I only need question 4, 6, and 8 to be answered. part 1 of 2 - Consider four vec
ID: 1532302 • Letter: I
Question
I only need question 4, 6, and 8 to be answered.
part 1 of 2
- Consider four vectors F~ 1, F~ 2, F~ 3, and F~ 4, where their magnitudes are F1 = 48 N, F2 = 23 N, F3 = 10 N, and F4 = 52 N. Let 1 = 130 , 2 = 140 , 3 = 17 , and 4 = 68 , measured from the positive x axis with the counter-clockwise angular direction as positive. What is the magnitude of the resultant vector F~ , where F~ = F~ 1 + F~ 2 + F~ 3 + F~ 4? Answer in units of N.
Question 4, part 2 of 2
-What is the direction of this resultant vector F~ ? Note: Give the angle in degrees, use counterclockwise as the positive angular direction, between the limits of 180 and +180 from the positive x axis. Answer in units of .
Question 5part 1 of 2
- Given two vectors F~ 1, and F~ 2. Where the magnitude of these vectors are F1 = 96 N , and F2 = 50 N . And where 1 = 298 , and 2 = 0 . The angles are measure from the positive x axis with the counter-clockwise angular direction as positive. What is the magnitude of the resultant vector kF~ k, where F~ = F~ 1 + F~ 2 ? Answer in units of N.
Question 6, part 2 of 2
-mNote: Give the angle in degrees, use counterclockwise as the positive angular direction, between the limits of 180 and +180 from the positive x axis. What is the direction of this resultant vector F~ ? Answer in units of .
Question 7, part 1 of 2
- A cannon fires a 0.677 kg shell with initial velocity vi = 8.4 m/s in the direction = 50 above the horizontal. The acceleration of gravity is 9.8 m/s 2 . x h 8.4 m/s 50 y y Shell’s trajectory curves downward because of gravity, so at the time t = 0.792 s the shell is below the straight line by some vertical distance. Your task is to calculate the distance h in the absence of air resistance. First, find what does h depend on (besides g): . . It depends only on the flight time t, and does not depend on the initial velocity vi or the initial angle .
Question 8, part 2 of 2
And now calculate the h. Answer in units of m.
Explanation / Answer
Breakdown each vector into its x and y components, where âx and ây
are unit vectors in the x and y directions
F1 = 48 < 130º48cos[130]•âx+ 48sin[130]•ây
F2 = 23 < -140º23cos[-140]•âx+ 23sin[-140]•ây
F3 = 10 < 17º10cos[17]•âx+ 10sin[17]•ây
F4 = 52 < -68º52cos[-68]•âx+ 52sin[-68]•ây
F1 = 48 < 130º=-30.85•âx+ 36.77•ây
F2 = 23 < -140º=-17.62•âx- 14.78•ây
F3 = 10 < 17º=9.56•âx+ 2.92•ây
F4 = 52 < -68º=19.48•âx 48.2•ây
<add>
F =-19.43•âx 23.29•ây
F= sqrt[(-19.43)^2 + (-23.29)^2]=29.57 Newtons.
= tan¹[(-23.29) (-19.43)]...[ ÷ ] 3rd quadrant
= 50º + 180º = 230º
= 50º 180º = -130º...since you prefer ±180º angles