(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
Student Solutions Manual For Katz's Physics For Scientists And Engineers: Foundations And Connections, Volume 1
- (a) For what values of the angle between two vectors is their scalar product positive? (b) For what values of is their scalar product negative?arrow_forwardUse the definition of the vector product and the definitions of the unit vectors i, j, and k to prove Equations 11.7. You may assume the x axis points to the right, the y axis up, and the z axis horizontally toward you (not away from you). This choice is said to make the coordinate system a right-handed system.arrow_forwardA 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_forward
- A small block of mass m = 200 g is released from rest at point along the horizontal diameter on the inside of a frictionless, hemispherical bowl of radius R = 30.0 cm (Fig. P7.45). Calculate (a) the gravitational potential energy of the block-Earth system when the block is at point relative to point . (b) the kinetic energy of the block at point , (c) its speed at point , and (d) its kinetic energy and the potential energy when the block is at point . Figure P7.45 Problems 45 and 46.arrow_forwardA small particle of mass m is pulled to the top of a friction less half-cylinder (of radius R) by a light cord that passes over the top of the cylinder as illustrated in Figure P7.15. (a) Assuming the particle moves at a constant speed, show that F = mg cos . Note: If the particle moves at constant speed, the component of its acceleration tangent to the cylinder must be zero at all times. (b) By directly integrating W=Fdr, find the work done in moving the particle at constant speed from the bottom to the top of the hall-cylinder. Figure P7.15arrow_forwardA small object is attached to two springs of the same length l, but with different spring constants k1 and k2 as shown in Figure P9.31. Initially, both springs are relaxed. The object is then displaced straight along the x axis from xi to xf. Find an expression for the work done by the springs on the object.arrow_forward
- In 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_forwardEstimate the kinetic energy of the following: a. An ant walking across the kitchen floor b. A baseball thrown by a professional pitcher c. A car on the highway d. A large truck on the highwayarrow_forward
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