Be a qt today lolololololol

Answers

Answer 1
Answer:

Answer:

you too

Explanation:

thank you for the free points


Related Questions

The compressibility factor provides a quick way to assess when the ideal gas law is valid. Use a solver to find the minimum temperature where the fluid has a vapor phase compressibility factor greater than 0.95 at 3 MPa. Report the value in oC, without units.
Logical variables: Running late? Complete the tunction RunningLate such that the logical variable on Time is true if no Traffic is true and gasEmpty is false. Ex >> n°Traffic = true ; >> gasEmpty-false; >onTime - Runninglate(noTraffic,gasEmpty) logical Your Function B Save C Reset EE MATLAB Documentation 1 function onTime - RunningLate (noTraffic,gasEmpty) 2 % complete the logical expression to the right of . using the variables n°Traffic and gasEmpty 4 onTime - 6 end Code to call your function C Reset 1 noTraffic true; gasEmpty true; 2 onTime-Runninglate(noTraffic, gasEmpty)
A tensile test is carried out on a bar of mild steel of diameter 20 mm. The bar yields under a load of 80 kN. It reaches a maximum load of 150 kN, and breaks finally at a load of 70 kN. Find (i) the tensile stress at the yield point (1i) the ultimate tensile stress; (iii) the average stress at the breaking point, if the diameter of the fractured neck is 10mm
The process in which the system pressure remain constant is called a)-isobaric b)-isochoric c)-isolated d)-isothermal
A 120-volt fluorescent ballast has an input current of 0.34 ampere and an input power rating of 22 watts. The power factor of the ballast is ____.

1. An automobile travels along a straight road at 15.65 m/s through a 11.18 m/sspeed zone. A police car observed the automobile. At the instant that the two
vehicles are abreast of each other, the police car starts to pursue the automobile at
a constant acceleration of 1.96 m/s². The motorist noticed the police car in his rear
view mirror 12 s after the police car started the pursuit and applied his brakes and decelerates at 3.05 m/s². (Hint: The police will not go against the law.)
a) Find the total time required for the police car to overtake the automobile.
b) Find the total distance travelled by the police car while overtaking the
automobile.
c) Find the speed of the police car at the time it overtakes the automobile.
d) Find the speed of the automobile at the time it was overtaken by the police car.

Answers

Answer:

a.) Time = 17.13 seconds

b.) 31.88 m

c.) V = 11.18 m/s

d.) V = 7.1 m/s

Explanation:

The initial velocity U of the automobile is 15.65 m/s.

 At the instant that the two vehicles are abreast of each other, the police car starts to pursue the automobile with initial velocity U = 0 at a constant acceleration of 1.96 m/s². Because the police is starting from rest.

For the automobile, let us use first equation of motion

V = U - at.

Acceleration a is negative since it is decelerating with a = 3.05 m/s² . And

V = 0.

Substitute U and a into the formula

0 = 15.65 - 3.05t

15.65 = 3.05t

t = 15.65/3.05

t = 5.13 seconds

But the motorist noticed the police car in his rear view mirror 12 s after the police car started the pursuit and applied his brakes and decelerates at 3.05 m/s².

The total time required for the police car to overtake the automobile will be

12 + 5.13 = 17.13 seconds.

b.) Using the third equation of motion formula for the police car at V = 11.18 m/s and a = 1.96 m/s²

V^2 = U^2 + 2aS

Where S = distance travelled.

Substitute V and a into the formula

11.18^2 = 0 + 2 × 1.96 ×S

124.99 = 3.92S

S = 124.99/3.92

S = 31.88 m

c.) The speed of the police car at the time it overtakes the automobile will be in line with the speed zone which is 11.18 m/s

d.) That will be the final velocity V of the automobile car.

We will use third equation of motion to solve that.

V^2 = U^2 + 2as

V^2 = 15.65^2 - 2 × 3.05 × 31.88

V^2 = 244.9225 - 194.468

V = sqrt( 50.4545)

V = 7.1 m/s

A horizontal curve on a single-lane highway has its PC at station 1+346.200 and its PI at station 1+568.70. The curve has a superelevation of 6.0% and is designed for 120 km/h. The limiting value for coefficient of side friction at 120 km/h is 0.09. What is the station of the PT? Remember that 1 metric station = 1000 m.

Answers

Answer:

The solution is given in the attachments.

Technician A says that the unitized structure of a hybrid vehicle is considerably different when compared to the same conventional model.Technician B says that hybrid vehicles have 12-volt and high voltage batteries.

Who is right?

Answers

Technician A says that the unitized structure of a hybrid vehicle is considerably different when compared to the same conventional model is right.

