in figure ex12.19, what magnitude force provides 5.0 nm | studysoup
in figure ex12.19, what magnitude force provides 5.0 nm net torque about the axle?
unifying the universe: reading assignment 1 (thursday, january 26 ): read preface, and chapter 1 to page 18 preface:  my aim in this book is to explore the first arena of science which emerged to reveal that the universe is  indeed fundamentally orderly and governed by natural laws.   before newton, it was generally thought that laws which operate in the heavens [outer space] are quite  distinct from laws valid in our earthly domain.   when different areas of science come together with the weaving of general principles there is a dramatic leap in our understanding of nature.   i aim to cover modern answers to the fascinating questions relating to the origin and unity of the  universe and its contents.  chapter 1: the shapes of n
coulomb's law:f=kq1q2r2, where k is the coulomb's constant and r is the separation between the two charges. the magnitude of the net force:fnet=fx2+fy2k = 9 × 109 n·m2/c2the force due to charge q1 is:f1a=kq1qar1a2=(9×109)(21.4×10-6)(13.5×10-6)(80.1×10-2)2[readmore]f1a = 4.05 nthe force due to charge, q2:f2a=kq2qar2a2=(9×109)(30.6×10-6)(13.5×10-6)[(80.1×10-2)2+(80.1×10-2)2]2f2a = 2.9 nforce due to charge, q3:f3a=kq3qar3a2=(9×109)(57.3×10-6)(13.5×10-6)(80.1×10-2)2f3a = 10.9 nthe net force along the x-axis:fx=f1a-f2acos(45°)=4.05-(2.9)cos(45°)fx = 2.0 nthe net force in the y-direction:fy=f3a-f2acos(45°)=10.9-(2.9)cos(45°)fy = 8.85 nthe magnitude of the net force:fnet=fx2+fy2=2.02+8.852fnet = 9.07 nthe magnitude of the net force on the charge at point a is 9.07 n. that depends on a lot of things. in particular, you would need to find the electric susceptibility of the paper. the field produced by static charge is inverse-square. the polarization of the paper varies directly with the external field, with a proportionality constant equal to the electric susceptibility. the dipole field created by the polarized paper is inverse-cube. overall, the force between the ruler and the paper will vary with the inverse fifth power of distance. to get any more detailed, you would need to know the geometry of the situation, and the charge distribution. The answer to the question is available in the PDF file https://www.assignmentexpert.com/https://www.assignmentexpert.com/homework-answers/physics-answer-3862.pdf Hi Gunleik\_Groven

In this PDF, para 2.4 you can find all the dimensions you are looking for: [https://www.intel.com/content/dam/support/us/en/documents/boardsandkits/NUC7i5BN\_NUC7i7BN\_TechProdSp...](https://www.intel.com/content/dam/support/us/en/documents/boardsandkits/NUC7i5BN_NUC7i7BN_TechProdSpec.pdf) [https://www.intel.com/content/dam/support/us/en/documents/boardsandkits/NUC7i5BN\_NUC7i7BN\_TechProdSp...](https://www.intel.com/content/dam/support/us/en/documents/boardsandkits/NUC7i5BN_NUC7i7BN_TechProdSpec.pdf)

Under this link you will find all detailed mechanical drawings: <https://www.intel.com/content/www/us/en/support/boards-and-kits/intel-nuc-kits/000006820.html> Mechanical Drawings and Custom Chassis Information for Intel® NUC 

Hope this will help

Leon




[View solution in original post](/t5/Intel-NUCs/Technical-drawing-of-the-NUC7i5BNK-board/m-p/417850#M21995) Here's the diagram you requested. All the best!  

