Answer:

Explanation:
According to the question
net force F = 2.20×10^6 N
displacement 
from figure , the horizontal forces are same in magnitude and opposite direction.
so , neglect these two forces.
we can take only vertical components of the force.
total force F' = F cos 19° + F cos 19°
= 2×F×cos 19° ................. (1
therefore , total work is
W = F'S
= (2F cos19)×S


Answer:
Oxygen cycle
Explanation:
The components of the reservoirs of oxygen that are exchange in our environment is the oxygen cycle
It suggests the movement of oxygen between the living and non-living parts.
- The cycle does not account for oxygen that is trapped and cannot be exchanged in nature.
- Oxygen is important component of the atmosphere.
- Gaseous exchange between living organisms and atmosphere involves oxygen to a very large extent.
All will have a dominant trait I can't see the following statments
Answer:
The ball doesn't strike the building because it strikes the ground at d=1.62 meters.
Explanation:
V= 5 m/s < 70º
Vx= 1.71 m/s
Vy= 4.69 m/s
h= Vy * t - g * t²/2
clearing t for the flying time of the ball:
t= 0.95 s
d= Vx * t
d= 1.62 m
Answer:
V_{average} =
, V_{average} = 2 V
Explanation:
he average or effective voltage of a wave is the value of the wave in a period
V_average = ∫ V dt
in this case the given volage is a square wave that can be described by the function
V (t) = 
to substitute in the equation let us separate the into two pairs
V_average = 
V_average = 
V_{average} = 
we evaluate V₀ = 4 V
V_{average} = 4 / 2)
V_{average} = 2 V