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given that 7/9 of a mass A is the same as 5/12 of a mass B, find the ration of mass A to mass B

given that 7/9 of a mass A is the same as 5/12 of a mass B, find the ration of mass A to mass B

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PHYSICS LAB FINAL Flashcards | Quizlet


Start studying PHYSICS LAB FINAL ... Momentum is defined as mass times velocity. You find the mass of ... you find the spring constant of a given spring to be 7.9 N ...
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PHYSICS 111 HOMEWORK SOLUTION #7 - NJIT SOS


PHYSICS 111 HOMEWORK SOLUTION #7 ... mass is 5.20 grams? a) ... N.B. The two masses are moving with the same speed due to the strained
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How to Calculate Mass: 10 Steps (with Pictures) - wikiHow


wiki How to Calculate Mass. ... than an object with less mass, if they're experiencing the same ... sub 1 and mass sub 2 if acceleration and force is given?
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PHYSICS 111 HOMEWORK SOLUTION, week 4, chapter 5, sec 1-7


PHYSICS 111 HOMEWORK SOLUTION, week 4, chapter 5, ... (12:00^i+ 8:00^j)N b) ... b)Find the magnitude of the acceleration of the objects. ) ...
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Chapter 3 Molar Mass Calculation of Molar Masses


1 C = 12.011 3 O = 3 x 15.9994 Molar mass of CaCO 3 = 100.087 g ... Given: 2.5 L of soln 0.10M HCl Find: ... the same units. Solution Preparation Chapter 5
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How do you calculate mass using density and volume? | Socratic


In order to find mass when you have density and volume, ... How do you calculate mass using density and volume? Chemistry Measurement Density. 2 Answers
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B) 4.0 m/s C) 7.9 m/s D) 5.6 m/s - Iowa State University


... 6.9 m/s B) 4.0 m/s C) 7.9 m/s D) 5.6 m/s E) ... (0,5)+24 60 (12,0)+26 60 (12,5)=(10,3) ... each of the same mass
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Chemistry Unit 2 Flashcards | Quizlet


Chemistry Unit 2. STUDY. PLAY. Create a ... One amu is equivalent to 1/12 of the mass of a ... Calculate the average atomic mass of argon to two decimal places given ...
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How to Calculate Mass Percent - wikiHow


How to Calculate Mass Percent. ... When you aren’t given masses, you can find the mass percent of an element within a compound using molar mass. ... C 6 H 12 O 6. 3.
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ChemTeam: Stoichiometry: Mass-Mass Problems


DON'T use the same molar mass in steps two and four. ... An unbalanced equation was given in the problem. It needs to be balanced: 2Al ... Example #5: How many grams ...
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Suggested Questions And Answer :


find the center of mass of the solid of uniform density biunded by the graphs of the equations

