The minute hand on a watch is 2.00 cm cm in length. What is the displacement vector of the tip of the minute hand Part A From

The minute hand on a watch is 2.00 cm cm in length. What is the displacement vector of the tip of the minute hand Part A From 8:00 to 8:20 a. m.? Express vector Δ r ⃗ Δr→ in the form Δ r x ΔrxDeltar_x , Δ r y ΔryDeltar_y , where the x and y components are separated by a comma. Δ r ⃗ Δr→ = nothing cm SubmitRequest Answer Part B From 8:00 to 9:00 a. m.? Express vector Δ r ⃗ Δr→ in the form Δ r x ΔrxDeltar_x , Δ r y ΔryDeltar_y , where the x and y components are separated by a comma.

Related Posts

This Post Has 7 Comments

  1. [tex]132|2 \\ 66 \ | 2 \\ 33 \ |3 \\ 11 \ |11 \\ 1 \\\\ 132=2*2*3*11 \\\\ 132=2p*q*r \\\\ 2p=2*2 \to\boxed{p=2} \\\\ \boxed{q=3} \ \ \ \ \ \ \ \ \ \ \ \ \boxed{r=11}[/tex]

  2. [tex]132|2 \\ 66 \ | 2 \\ 33 \ |3 \\ 11 \ |11 \\ 1 \\\\ 132=2*2*3*11 \\\\ 132=2p*q*r \\\\ 2p=2*2 \to\boxed{p=2} \\\\ \boxed{q=3} \ \ \ \ \ \ \ \ \ \ \ \ \boxed{r=11}[/tex]

  3. Check the explanation

    Explanation:

    Kindly check the attached image below to see the step by step explanation to the question above.

    [tex]The minute hand on a watch is 2.00 cm cm in length. What is the displacement vector of the tip of th[/tex]

  4. Your first job is to create a Java program that repeatedly asks the user whether they wish to calculate another square root. If the response is "y", then the program should proceed; if it is anything else, then the program should quit. Whenever it proceeds, the program should prompt the user for a number (a positive double, and your program may simply assume the input is consistent with this requirement) and then report the square root of that number to within a relative error of no more than 0.01%. The computation must be done using Newton iteration.

    The intuitive idea of Newton iteration for computing square roots is fairly straightforward. Suppose you have a guess r for x1/2 that is too large; the argument is similar if it is too small. If r is too large to be the square root of x, then x/r must be too small, so the average of r and x/r should be a better guess than either r or x/r. This suggests that if you repeatedly replace your current guess r by (r + x/r)/2, then your sequence of guesses should converge to x1/2. And indeed it can be proved that it does. A good initial guess for x1/2 is simply r = x. If you continue updatingr until |r2 – x |/x < ε2, then the relative error of the guess r will be less than ε.

    After your initial program works, there are a number of other requirements to change it slightly, one step at a time, as explained below.

    Open the src folder of this project and then open (default package). As a starting point you should .java. Rename it Newton1 and delete the other files from the project (if needed, see Creating a Program from a Skeleton (also Renaming a Java Program) for details).

    Edit Newton1.java to satisfy the problem requirements stated above, including updating comments appropriately. Estimating the square root should be done in a static method declared as follows:

    1

    2

    3

    4

    5

    6

    7

    8

    9

    10

    /**

    * Computes estimate of square root of x to within relative error 0.01%.

    *

    * @param x

    *            positive number to compute square root of

    * @return estimate of square root

    */

    private static double sqrt(double x) {

       ...

    }

    Copy Newton1.java to create Newton2.java. Change sqrt (including its Javadoc comments) so it also works when x = 0.

    Copy Newton2.java to create Newton3.java. Change it so the main program prompts the user to input the value of ε (rather than assuming it is 0.0001), just once as the program begins, and so this value is also passed to sqrt.

    Copy Newton3.java to create Newton4.java. Change it so the main program does not ask the user whether they wish to calculate another square root, but instead simply asks for a new value of x and interprets a negative value as an indication that it's time to quit.

    Select your Eclipse project Newton (not just some of the files, but the whole project), create a zip archive of it, and submit the zip archive to the Carmen dropbox for this project, as described in Submitting a Project.

  5. Hey there!

    This is showing the associative property because it shows that no matter the order that something is multiplied in, it will remain the same.

    I hope this helps!

  6. IT IS ASYMMETRIC.

    Step-by-step explanation:

    Here we will consider all options to determine which of our options holds true.

    Given set A;

    А - {0, 1, 2}

    R {(0,0), (0, 1), (0, 2), (1,2))

    Let us not fail to keep the conditions given in the question at heart;

    To Check for intransitive;

    This condition holds true for all x, y, z belonging to A has -

    (1) (x, y)and (y, z) but not (x, z) in R5OR

    (2) don't have (x, y) in R5 OR

    (3) (x, y)but don't have (y, z) in R5,

    for 0, 1, 2 we have (0, 1) and (1, 2) but also have (0, 2) in R5 .

    Condition failed here so, no need to check ahead.

    ⇒ So, It is NOT Intransitive.

    To Check for irreflexivity;

    This condition holds true for all x belonging to A -

    (1) don't have (x, x) in R5,

    for 0 we have (0, 0) in R5 so the condition is failed right here So, no need to check further for irreflexivity.

    ⇒ So, It is NOT Irreflexive.

    Now let's Check for asymmetric,

    This condition holds true for all x, y belonging to A has -

    (1) (x, y) in R5 but not (y, x) OR

    (2) don't have (x, y) in R5,

    for 0, 1 we have (0, 1) in R5 but do not have (1, 0),

    for 1, 0 we don't have (1, 0) in R5

    for 1, 2 we have (1, 2) but not (2, 1) in R5

    for 2, 1 we don't have (2, 1) in R5

    for 0, 2 we have (0, 2) in R5 but not (2, 0)

    finally for 2, 0 we don't have (2, 0) in R5

    So, Condition satisfied for every г.уеА

    ⇒ IT IS ASYMMETRIC.

    cheers i hope this helps

Leave a Reply

Your email address will not be published. Required fields are marked *