The magnitude of the vector C is 96.32m
<h3>How to solve for the magnitude of vector c</h3>
Ax = AcosθA
= 40 cOS 20
= 37.59
Ay = AsinθA
-40sin20
= -13.68
Bx = B cos θ B
= 75Cos50
= 48.21
By = BsinθB
= 75sin50
= 57.45
Cx = AX + Bx
= 37.59 + 48.21
= 85.8
Cy = Ay + By
= -13.65 + 57.45
= 43.77
The magnitude is solved by
|c| = 
= √85.8² + 43.77²
= 96.32m
The magnitude of the vector c is 96.32m
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Answer:
Field, In physics, a region in which each point is affected by a force.
Explanation:
Newton's 2nd law of motion:
Force = (mass) x (acceleration)
= (0.314 kg) x (164 m/s²)
= 51.5 newtons
(about 11.6 pounds).
Notice that the ball is only accelerating while it's in contact with the racket. The instant the ball loses contact with the racket, it stops accelerating, and sails off in a straight line at whatever speed it had when it left the strings.
~ I hope this helped, and I would appreciate Brainliest. ♡ ~ ( I request this to all the lengthy answers I give ! )
<span>Answer:
So this involves right triangles. The height is always 100. Let the horizontal be x and the length of string be z.
So we have x2 + 1002 = z2. Now take its derivative in terms of time to get
2x(dx/dt) = 2z(dz/dt)
So at your specific moment z = 200, x = 100âš3 and dx/dt = +8
substituting, that makes dz/dt = 800âš3 / 200 or 4âš3.
Part 2
sin a = 100/z = 100 z-1 . Now take the derivative in terms of t to get
cos a (da./dt) = -100/ z2 (dz/dt)
So we know z = 200, which makes this a 30-60-90 triangle, therefore a=30 degrees or π/6 radians.
Substitute to get
cos (Ď€/6)(da/dt) = (-100/ 40000)(4âš3)
âš3 / 2 (da/dt) = -âš3 / 100
da/dt = -1/50 radians</span>