x^2+y^2=a^2 is a cylinder with its circular cross-section in the x-y plane; the cylinder is sliced at an angle by the plane z=cy, forming a wedge. y=0 is the x-z plane and z=0 is the x-y plane; these planes and the conditions that y and z are positive impose further constraints on the shape. The cylinder is sliced in half by the x-z plane so the cross-section will be a semicircle and only the upper half is required. The semicircular end rests on the z=0 plane, and doesn't extend beyond the vertex of the wedge. The length of the wedge is ac, because the slope of the angle of the wedge is tan^-1(c) to the y axis and the radius of the wedge is a. Viewed from the side, along the x axis, the wedge is a right-angled triangle with sides a (height), ac (length) and hypotenuse a√(1+c^2). In everyday terms, the wedge looks like the end of a lipstick that has been sliced across. Going back to the side view of the wedge, we can see that the flat end has a height a. As we move along the wedge towards the point the height changes and shrinks to zero when we reach the vertex. The length at a point P(y,z) is a-y. This height is the distance of a chord to the circumference, starting at height a. The first task is to work out the area of a segment. We do this by subtracting the area of a triangle from the area of a sector, given the distance of the chord from the circumference. This distance is a-y, so the distance from the centre of the circle is, conveniently, y. y/a=cosø where ø is half the angle subtended by the arc of the sector. So ø=cos^-1(y/a) and the area of the sector is given by 2ø/2π=ø/π=A/πa^2, or A=a^2ø (ø is measured in radians). The area of the isosceles triangle formed by joining the radial ends of the arc of the sector is B=aysinø=y√(a^2-y^2), and the area of the segment is A-B=a^2cos^-1(y/a)-y√(a^2-y^2). (Note that when y=0, this expression becomes πa^2/2, the area of the semicircle.) The volume of the segment is (a^2cos^-1(y/a)-y√(a^2-y^2))dz where infinitesimal dz is the thickness of the segment. The mass of the segment is directly proportional to its volume because the density is uniform, so the volume is sufficient to represent the mass. The centre of gravity (COG) of the wedge can be found by summing moments about a point G(x,y,z) which is the COG. This sum has to be zero. Since a semicircle is symmetrical in the x-y plane about the y axis, we know the x coord of G must be zero. We need G(0,g,h) where g is the height along the z axis of the COG and h the z position. We need to find the COG of a segment first. The length of the chord of a segment is 2√(a^2-y^2). Earlier we discovered this when we were finding out the areas of the sector and isosceles triangle. Consider just two dimensions. The length of the chord is one dimension and if we create a rectangle from this length and an infinitesimal width dy, the area will be 2√(a^2-y^2)dy. If we think of the rectangle as having mass, its mass is proportional to its area, so area is a sufficient measure for mass. Let's call the COG of the segment point C(x,y)=(0,q), then the moment of the rectangle about the COG=C(0,q) is mass times distance from C=2(q-y)√(a^2-y^2)dy. If we sum the rectangles over the interval y to a we have 2∫((q-y)√(a^2-y^2)dy)=0 for y≤y≤a. The interval looks strange, but it means that y is just a general value, which is determined when we return to the wedge. What we are trying to do is to find q relative to y. We need to evaluate the definite integral, so we split it: 2q∫(√(a^2-y^2)dy)-2∫(y√(a^2-y^2)dy. Call these (a) and (b). To evaluate (a), let y=asinu, then dy=acosudu; √(a^2-y^2)=acosu and the integral becomes 2a^2q∫(cos^2(u)du). Since cos(2u)=2cos^2(u)-1, cos^2(u)=½(cos(2u)+1), so we have a^2q∫((cos(2u)+1)du)=a^2q(sin(2u)/2+u). Since sin(2u)=2sinucosu, this becomes a^2q(sinucosu+u)=a^2q(y/a * √(1-(y/a)^2)+sin^-1(y/a))=qy√(a^2-y^2)+a^2qsin^-1(y/a). Now we apply the limits for y: πa^2q/2-qy√(a^2-y^2)+a^2qsin^-1(y/a). To evaluate (b), let u=a^2-y^2, then du=-2ydy and the integral is +∫√udu = (2/3)u^(3/2)=(2/3)(a^2-y^2)^(3/2). Applying the limits we have: -(2/3)(a^2-y^2)^(3/2). (a)+(b)=πa^2q/2-qy√(a^2-y^2)+a^2qsin^-1(y/a)-(2/3)(a^2-y^2)^(3/2)=0. Multiply through by 6: 3πa^2q-6qy√(a^2-y^2)+6a^2qsin^-1(y/a)-4(a^2-y^2)^(3/2)=0. From this q=4(a^2-y^2)^(3/2)/(3πa^2-6y√(a^2-y^2)+6a^2sin^-1(y/a)). Fearsome though this looks, when y=0 and z=0, q=4a^3/(3πa^2)=4a/(3π), which is in fact the COG of a semicircle. When y=a and z=ac, q=0. Strictly speaking, we should write q(y) instead of just q, showing q to be a dependent variable rather than a constant. To find the COG of the wedge, we need to find the moments of the COGs of the segments. For a segment distance y above and cy along the z axis, we have the above expression for q. So q(y) is the y coord of the COG of the segment and cy is the z coord. The x coord is 0. The mass of the segment is (a^2cos^-1(y/a)-y√(a^2-y^2))dz, as we saw earlier, and this mass can be considered to be concentrated or focussed at the COG of the segment (0,q(y),cy). The COG of the wedge is at G(0,g,h) so the distance of the mass of the segment from G is g-q(y) and h-cy in the y and z directions. The y-moment is (g-4(a^2-y^2)^(3/2)/(3πa^2-6y√(a^2-y^2)+6a^2sin^-1(y/a))(a^2cos^-1(y/a)-y√(a^2-y^2))dz and the z-moment is (h-cy)(a^2cos^-1(y/a)-y√(a^2-y^2))dz. The sums of these individual components are zero. These lead to rather complicated integrals, requiring considerably more space than is available to work out, even if I knew how to do so!
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given that 7/9 of a mass A is the same as 5/12 of a mass B, find the ration of mass A to mass B

start: (7/9)a=(5/12)b a=(9/7)*(5/12)b a=(45/84)b   =(15/28)b
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3+i, 3 Lowest Degree

