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- K1 = (60/2)+3=33Let the region R be the area enclosed by the function f (x) = x³ – 1 and g(x) = 2x – 1.Find the volume of the solid generated when the region R is revolved about the line y = -3. You may use a calculator and round to the nearest thousandth. 2 1 -0.5 0.5 1 1.5 -1.5 -1 -2 -4 -5 -6 -7 -8 6- 2.By using Sarrus' Method 15 2| 4 -2 3 13 1. -1 21
- a = 1 m b = 2 m c = 3 m Force AB = 500 Newtons FAB Find the y-component of unit vector AC. (Hint: the value should be less than 1). Find the z-component of unit vector AC. (Hint: the value should be less than 1). Find the x-component of Force AB in N. If the calculated answer is not whole number, express it in 4 significant figures. If the value is negative, include a negative sign.The resultant of the concurrent forces in Figure 1-33 is 200 N up to the left at 0, = 60°. Compute the values of F and 0 required to give this resultant. F 500 N 30 240 N Figure 1-33Q3: Draw the geometry shown in figure below. 38 88 12p 56 -2812 50
- 3m Find the rectangular representation of the force F, given that its magnitude 240 N. * F = -135.76 i+ 169.71 j– 101.82 k F = 135.76 i + 169.71 j+ 101.82 k F = 135.76 i - 169.71 j+ 101.82 k F = -135.76i+ 169.71 j+ 101.82 kQ5/ Let f(x) = 3√3x. Consider the portion of the curve where 0 ≤ y ≤ 2. Find the surface area of the surface generated by revolving the graph of f(x) around the y- axis.3 degree indeterminate Replace both 10KN with 101 KN
- Given in table6 X1=3 Y1=5Three forces intersect at point (1, 1, 1). Their magnitude and respective second point of incidence are as follows: 450 lbs (4, -3, 7), 600lbs (-1, 5, -4), 300lbs (-3, 6, -2) 1. Compute the direction cosine on the x-axis. a. 0.394 b. 0.124 c. 0.762 d. 0. 513 2. Compute the direction cosine on the y-axis. a. 0.394 b. 0.124 c. 0.762 d. 0. 513 3. Compute the direction cosine on the x-axis. a. 0.394 b. 0.124 c. 0.762 d. 0. 513The following diagram corresponds to questions 1 to 3. A beam rests on two sharp edges as shown on the image. The beam has a length of 7,50 m and a mass of 4,25 kg. Object 1 has a mass of 1,50 kg; object 2 has a mass of 2,35 kg. Point P is 0,50 m form the center of the beam. CG 1. How much is the torque done by object 1 around point P? A. 55,2 N*m B. 7,36 N*m C. 62,5 N*m D. Object 1 doesn't exert a torque. 2. How much is the torque done by the force of gravity of the beam around point P? A. 20,8 N*m N-m ך177 .B C. 156 N*m D. 313 N*m 3. If you needed to cancel the nomal forces of the two objects, where you should place object 2? The axis of rotation is point P. A. 3,30 m from point B. 9,18 m from point P C. 5,69 m from point P D. 3,62 m from point P