B4 the tackle:
<span>The linebacker's momentum = 115 x 8.5 = 977.5 kg m/s north </span>
<span>and the halfback's momentum = 89 x 6.7 = 596.3 kg m/s east </span>
<span>After the tackle they move together with a momentum equal to the vector sum of their separate momentums b4 the tackle </span>
<span>The vector triangle is right angled: </span>
<span>magnitude of final momentum = √(977.5² + 596.3²) = 1145.034 kg m/s </span>
<span>so (115 + 89)v(f) = 1145.034 ←←[b/c p = mv] </span>
<span>v(f) = 5.6 m/s (to 2 sig figs) </span>
<span>direction of v(f) is the same as the direction of the final momentum </span>
<span>so direction of v(f) = arctan (596.3 / 977.5) = N 31° E (to 2 sig figs) </span>
<span>so the velocity of the two players after the tackle is 5.6 m/s in the direction N 31° E </span>
<span>btw ... The direction can be given heaps of different ways ... N 31° E is probably the easiest way to express it when using the vector triangle to find it</span>
Answer:
I think he answer is C but I could be wrong
Explanation:
brainliest plz
Answer:
- 5436 J
Explanation:
mass of car, m = 120 kg
radius of loop, r = 12 m
velocity at the bottom (A) = Va = 25 m/s
Velocity at the top(B) = Vb = 8 m/s
Vertical distance from A to B = diameter of loop, h = 2 x 12 = 24 m
by use of Work energy theorem
Work done by all the forces = change in kinetic energy of the body
Work done by the force + Work done by the friction = Kinetic energy at B - kinetic energy at A
- m x g x h + Work done by friction = 0.5 x 120 x (Vb^2 - Va^2)
- 120 x 9.8 x 24 + Work done by friction = 60 x (64 - 625)
- 28224 + Work done by friction = - 33660
Work done by friction = -33660 + 28224 = - 5436 J