

B) The daily average asphalt content has to obtained in order to determine the upper and lower control limits using an average asphalt content of 5.5% +/- 0.5% everyday

The upper limit : 14 may = ( 5.8 + 5.1 ) / 2 = 5.85

The lower limit : 16 may = ( 5.2 + 4.8 ) / 2 = 5.0

attached below is the required plot

C ) Not all the samples meet the contract specifications and the samples that do not meet up are samples from :

15 may and 16 may . this is because their Asphalt contents are 6.2 and 4.8 respectively and sample number 18 and 20

D ) what can be observed is that the ASPHALT content fluctuates between the dates while the contract specification is fixed