Express F = 10O N (shown in the figure below) as a Cartesian vector. TOON. 60° 45 35.4i + 35.4j –- 86.6k 35.4i - 35.4j – 86.6k 35.4i + 35.4j + 86.6k 35.4i – 35.4j + 86.6k -
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A: Note: As per our guidelines we are supposed to answer only the first 3 sub-parts. Kindly repost…
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Q: Consider the following values: - F5 a = 12 m; b = 10 m; c = 10 m; d = 8 m; F1 = 6 kN; F2 = 7 kN; F3…
A: Note: As per our guidelines we are supposed to answer only the first 3 sub-parts. Kindly repost…
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Q: Consider the following values: - F5 a = 25 m; b = 5 m; c = 25 m; d 5 m; F1 = 5 kN; F2 = 6 kN; F3 = 7…
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Q: ᴀ ꜰᴀɴ ᴅᴇꜱᴄʀɪʙᴇᴅ ɪɴ ᴀ ᴍᴀɴᴜꜰᴀᴄᴛᴜʀᴇʀ’ꜱ ᴛᴀʙʟᴇ ɪꜱ ʀᴀᴛᴇᴅ ᴛᴏ ᴅᴇʟɪᴠᴇʀ 500 ᴍ3/ᴍɪɴ ᴀᴛ ᴀ ꜱᴛᴀᴛɪᴄ ᴘʀᴇꜱꜱᴜʀᴇ ᴏꜰ 254…
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- The line of action of force F passes through points A and B. Find the moment of force F about axis CD in Cartesian vector notation, given : F = 45 N AX = 7 m, AY = 8 m, AZ = 10 m BX = 3 m, BY = 5 m, BZ = 5 m CX = 7 m, CY = 7 m, CZ = 8 m DX = 6 m, DY = 5 m, DZ = 4 m Answer Selection: MCD = <288, -74.5, -371> N·m <485, -61.9, -185> N·m <334, -51.3, -308> N·m <201, -44.3, -447> N·m <402, -30.9, -266> N·mFind the Following: 1.Keq 2. Ceq 3. Natural angular velocityTwo variable quantities A and B are found to be related by the equation given below. What is the rate of change dA/dtat the moment when A=3 and dB/dt=5? A3+B3=35
- ᴀ ꜰᴀɴ ᴅᴇꜱᴄʀɪʙᴇᴅ ɪɴ ᴀ ᴍᴀɴᴜꜰᴀᴄᴛᴜʀᴇʀ’ꜱ ᴛᴀʙʟᴇ ɪꜱ ʀᴀᴛᴇᴅ ᴛᴏ ᴅᴇʟɪᴠᴇʀ 500 ᴍ3/ᴍɪɴ ᴀᴛ ᴀ ꜱᴛᴀᴛɪᴄ ᴘʀᴇꜱꜱᴜʀᴇ ᴏꜰ 254 ᴄᴍ ᴏꜰ ᴡᴀᴛᴇʀ ᴡʜᴇɴ ʀᴜɴɴɪɴɢ ᴀᴛ 205 ʀᴘᴍ ᴀɴᴅ ʀᴇQᴜɪʀɪɴɢ 3.6 ᴋᴡ. ɪꜰ ᴛʜᴇ ꜰᴀɴ ꜱᴘᴇᴇᴅ ɪꜱ ᴄʜᴀɴɢᴇᴅ ᴛᴏ 500 ʀᴘᴍ ᴀɴᴅ ᴀɪʀ ʜᴀɴᴅʟᴇᴅ ᴡᴇʀᴇ ᴀᴛ 65ᴅᴇɢ ᴄ ɪɴꜱᴛᴇᴀᴅ ᴏꜰ ꜱᴛᴀɴᴅᴀʀᴅ 21ᴅᴇɢ ᴄ, ꜰɪɴᴅ ᴛʜᴇ ᴘᴏᴡᴇʀ ɪɴ ᴋᴡ. ʀᴏᴜɴᴅ ʏᴏᴜʀ ᴀɴꜱᴡᴇʀ ᴛᴏ 2 ᴅᴇᴄɪᴍᴀʟ ᴘʟᴀᴄᴇꜱ.Two forces in the x and y directions are given as 5.451 kN and 3.279 kN, respectively. What is the resultant force? Question 15Answer a. 7.45 N b. 7458 N c. 4.341 N d. 6361 N e. 6.31 NWhich of the following is a scalar quantity A) electric current B) electric field C) Acceleration D) linear momentum
- (The complete question is in the picture) The force acting on an object is F = 3.50 [N]ˆi−5.20 [N]ˆj−2.30 [N]kˆ. The vector from the origin to the point where the force is applied is r = 3.50 [m]ˆi + 2.50 [N]ˆj. What is the vector torque with respect to the origin produced by this force?A. 10.9 [N.m]ˆi + 11.5 [N.m]ˆj − 27.0 [N.m]kˆB. −18.2 [N.m]ˆi + 8.75 [N.m]ˆjC. −5.75 [N.m]ˆi + 8.05 [N.m]ˆj − 27.0 [N.m]kˆD. −5.75 [N.m]ˆi − 8.05 [N.m]ˆj + 9.45 [N.m]kHow to solve these two equations, to find IJ (kN) and ID (kN)? x: -IJ*cos37.5º -ID*cos56.9º+(-26.3kN)*cos37.5º)=0 y: IJ*sin37.5º -ID*sin56.9º-10kN-(-26.3kN)*sin37.5º=0A particle rotates with constant speed in a circle. Let be the net torque on the particle and F the net force on the particle. Then: a. tau > 0 and F > 0 O b. tau > 0 and F = 0 . tau = 0 and F = 0 Od. tau = 0 and F > 0
- Using graphical approach, find the actual magnitude and direction of the resultant vector of D, E, F, G, H, I, J vectors. If 1 cm in the paper represents= 10ft/s, what is the magnitude of the resultant vector in ft/s? D= 3.3 cm E= 5.1 cm F= 3.6 cm G= 4.3cm H= 1.4 cm I= 2.35 cm J= 2.5 cmGiven uAF=1.5/6.5i+2/6.5j−6/6.5k, uAE=−1/3i−2/3j−2/3k, uAG=2/7i+3/7j−6/7k, and magnitude of the force FAF=1083 N. Find the projection on line AE of the force in cable AF.As shown in the figure below two boats are traveling in circular paths. Boat A is traveling at a constant vA = 19m/s and Boat B is traveling at a constant vB = 17m/s but in the opposite direction. Radius of Curvature for boat A is ρA = 199m, and for boat B is ρB = 110m. Find the e^tA component of v¯B/A in m/s: Find the e^nA component of v¯B/A in m/s: Find the e^tA component of a¯B/A in m/s^2: Find the e^nA component of a¯B/A in m/s^2: (Remember: Boat A is traveling at a constant vA = 19m/s and Boat B is traveling at a constant vB = 17m/s but in the opposite direction. Radius of Curvature for boat A is ρA = 199m, and for boat B is ρB = 110m) e^nA=−e^nB at this moment Select one: True False