Ke = (1/2)mv²
m = 100kg, v = 10 km/s = 10*1000 = 10000m/s
Ke = (1/2)*100*10000
Ke = 500000 Joules
Centre of Mass then axis of rotation and then moment of inertia. This was the toughest question for your level... happy to help ^_^. It was purely experimental question.
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:
The correct answer is option D i.e. A and C
Explanation:
The correct answer is option D i.e. A and C
for proficient catching player must
- learn to absorbed the ball force
- moves the hang according to ball direction to hold the ball
- to catch ball at high height move the finger at higher position
- to catch ball at low height move the finger at lower position
Answer:
yes you are totally right
