(a)
The angle between the pair of vectors.
(a)
Answer to Problem 67PQ
The angle between the pair of vectors is
Explanation of Solution
Write the expression for the magnitude of vector
Here,
Write the expression for the magnitude of vector
Here,
Write the expression for dot product between
Write the expression for angle between
Conclusion:
Substitute
Substitute
Substitute
Substitute
Therefore, the angle between the pair of vectors is
(b)
The angle between the pair of vectors.
(b)
Answer to Problem 67PQ
The angle between the pair of vectors is
Explanation of Solution
Write the expression for the magnitude of vector
Here,
Write the expression for the magnitude of vector
Here,
Write the expression for dot product between
Write the expression for angle between
Conclusion:
Substitute
Substitute
Substitute
Substitute
Therefore, the angle between the pair of vectors is
(c)
The angle between the pair of vectors.
(c)
Answer to Problem 67PQ
The angle between the pair of vectors is
Explanation of Solution
Write the expression for the magnitude of vector
Here,
Write the expression for the magnitude of vector
Here,
Write the expression for dot product between
Write the expression for angle between
Conclusion:
Substitute
Substitute
Substitute
Substitute
Therefore, the angle between the pair of vectors is
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Chapter 9 Solutions
Physics for Scientists and Engineers: Foundations and Connections
- A particle moves in the xy plane (Fig. P9.30) from the origin to a point having coordinates x = 7.00 m and y = 4.00 m under the influence of a force given by F=3y2+x. a. What is the work done on the particle by the force F if it moves along path 1 (shown in red)? b. What is the work done on the particle by the force F if it moves along path 2 (shown in blue)? c. What is the work done on the particle by the force F if it moves along path 3 (shown in green)? d. Is the force F conservative or nonconservative? Explain. FIGURE P9.30 In each case, the work is found using the integral of Fdr along the path (Equation 9.21). W=rtrfFdr=rtrf(Fxdx+Fydy+Fzdz) (a) The work done along path 1, we first need to integrate along dr=dxi from (0,0) to (7,0) and then along dr=dyj from (7,0) to (7,4): W1=x=0;y=0x=7;y=0(3y2i+xj)(dxi)+x=7;y=0x=7;y=4(3y2i+xj)(dyj) Performing the dot products, we get W1=x=0;y=0x=7;y=03y2dx+x=7;y=0x=7;y=4xdy Along the first part of this path, y = 0 therefore the first integral equals zero. For the second integral, x is constant and can be pulled out of the integral, and we can evaluate dy. W1=0+x=7;y=0x=7;y=4xdy=xy|x=7;y=0x=7;y=4=28J (b) The work done along path 2 is along dr=dyj from (0,0) to (0,4) and then along dr=dxi from (0,4) to (7,4): W2=x=0;y=0x=0;y=4(3y2i+xj)(dyj)+x=0;y=4x=7;y=4(3y2i+xj)(dyi) Performing the dot product, we get: W2=x=0;y=0x=0;y=4xdy+x=0;y=4x=7;y=43y2dx Along the first part of this path, x = 0. Therefore, the first integral equals zero. For the second integral, y is constant and can be pulled out of the integral, and we can evaluate dx. W2=0+3y2x|x=0;y=4x=7;y=4=336J (c) To find the work along the third path, we first write the expression for the work integral. W=rtrfFdr=rtrf(Fxdx+Fydy+Fzdz)W=rtrf(3y2dx+xdy)(1) At first glance, this appears quite simple, but we cant integrate xdy=xy like we might have above because the value of x changes as we vary y (i.e., x is a function of y.) [In parts (a) and (b), on a straight horizontal or vertical line, only x or y changes]. One approach is to parameterize both x and y as a function of another variable, say t, and write each integral in terms of only x or y. Constraining dr to be along the desired line, we can relate dx and dy: tan=dydxdy=tandxanddx=dytan(2) Now, use equation (2) in (1) to express each integral in terms of only one variable. W=x=0;y=0x=7;y=43y2dx+x=0;y=0x=7;y=4xdyW=y=0y=43y2dytan+x=0x=7xtandx We can determine the tangent of the angle, which is constant (the angle is the angle of the line with respect to the horizontal). tan=4.007.00=0.570 Insert the value of the tangent and solve the integrals. W=30.570y33|y=0y=4+0.570x22|x=0x=7W=112+14=126J (d) Since the work done is not path-independent, this is non-conservative force. Figure P9.30ANSarrow_forwardIn three cases, a force acts on a particle, and the particle is displaced from an initial position to a final position. Figure 9.11 (page 255) shows the position-versus-force graphs, indicating the initial and final positions of the particle in each case. Find the work done by the force on the particle and sketch the force and displacement vectors along with the appropriate axis in each case.arrow_forwardIn each situation shown in Figure P8.12, a ball moves from point A to point B. Use the following data to find the change in the gravitational potential energy in each case. You can assume that the radius of the ball is negligible. a. h = 1.35 m, = 25, and m = 0.65 kg b. R = 33.5 m and m = 756 kg c. R = 33.5 m and m = 756 kg FIGURE P8.12 Problems 12, 13, and 14.arrow_forward
- +The dot product of any two vectors as and r can be written as: A: 4. A,4,4, cos0 c: 4 x 4.= 4,4, sine B: 4 x4, =(4,4, sin 0)n D: non of these.arrow_forwardTwo vectors have a magnitude of 3 and 2 and their dot product has a value of -5. Which of the following is the angle between the two vectors? O 146.44 degrees O None of the given choices. O 33.6 degrees O 99.6 degrees O -56.4 degreesarrow_forwardDetermine the dot product of vectors P & Q. P = 13i −50j + 21k m Q = 12i-3j+4k m Select one: a. 59 m2 b. 390 m2 c. 79 m2 d. 420 m2arrow_forward
- The vectors k and in the figure have a magnitude of 10.00 [m-¹] and 5.00 [m], respectively. The vector k makes an angle of 35.0° with respect to the y-axis, while the vector makes an angle of 15.0° with respect to the x-axis. What is the scalar product k.x? y 15.0° x 35.0° 5.00 [m] k 10.0 [m-¹]arrow_forwardQuestion 6 Scalar ng bayan! The vectors and in the figure have magnitudes of 5.00 [m-¹] and 15.0 [m], respectively. The vector * makes an angle of 10.0° with respect to the y-axis, while the vector makes an angle of 35.0° with respect to the x- axis. What is the scalar product K. ? y 35.0° x x 10.0⁰ 15.0 [m] 5.00 [m ¹] 75.0 O 73.9 53.0 O 61.4 Tarrow_forward
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