Question: 3+i, 3 Lowest Degree. Find a polynomial function of lowest degree with rational coefficients that has the given numbers as some of its zeros. When a polynomial has complex roots, they always come as a pair. So if x = 3+i is one root, then x = 3 - i is another root. (Two complex roots will always be of the form a + ib, a - ib. ) So we have at least three roots: x = 3, x = 3 + i, x = 3 - i. Creating the polynomial from these three roots. (x - 3)(x - (3 + i))(x - (3 - i)) = 0 (x - 3)(x^2 - 3x + 3i - 3x - 3i + 9 + 1) = 0 (x - 3)(x^2 - 6x + 10) = 0 x^3 - 6x^2 + 10x - 3x^2 + 18x - 30 = 0 x^3 - 9x^2 + 28x - 30 = 0
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problem solving in integral calculus about centoid

The centroid lies on the y-axis (x=0), because the semicircle is symmetrical about this axis. We need to find the y coord. The centroid is the "centre of mass" of the semicircle, and it can be considered to be made up of a number of very thin rectangles of width dx and length y or width dy and length x. The area of each rectangle is ydx (vertical) or xdy (horizontal) and the mass of the rectangle is proportional to this area. The area of the semicircle is 1/2 the area of its circle, which is (pi)r^2 where r is the radius, which is given by the equation of the circle x^2+y^2=4=r^2. The area of the semicircle is therefore 2(pi) and its mass is proportional to this area. The centroid is the average of the "moments" of the rectangles divided by the mass of the semicircle. The moment is mass*distance from the origin=xydx or yxdy for x and y cords respectively. The sum of these moments is the integral(xydx) (x coord) and integral(yxdy) (y coord) between the limits -2 and 2 for x, and 0 and 2 for y. These points on the axes are where the semicircle intersects. The centroid is the point (integral(xydx)/2(pi),integral(yxdy)/2(pi)). y=sqrt(4-x^2) or x=sqrt(4-y^2), and the integral(ysqrt(4-y^2)dy)/2(pi)=-(2/3)(4-y^2)^(3/2)/4(pi). This will give us the y coord of the centroid, and we already know the x coord is zero because of symmetry. The limits for y are 0 to 2: (2/3)4^(3/2)/4(pi)=(2/3)8/4(pi)=4/(3(pi))=0.4244. So the centroid is (0,0.4244). [integral(ysqrt(4-y^2)dy)/2(pi): let u=4-y^2, then du=-2ydy, so ydy=-du/2. The integral becomes integral(-u^(1/2)du)/4(pi)=-(2/3)u^(3/2)/4(pi)=-(2/3)(4-y^2)^(3/2)/4(pi).]  
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is there a formula for finding a rational function of a given graph?

First you have to graph the lines.  Then you can see if there is any function to discribe them.     Since these lines do not have limits as the domain and range I would say that there is not a feunction to describe them.  Due to the fact that y has 2 values at all negative numbers and on the y-axis. f(x) = {y = -1/4x,   -3 <= x < 1}           {y = x - 5,     1 <= x < infinity.
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find a third degree polynomial equation with rational coefficients that has roots -4 and 6+i

A cubic equation always has three roots. These three roots are: 1) three real roots or, 2) 1 real root and two complex roots If one of the two complex roots is a + ib, then the other complex root is a - ib. We are given two of the roots as -4 and 6+i. Since one of the roots is complex and equals 6+i, the the other complex root is 6-i. Our three roots then are: -4, 6+i, 6-i. Our three solutions to the cubic equation are: x = -4, x = 6+i, x = 6-i. Which can be rewritten as: x + 4 = 0, x - (6+i) = 0, x - (6-i) = 0 Multiplying these together gives us the original cubic equation. (x + 4)(x - (6+i))(x - (6-i)) = 0 Multiplying this out, (x + 4)(x^2 - (6-i)x - (6+i)x + (6-i)(6+i)) = 0 (x + 4)(x^2 - 12x + 6^2 - i^2) = 0 (x + 4)(x^2 - 12x + 37) = 0 x^3 - 8x^2 - 11x +148 = 0  
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A standard Hyperbola has its vertices ar 5,7 and 3,-1.Find its equation and its asymptotes