Greg  

![I need a serpentine belt routing diagram for a - c528ac5.gif](http://i.fixya.net/uploads/images/c528ac5.gif) no, the moment of force is a vector quantity. Here's an example of one possibility. Just season to taste... 



```
extends Node2D
var x_delta = 0

func _physics_process(delta):
    var input_velocity = Vector2.ZERO

    if Input.is_action_just_pressed("ui_right"):
        x_delta = 1

    if Input.is_action_just_pressed("ui_left"):
        x_delta = -1

    if Input.is_action_just_released("ui_right") && x_delta > 0:
        x_delta = 0

    if Input.is_action_just_released("ui_left") && x_delta < 0:
        x_delta = 0

    input_velocity.x = x_delta

    print(input_velocity.x)

``` hello, 

this code checks to see if the diagonal elements of a given matrix a (assuming n x n) are larger in magnitude than the sum of the magnitude of the non-diagonal elements in its row.

line by line explanation:

the first line loops through all the rows of a.

 for i = 1:n

the second line is defining j = [1 2 3 ... n], so a vector containing 1..n,
 
 j = 1:n;

the third line is removing the diagonal element, aka, the ith row and this removes the ith column, so the i,i th element,
 
 j(i) = [];

the fourth line takes the absolute value of all the elements in the ith row of a, excluding the diagonal element: a(i,j) = a(ith row, elements 1,2,...,i-1,i+1,...,n) since we removed the ith element from j previously.
 
 b = abs(a(i,j));

the fifth line computes the absolute value of the diagonal element minus the sum of the absolute value of all the other elements in the row. if check is positive, we know the magnitude of the diagonal element was greater than the sum of all the magnitudes of the other elements in its row.
 
 check(i) = abs(a(i,i)) - sum(b);

the sixth line checks if it's less than zero, in which case the diagonal elements magnitude is not greater than or equal to the sum of the magnitudes of the other elements.
 
 if check(i) < 0

if this is the case, the matrix is not strictly diagonally dominant, which the seventh line prints.
 
 fprintf('the matrix is not strictly diagonally dominant at row %2i

',i)

the eighth and ninth lines end the cycle.
 
 end
 end

sorry if this was too much detail,

hope this helps! in the figure below, the angle  is shown in the standard position where the point  is on the terminal side of , and  is the distance from the origin to . 

if  is the distance from the origin to  then the trigonometric function of  are
  
  
     
therefore, the trigonometric functions are , , and . the graph of this equation is a line.  we know that because each term is either a constant (1.5) or the product of a constant and the first power of a variable (d, 0.8t).
since we have two variables (d and t), the line sits in a two-dimensional plane, with a vertical axis and a horizontal axis.  d stands alone in the equation, so its value depends on the value oft.  therefore, the vertical--or dependent--axis will represent values of d, and the horizontal--or independent--axis will represent values of t.
at this point, then, we can set up a coordinate plane with a vertical axis labeled d and a horizontal axis labeled t.  all of the points on our plane have a coordinate (t, d).
to graph a line, we need two points.  since the question asks for the axial intercepts anyway, we can use those.
we'll start with the d-intercept, where the line crosses the d-axis.  all of the points on the d-axis have a t-coordinate of zero, so to find where the line crosses the axis, we substitute zero for t in our equation and solve.
d = 1.5 + 0.8 × t
d = 1.5 + 0.8 × (0)
d = 1.5 + 0
d = 1.5
so our d-intercept has the coordinates (0, 1.5), and we have our first point and our first intercept.
similarly, to find the t-intercept, we note that all of the points on the t-axis have a d-coordinate of zero, so we set d equal to zero and solve.
d = 1.5 + 0.8 × t
0 = 1.5 + 0.8 × t
0 - 1.5 = 1.5 - 1.5 + 0.8 × t
-1.5 = 0.8 × t
-1.5 ÷ 0.8 = (0.8 ÷ 0.8) × t
-1.875 = 1 × t
-1.875 = t
and the d-intercept is (-1.875, 0).
now we have both axial intercepts, so that part of the question is complete.  we can plot those two points and draw the line that passing through them (extending on, at either side, to infinity).
all that remains is to find the turning points.  a turning point, also called a local maximum or local minimum, is where the value of a function goes from decreasing to increasing or from increasing to decreasing.  (the graph shown at http://en.wikipedia.org/wiki/file:parabola\_with\_focus\_and\_directrix.svg has a minimum at the origin, for example.)
a line has a constant slope.  to say that differently, the angle between any two points on a line is the same as the angle between any two other points on the line.  if a line is increasing in one location (like our line), it is increasing everywhere, and it is increasing everywhere at the same rate.  therefore, a line has no turning points.