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:
Since strong nuclear forces involve only nuclear particles (not electrons, bonds, etc) items 3 and 4 are eliminated.
Again item 2 refers to bonds between atoms and is eliminated.
This leaves only item 1.
Nuclear forces are very short range forces between components of the nucleus.
Weak nuclear forces are trillions of times smaller than strong forces.
Gravitational forces are much much smaller than the weak nuclear force.
Here We can use principle of angular momentum conservation
Here as we know boy + projected mass system has no external torque
Since there is no torque so we can say the angular momentum is conserved

now we know that
m = 2 kg
v = 2.5 m/s
L = 0.35 m
I = 4.5 kg-m^2
now plug in all values in above equation

![1.75 = [4.5 + 0.245]\omega](https://tex.z-dn.net/?f=1.75%20%3D%20%5B4.5%20%2B%200.245%5D%5Comega)


so the final angular speed will be 0.37 rad/s
The answer is D) Velocity
Solution :
Speed of the air craft,
= 262 m/s
Fuel burns at the rate of,
= 3.92 kg/s
Rate at which the engine takes in air,
= 85.9 kg/s
Speed of the exhaust gas that are ejected relative to the aircraft,
=921 m/s
Therefore, the upward thrust of the jet engine is given by

F = 85.9(921 - 262) + (3.92 x 921)
= 4862635.79 + 3610.32
= 
Therefore thrust of the jet engine is
.