this equation expands toThe vertices share no common coordinate so the hyperbola cannot be orientated in the usual x-y form. The hyperbola must therefore be skew (tilted axis) and the line joining the vertices is sloping. This line is given by the equation y=((7+1)/(5-3))x+c where c is to be found. y=4x+c, and -1=12+c, plugging in (3,-1). c=-13. Therefore the axis of the hyperbola is y=4x-13. The centre or origin of the hyperbola lies midway between them at ((1/2)(5+3),(1/2)(7-1))=(4,3) and the equation of the conjugate axis is given by y=-x/4+k where k is found by plugging in the midpoint: 3=-4/4+k, so k=4 and y=4-x/4 is the equation of the conjugate axis. Let's define the axes of the hyperbola as X and Y instead of x and y. We can place the centre of the hyperbola at (0,0) in the X-Y plane so that in this plane the equation of the hyperbola is Y^2/A^2-X^2/B^2=1. The vertices are found by finding Y when X=0, so (0,A) and (0,-A) are the vertices. The asymptotes are given by (Y/A+X/B)(Y/A-X/B)=0. So Y=AX/B and Y=-AX/B are their equations. The distance between the vertices is 2A so we can work this out from the x-y coordinates of the vertices using Pythagoras' Theorem: 2A=sqrt((7+1)^2+(5-3)^2)=sqrt(68). A=sqrt(68)/2, A^2=68/4=17. Since the hyperbola is rectangular, B=A, and the equation of the hyperbola is Y^2-X^2=17. The asymptotes then become Y=X and Y=-X. Now we have to transform X-Y to x-y, so that the equation can be expressed in the x-y plane. We use geometry and trigonometry  for this. Let O'(4,3) be the location of the origin of the X-Y plane. Let P(x,y) be any point. In the X-Y plane this is P'(X,Y). Join O'P'=O'P. This line can be represented in both reference frames. O'P^2=X^2+Y^2=(x-4)^2+(y-3)^2 by Pythagoras. The gradient of the Y axis with respect to the x-y frame is 4, and we can represent this as tan(w) so w=tan^-1(4) or tan(w)=4; sin(w)=4/sqrt(17); cos(w)=1/sqrt(17). The angle z that O'P makes with the line y=3 is given by tan(z)=(y-3)/(x-4). sin(z)=(y-3)/O'P; cos(z)=(x-4)/O'P. Let v=90-w. sin(v+z)=Y/O'P; cos(v+z)=X/O'P. v+z=90-w+z=90-(w-z), so sin(v+z)=cos(w-z)=cos(w)cos(z)+sin(w)sin(z)=(x-4)/(O'Psqrt(17))+4(y-3)/(O'Psqrt(17))=(x+4y-16)/(O'Psqrt(17)). cos(v+z)=sin(w-z)=sin(w)cos(z)-cos(w)sin(z)=4(x-4)/(O'Psqrt(17))-(y-3)/(O'Psqrt(17))=(4x-y-13)/(O'Psqrt(17)). Y=O'P*sin(v+z)=(x+4y-16)/sqrt(17); X=O'P*cos(v+z)=(4x-y-13)/sqrt(17). So P maps to the x-y plane using these transformations. Y^2-X^2=17 becomes (x+4y-16)^2-(4x-y-13)^2=289, the equation of the hyperbola. This expands to x^2+8x(y-4)+16y^2-128y+256-(16x^2-8x(y+13)+y^2+26y+169)=289; -15x^2+16xy+15y^2+72x-154y-202=0. The asymptotes are x+4y-16+4x-y-13=0; 5x+3y-29=0 and x+4y-16-4x+y+13=0; 5y-3x-3=0.
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f(1)=x³+bx²-9x+c find the values of b and c

f(x)=x^3+bx^2-9x+c; f(1)=1+b-9+c. We are not given the value of f(1) so let's start by assuming f(1)=0 as a slight addition to the original question. This means that 1+b-9+c=0, b+c=8. f(1)=0 means that 1 is a zero of the expression and x^3+bx^2-9x+c=(x-1)(x^2+ax-c) where a needs to be found. This expands to: x^3+(a-1)x^2-(c+a)x+c so we can write b=a-1 and c+a=9. The quadratic x^2+ax-c can be written x^2+(9-c)x-c, so x=(c-9+sqrt(81-18c+c^2+4c))/2=(c-9+sqrt(c^2-14c+81))/2. The question offers no more information to determine unique values for b and c, but we can put in some constraints because b and c must be rational. c^2-14c+81=n^2, a perfect square. Completing the square for the left-hand side: (c^2-14c+49)-49+81=n^2; (c-7)^2=n^2-32; c-7=+sqrt(n^2-32). When n=6, c=7+2, so c=9 or 5 and b=-1 or 3, because b+c=8. Therefore f(x)=x^3-x^2-9x+9=(x-1)(x^2-9)=(x-1)(x-3)(x+3) or x^3+3x^2-9x+5=(x-1)(x^2+4x-5)=(x+5)(x-1)^2. Therefore the solutions are (b,c)=(-1,9) or (3,5).
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Algebra 2 homework, please help.

