Three forces act on a particle, but it remains stationary. Two of the forces are in newtons. What is the magnitude (N) of the third force? = 12.31 -6.50 and F₂ = -9.60 +8.707, both
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- The resultant of the concurrent forces has a magnitude of 1000 N passing through points (x₁ = 0, y₁ = 0, z₁ = 0) and (x₂ = 2, y₂ = 3, z₂ = 4). 1. Compute the x- component of the resultant force. a. 371.1 N b. 392.3 N c. 422.6 N d. 245.9 N 2. Compute the y – component of the resultant force. a. 512.9 N b. 557.1 N c. 435.8 N d. 634.7 N 3. Compute the z – component of the resultant force. a. 632.7 N b. 642.8 N c. 742.1 N d. 791.3 NTwo masses m,=9 Kg and m,=9 Kg are connected with string and pulled with a force F=25 N that makes an angle of 53° with a smooth horizontal surface as shown in the figure. Find the tension in the string in a unit of Newton. F m2 A. 10.64 B. 7.523 C. None D. 15.05 E. 3.761 O d O O Uodou47.The component of the force (3i – j+2k)N in the direction the vector(-4j+3k) is A 10 в 9 A C 9 /5 D 2
- 5m 4m 3 m B. Figure 2-215. Gotta stay boosted. A 1.50 [kg] box moves along a frictionless horizontal surface with an initial speed of 5.00 [m/s] going to the right. A constant 3.00 [N] force is then applied to the box at an angle 0 = 35.0° with respect to the displacement vector. What is its final speed after moving 2.00 [m]? A. 4.91 [m/s] B. 5.62 [m/s] C. 5.74 (m/s] D. 6.00 [m/s]Two forces act on a 2.90 kg object, the gravitational force and a second, constant force. The object starts from rest and in 1.20 s is displaced (4.10 3.30ĵ) m. Write the second force in unit vector notation. (Enter your answer in kg. m/s². Assume the gravitational force acts in the -ĵ direction.) kg. m/s² F
- A physicist has just discovered a new kind of particle. The force between two of them is F = –b/r, where r is the distance between the particles. What are the SI units of b? kg/s? kg · m/s? ,2 kg · m?/s? .2 kg · m³/s? .2 kg · m4/s² none of theseDETAILS PREVIOUS ANSWERS SERPSE10 7.4.P.016. MY NOTES ASK YOUR TEACHER The force acting on a particle is F, = (10x – 17) N, where x is in meters. (a) Make a plot of this force versus x from x = 0 to x = 3.00 m. F (N) F (N) 50 20 40 10 3어 20 0,5 1.0 15 20 2,5 x (m) 30 10 -10 x (m) 3.0 0.5 1.0 1.5 2.0 2.5 F (N) F (N) 0,5 1.0 + x (m) 2,5 30 1아 1.5 2,0 -1어 2,5 x (m) 30 -20 0,5 1,0 15 -30 -10 - 40 -20 - 50 (b) From your graph, find the net work done by this force on the particle as x = 0 to x = 2.55 m. (Include the correct sign.) it moves from Need Help? Read ItSent Mall Daniella messaged 8 Timer for Google F X + с ☐ Apps M Gmail ►YouTube Maps 2 h 12 min zFVLOdcF What must be the direction of the fourth force of the system below to make the magnitude zero? B (80.0 N) A (100.0 N) 30.0° X 53.0° C (40.0 N) O75.5 deg south of west 15.5 deg east of south 15.5 deg south of east 75.5 east of south The string supporting 6 kg block is cut. Find the magnitude of the net torque of form-timer.com/app 30.0° 0 :
- Two forces act on a 2.50 kg object, the gravitational force and a second, constant force. The object starts from rest and in 1.20 s is displaced (5.50î – 3.30j) m. Write the second force in unit vector notation. (Enter your answer in kg · m/s. Assume the gravitational force acts in the -ĵ direction.) kg · m/s2 II1. A proton of mass 1.67 x 10-27 kg is initially at rest. It then moves through 1.80 cm along a straight line, reaching a speed of 3.00 x 106 m/s. During this motion, the net force acting on the proton is A. None of the other answers B. 4.18 x 10-13 N С. C. 1.39 x 10-19 N D. 4.18 x 10-16 N Question 4 of 10 A AMovina to th enovt NUstio n preu ents2.6.2 Force Resultant in 3-D F₂ = 2 KN 60 60° Express each force in Cartesian vector form. 45 F₁ = 5 KN O F₁ = (2.50 i +3.54 j +2.50 k) kN, F₂ = −2 j kN Ο O F₁ = (2.17i+3.75 j +4.33 k) kN, F₂ = -2 j kN F1 O F₁ = (4.33 i +3.75j +2.17 k) kN, F₂ = -2 j kN O F₁ = (4.33 i +3.54j+4.33 k) kN, F₂ = -2 j kN