There are some sections of the SDS that are not mandatory.

Which of the following is not a reason to give yourself extra "cushion" when driving?

damaskus [11]

Answer:

The question is incomplete, the complete question is:

Which of the following is not a reason to give yourself extra "cushion" when driving?

A. Poor visibility B. Poor road conditions C. Inclement weather D. None of these.

The correct answer is D. None of these.

Explanation:

All the options are not reasons to give yourself an extra cushion when driving, rather they are reasons that are not favorable to driving at all.

A cushion is a certain amount of distance you are supposed to keep between you and the car in front of you to allow easy maneuvering in any condition.

A typical cushion is 3 seconds between you and the car in front of you, in less than perfect conditions like bad weather or poor road conditions an additional second must be added to it.

7 0

3 years ago

What are the steps to execute an instruction by cpu?What is the function of DMA controller.

JulijaS [17]

Explanation:

1. A sequence of instructions is stored in memory.

2. The memory address wherever the first instruction is found is copied to the instruction pointer.

3. The CPU sends the address within the instruction pointer to memory on the address bus.

4. The CPU sends a “read” signal to the control bus.

5. Memory responds by sending a copy of the state of the bits at that memory location on the

data bus, that the CPU then copies into its instruction register.

6. The instruction pointer is automatically incremented to contain the address of the next

instruction in memory.

7. The CPU executes the instruction within the instruction register.

8. Go to step 3

Steps 3, 4, and 5 are called an instruction fetch. Notice that steps 3 – 8 constitute a cycle, the instruction execution cycle. It is shown graphically below.

A DMA controller can generate memory addresses and initiate memory read or write cycles. It contains several hardware registers that can be written and read by the CPU. These include a memory address register, a byte count register, and one or more control registers.

4 0

3 years ago

1 // Lab 2 tryIt2A 2 #include 3 using namespace std; 4 5 int main() 6 { int x = 1, y = 3; 7 int X = 2, Y = 4; 8 9 cout <<

padilas [110]

Answer:

Here is the complete program:

#include <iostream>

using namespace std;

int main()

{ int x = 1, y = 3;

int X = 2, Y = 4;

cout << "tryIt 2A" <<endl;

cout << x << y << endl;

cout << "x" << "y" << endl;

cout << X << " " << Y << endl;

cout << 2 * x + y << endl;

cout << 2 * X + Y << endl;

//cout << x + 2*y << endl;

cout << "x = ";

cout << x;

cout << " y = ";

cout << y;

return 0;

}

Explanation:

I will explain the code line by line in the comment with each line of code and the output of each cout statement.

int x = 1, y = 3;

This statement assigns value 1 to integer variable x and 3 to int variable y

int X = 2, Y = 4;

This statement assigns value 2 to integer variable X and 4 to int variable Y As C++ is a case sensitive language so variable x and y are different from variables X and Y.

cout << "tryIt 2A" <<endl;

This statement has cout which is used to display output on the screen. So the output displayed by this cout statement is:

tryIt2A

cout << x << y << endl;

This statement will print the values stored in x and y variables. So output displayed by cout statement here is 1 and 3. As there is not space or next line specified in the statement so output displayed will look like this:

13

cout << "x" << "y" << endl;

This statement will display x and y but these are not the variable x and y. They are enclosed in double quotation marks so they are treated as strings not variables so the output displayed is:

xy

cout << X << " " << Y << endl;

This statement will print the values stored in X and Y variables. So output displayed by cout statement here is 2 and 4. As there is space " " specified in the statement so 2 and 4 are displayed with a space between them so the output displayed will look like this:

2 4

cout << 2 * x + y << endl;

This statement has an arithmetic operation in which 2 is multiplied by the values stored in variable x and then the result is added by value of y. So 2*1 = 2 and 2 + 3 = 5. So the result produced by this cout statement is:

5

cout << 2 * X + Y << endl;

This will work same as above cout statement but the only difference is that the values of capital X and Y variables are calculated here. So 2 * 2 = 4 and then 4 + 4 = 8. The result produced by this cout statement is:

8

//cout << x + 2*y << endl;

This is a comment because before this statement // is written which is used for single line comment. So compiler ignores comments and will not compile this statement.

cout << "x = ";

This will display "x = " as it is not variable but it is treated as a line to be displayed on the screen. So cout statement displays:

x =

cout << x;

This will print the value stored in x variable as there are no double quotes around x so it is a variable which contains value 1. In the above statement there is no endl so the output of this cout statement is displayed with the output of previous cout statement. So the following line is displayed on screen:

x = 1

cout << " y = ";

This will display "y = " as it is not variable but it is treated as a line to be displayed on the screen. In the above statement there is no endl so the output of this cout statement is displayed with the output of previous cout statement. So the following line is displayed on screen

x = 1 y =

cout << y;

This will print the value stored in y variable as there are no double quotes around y so it is a variable which contains value 3. In the above statement there is no endl so the output of this cout statement is displayed with the output of previous cout statement. So the following line is displayed on screen:

x = 1 y = 3

So the output of the entire program along with the program is attached as screenshot.

6 0

4 years ago

an object of mass 2kg is released from a top of inclined plane 30° and height 6m. The coefficient of kinetic friction of the sur

mel-nik [20]

Explanation:

1) Work done = force x distance x cos(θ)

= 0.15 x 6 x cos(30)

= 0.779

2) Ek = ½mv²

v = acceleration due to gravity so 9.81

Ek = ½(2)(9.81)²

Ek = 96.2361

3) v = (√(2em)) / m

= (√(2(96.2361)(2)) / 2

= 9.81 so especially with no time given, I can only assume the acceleration due to gravity but take it with a pinch of salt.

5 0

2 years ago

Steam enters an adiabatic turbine at 800 psia and9008F and leaves at a pressure of 40 psia. Determine themaximum amount of work

Naily [24]

Answer:

$w_{out}=319.1\frac{BTU}{lbm}$

Explanation:

Hello,

In this case, for the inlet stream, from the steam table, the specific enthalpy and entropy are:

$h_1=1456.0\frac{BTU}{lbm} \ \ \ s_1=1.6413\frac{BTU}{lbm*R}$

Next, for the liquid-vapor mixture at the outlet stream we need to compute its quality by taking into account that since the turbine is adiabatic, the entropy remains the same:

$s_2=s_1$

Thus, the liquid and liquid-vapor entropies are included to compute the quality:

$x_2=\frac{s_2-s_f}{s_{fg}}=\frac{1.6313-0.39213}{1.28448}=0.965$

Next, we compute the outlet enthalpy by considering the liquid and liquid-vapor enthalpies:

$h_2=h_f+x_2h_f_g=236.14+0.965*933.69=1136.9\frac{BTU}{lbm}$

Then, by using the first law of thermodynamics, the maximum specific work is computed via:

$h_1=w_{out}+h_2\\\\w_{out}=h_1-h_2=1456.0\frac{BTU}{lbm}-1136.9\frac{BTU}{lbm}\\\\w_{out}=319.1\frac{BTU}{lbm}$

Best regards.

3 0

3 years ago