What are hybrid vehicle?

Hybrid vehicle are defined as a powered by a combustion engine and/or a number of electric motors that draw power from batteries. A gas-powered car simply has a traditional gas engine, but a hybrid car also features an electric motor.

One important advantage of hybrid cars is their capacity to reduce the size of the main engine, which improves fuel efficiency. Many hybrid vehicles employ electric motors to accelerate slowly at first until they reach higher speeds. They then use gasoline-powered engines to increase fuel efficiency.

Thus, technician A says that the unitized structure of a hybrid vehicle is considerably different when compared to the same conventional model is right.

To learn more about hybrid vehicle, refer to the link below:

brainly.com/question/14610495

#SPJ5

A train was right get it

//This method uses the newly added parameter Club object //to create a CheckBox and add it to a pane created in the constructor //Such check box needs to be linked to its handler class

Answers

Comments on Question

Your questions is incomplete.

I'll provide a general answer to create a checkbox programmatically

Answer:

// Comments are used for explanatory purpose

// Function to create a checkbox programmatically

public class CreateCheckBox extends Application {

// launch the application

public void ClubObject(Stage chk)

{

// Title

chk.setTitle("CheckBox Title");

// Set Tile pane

TilePane tp = new TilePane();

// Add Labels

Label lbl = new Label("Creating a Checkbox Object");

// Create an array to hold 2 checkbox labels

String chklbl[] = { "Checkbox 1", "Checkbox 2" };

// Add checkbox labels

tp.getChildren().add(lbl);

// Add checkbox items

for (int i = 0; i < chklbl.length; i++) {

// Create checkbox object

CheckBox cbk = new CheckBox(chklbl[i]);

// add label

tp.getChildren().add(cbk);

// set IndeterMinate to true

cbk.setIndeterminate(true);

}

// create a scene to accommodate the title pane

Scene sc = new Scene(tp, 150, 200);

// set the scene

chk.setScene(sc);

chk.show();

}

How much heat is lost through a 3’× 5' single-pane window with a storm that is exposed to a 60°F temperature differential?A. 450 Btu/hB. 900 Btu/hC. 1350 Btu/hD. 1800 Btu/h

Answers

Answer:

A. 450 btu/h

Explanation:

We solve this problem by using this formula:

Q = U x TD x area

U = U value of used material

TD = Temperature difference = 60°

Q = heat loss

Area = 3x5 = 15

We first find U

R = 1/u

2 = 1/U

U = 1/2 = 0.5

Then when we put these values into the formula above, we would have:

Q = 0.5 x 15 x 60

Q = 450Btu/h

Therefore 450btu/h is the answer

The substance called olivine may have any composition between Mg2SiO4 and Fe2SiO4, i.e. the Mg atoms can be replaced by Fe atoms in any proportion without altering the crystal structure except by expanding it slightly: this is an example of a binary solid solution series. For different compositions, the lines in the powder diffraction patterns are in slightly different positions, because of the cell expansion, but the overall pattern remains basically the same. The spacing of the lattice planes varies linearly with composition, and this can be used in a rapid and non- destructive method of analysis. a. The (062) reflection from olivine is strong and well resolved from other lines. Calculate d062 for an olivine that displays its (062) reflection at a Bragg angle of 37.21° (i.e., a diffraction angle of 74.42°) when x-rays with a wavelength of 0.1790 nm are used. b. The d062 spacing as measured accurately for synthetic materials is 0.14774 nm for Mg2SiO4 and 0.15153 nm for Fe2SiO4. What would be the approximate composition, expressed in mol.% Mg2SiO4, of an olivine material for which do62 has the value obtained in part 2.1 above?

Answers

Answer:

The answer is "0.147 nm and  99.63 mol %"

Explanation:

In point (a):

\to nk1 = 062

\to \text{Bragg angle}\theta =37.21^(\circ)

\to \text{diffraction angle}2 \theta = 74.42^(\circ)

\to \lambda = 0.1790 nm

find:

d(062)=?

formula:

\to nx = 2d \sin  \theta

\to  d(062) = (1 * 0.1790^(\circ))/(2 * \sin 37.21^(\circ))\n

               = (0.1790^(\circ))/(2 * 0.604738126)\n\n= (0.29599589)/(2)\n\n= 0.147 \n

In point (b):

\to Mg_2SiO_4\longleftrightarrow  Fe_2SiO_4

d= 0.14774  \ \ \ \ \ olivine = 0.147 \ \ \ \  \ 0.15153

formula:

\to d=(a)/(√(n^2+k^2+i^2))\n

that's why the composition value equal to 99.63 %