  First set of questions is all roots, or solutions, of the equations and second set is to find where the functions are zero, the rational roots only. The methods for roots and zeroes are basically the same. 1. 3x^2+10x+3=3x^2+9x+x+3=3x(x+3)+(x+3)=(3x+1)(x+3)=0; x=-1/3, -3 2. 5x^2+5x-x-1=5x(x+1)-(x+1)=(5x-1)(x+1)=0; x=1/5, -1 3. Cubic equation. Trial shows that x=-1 is a solution, so divide cubic by factor (x+1). Use synthetic division: -1 | 3 11  5 -3      | 3 -3 -8  3        3  8 -3 | 0 and  (x+1)(3x^2+8x-3)= (x+1)(3x^2+9x-x-3)=(x+1)(3x(x+3)-(x+3))=(x+1)(3x-1)(x+3)=0; x=-1, 1/3, -3 4. 3x^4-2x^2-5=(3x^2-5)(x^2+1)=0; 3x^2=5, so x=+/-sqrt(5/3) 5. Trial shows that 3x=-1 is a solution, i.e., x=-1/3. Use synthetic division: -1/3 | 3 10 30   9         | 3  -1 -3  -9           3   9 27 | 0      (3x+1)(x^2+3x+9)=0; x=-1/3 is the only real solution 6. (5x^2+7)(x^2-2)=0; x=+/-sqrt(2) 7. No real roots 8. (5x^2+2)(x^2-8)=0; x=+/-sqrt(8) or +/-2sqrt(2) 9. (5x^2+9)(x^2-4)=(5x^2+9)(x+2)(x-2)=0; x=+/-2 10. (5x^2+3)(x^2-8)=0; x=+/-sqrt(8) or +/-2sqrt(2) 11. f(x)=5x(x^2-1)-(x^2-1)=(5x-1)(x-1)(x+1); x=1/5, 1, -1 12. Try x=1/2, f(x)=0. So f(x)=(2x-1)(x^2-11x-1) has no more rational roots 13. f(x)=3x(x^2-1)-(x^2-1)=(3x-1)(x-1)(x+1); zeroes at x=1/3, 1, -1 14. f(x)=(x+2)(x^2+7x+3); only rational zero at x=-2 15. (x-3)(2x^2+7x-3); rational zero at x=3 16. (x-1)^2(4x-1); x=1, 1/4 17. (x+1)(2x-1)(2x+1); x=-1, +/-1/2 18. (x+1)^2(3x-1); x=-1, 1/3 19. (x-5)(3x^2-7x-75); x=5, no other rational roots 20. (x-3)(2x^2+11x-4); x=3, other roots irrational
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Find the center of the mass of a solid with constant density & bounded below by the disk R : x^2+y^2≤4 in the plane z=0 and above by the parabolaid z=4-x^2-y^2

The paraboloid equation can be written x^2+y^2=4-z. At z=0 this is the circle x^2+y^2=4 (centre at the origin and radius=2). As we move along the z axis the cross-section parallel to the x-y plane remains as a circle (x^2+y^2≤4) which reduces to a point at the vertex of the paraboloid at (0,0,4). The centre of gravity (COG) is therefore on the z axis at P(0,0,p) because the paraboloid is symmetrical, with the z axis acting like a spindle. We have to find p. We can consider the solid as a set of discs pierced through their centres by the z axis as a spindle. The volume of each disc is related to its mass by the constant density. The thickness of the disc is the infinitesimal dz. The radius of a disc is √(4-z), so its volume is π(4-z)dz. This expression can be taken to represent its mass. The mass of each disc is concentrated at its centre because of symmetry, so its COG is at the centre on the z axis. We need to consider the COG of all the discs together, so we need to define the moment of each disc about P. This is the product of mass and distance from P, which is p-z. The negative moments have to balance the positive moments. The sum total of the moments is zero. The expression for the moment of each disc is π(4-z)(p-z)dz and the sum of these must be zero, so π∫((4-z)(p-z)dz)=0 where the limits are 0≤z≤4. π∫((4p-z(p+4)+z^2)dz)=0; π[4pz-z^2(p+4)/2+z^3/3]=0, 0≤z≤4.  π(16p-8(p+4)+64/3)=0; 8p-32+64/3=0; 8p=32/3, p=4/3, so the COG is P(0,0,4/3).
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