B. transition metals for sure
Answer:
Approximately 
(assuming that the projectile was launched at angle of 
above the horizon.)
Explanation:
Initial vertical component of velocity:
.
The question assumed that there is no drag on this projectile. Additionally, the altitude of this projectile just before landing 
is the same as the altitude 
at which this projectile was launched: 
.
Hence, the initial vertical velocity of this projectile would be the exact opposite of the vertical velocity of this projectile right before landing. Since the initial vertical velocity is 
(upwards,) the vertical velocity right before landing would be 
(downwards.) The change in vertical velocity is:
.
Since there is no drag on this projectile, the vertical acceleration of this projectile would be 
. In other words, 
.
Hence, the time it takes to achieve a (vertical) velocity change of 
would be:
.
Hence, this projectile would be in the air for approximately .
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.
<span> d = r*t is the basic distance equation
d = 6000 km
t with the tail wind = 6 hr
r with the tail wind = speed of the plane + wind speed = s + w
t with the head wind = 7.5 hr
r with the head wind = speed of the plane - wind speed = s-w
(s+w)*6 = 6000
(s-w)*7.5 = 6000
s + w = 1000
s - w = 800
</span><span> 2s = 1800
s = 900 km/h
s + w = 1000
w = 100
Check the anwer by calculating the return trip.
(900-100) * 7.5 = 800 * 7.5
800 * 7.5 = 6000 km
Answer: The rate of the jet in still air is 900 km/h. The rate of the wind is 100 km/hr.</span>
