Questions and Answers
Your Questions About X1
Find intersection coordinates given the distance of two points?
I have the map coordinates of point A and B.
I have to find the coordinate of point C knowing that this point is X1 meters distant from point A and X2 meters distant from point B.
I am sure this can only be done using UTM coordinates, but how?
Thanks for any help.
- No drawing with compass
Us a compass to draw a circle that is x1 meters in radius around point A, and another that is x2 meters in radius around point B. They’ll intersect at 2 points, so you’ll have to pick one based on other info (if they gave you any!)
What is the equilibrium concentration of hydronium ion in a solution?
HX is a weak acid that reacts with water according to the following equation:
HX(aq) + H2O(l) H3O1+(aq) + X1-(aq)Ka = 1.58e-05
What is the equilibrium concentration of hydronium ion in a solution that is 0.0944 M in HX and 0.136 M in X1- ion?
[H3O1+] = ____M
Set up your equation
Ka = Products / Reactants
Ka = [H3O+][X-] / [HX]
Note: (H20 can be ignored)
Plug in your values:
1.58×10^-5 = [H3O+][.136M] / [.0944M]
1.49×10^-6 = [H3O+][.136M]
[H3O+]eq = 1.10×10^-5M
How do you find the slope and y-intercept in this equation: x = -2 and graph it?
Also, there is a question “What does x1 stand for?” and “What does y1 stand for?”
Only the 1′s are little 1′s that are a little below the number. It’s what they use in equations where you substitute in a value for x or y, but what would it ‘stand’ for?
X = -2 is a vertical line. It is a line going up and down through -2 on the x-axis.
There is no y-intercept b/c it will not cross the y-axis
For a vertical line, there is no slope. Your answer would be: “no slope”
(Do not say you have zero slope b/c that would be a horizontal line with y = (any number) like y = 4)
As for the x1 and y1 question…there is a formula for finding slope and it is:
y2 – y1
x2 – x1
You use this formula if you are given two points on a line:
Example: (4, 2)(5,
y2 = 8
y1 = 2
x2 = 5
x1 = 4
You are numbering the coordinates really to work out the formula.
( 4 , 2 )…..(5 , (x1,y1)……(x2,y2)
(8 – 2)
(5 – 4)
so my slope would be 6 / 1 or 6
You label them so you know how to plug them into the formula to get your slope.
Hope this helps!
What is the coefficient of friction between the block and the table?
A block weighing 2 lb is forced against a horizontal spring of negligible mass, compressing the spring an amount x1 = 6 inches. Upon releasing the block, it moves on a horizontal table top a distance x2 = 2ft before coming to rest. The spring constant k is 8 lb/ft.
The potential energy of the spring must equal the work done to slide the block.
The potential energy of the spring E is ½ k x², where k = 8 lb/ft and x = 0.5 ft.
The work done to move the block is W = F x, where F is the frictional force and x is the distance the block moved, 2 ft. The friction force F is the normal force N multiplied by the coefficient of friction ?. Here N = 2 lb.
Equating Energy to Work,
½ (8 lb/ft) (0.5 ft )² = ? (2 lb) (2 ft)
? = 0.25
How do I find the eigenvectors of this matrix?
[2 3 0
3 2 0
0 0 1]
I can find the eigenvalues (5,-1,1) easily but I seem to be doing something wrong to find the vectors. (3-u)x1+2y1=0 and I should be able to find my vectors by plugging in the eigenvalues but! when I check my work (plug in the eigenvalues into another equation) I get an extra eigenvector.
Subtract your matrix from xI and get
x-2 -3 0
-3 x-2 0
0 0 x-1
Now, find the determinant:
[(x-2)^2 - 9](x – 1) = (x – 5)((x + 1)(x – 1)
5, -1 and 1 are eigen values.
Now multiply your matrix by a vecorr (x1 x2 x3)
For the eigenvalue 1:
2×1+3×2 = x1
3×1+2×2 = x2
x3 = x3
x1 + 3×2 = 0
3×1 + x2 = 0
You don’t have to work too hard to find that x1=x2=0, and x3 is any real number.
For the eigenvalue 5
2×1+3×2 = 5×1
3×1+2×2 = 5×2
x3 = 5×3
Well, x3 = 0
Let’s find the dependency between x1 and x2.
Take (1,1,0) as an eigenvector of 5
How do I answer this question about midpoint?
The question is,
A line segment has (x1, y1) as one endpoint and (xm, ym) as its midpoint. Find the other endpoint (x2,y2) of the line segment in terms of x1, y1, xm, and ym.
I’ve used mid point before but I don’t know what to do when I’m not given the values of x and y.
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