Convection can best be observed as she blows the warm steam air that rises.
As the warm steam rises, she forces displaces it with cool air from her mouth. Because the warm steam is less dense it rises and because the cool air is more dense, it displaces the warm air.
This scenario is an example of convection.
This is a projectile motion problem, so, we use the formula for trajectory:
y =xtanα + gx^2/2v^2(cosα)^2
where
y is the vertical distance (y = 50 m)
x is the horizontal distance (x=90 m)
α is the angle of trajectory; since it levels of HORIZONTALLY, α = 0°
v is the initial velocity
g is the acceleration due to gravity which is 9.81 m/s^2
Substituting to the formula,
50 =90tan(0°) + (9.81)(90)^2/2v^2(cos0°)^2
v = 28.2 m/s
Answer:
The vacuum of space
Heat management problems
The difficulty of re-entry
Orbital mechanics
Micrometeorites and space debris
Cosmic and solar radiation
The logistics of having restroom facilities in a weightless
Explanation:
According to the elastic conservation momentum
m1v1 + m2v2 = m1V1 +m2V2
v: velocity before collision, V after collision
<span>the magnitude of the final velocity of the red marble
</span>m1v1 + m2v2 = m1V1 +m2V2
V2 is the final velocity we must find
V2 = 1 / m2 ( m1v1 + m2v2 - m1V1)
= 1/1.2 (3.5x15 + 1.2x 3.5 - 3.5x 5.5)= 63.29 cm/s
