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:
649kg/m^3
Explanation:
Let p be the density of this particular object.
Formula for density:
We can substitute the givenmass and volume to find density of the object.
Therefore the density of this object is 649kg/m^3.
The wave diagramed in blue.
Answer:
0.000001 kg
Explanation:
because 1 kg equal 1,000,000 milligrams
we take 
which equals 0.000001 kg
Answer:
The answers are options B,D and E
Explanation:
B) The particles in the liquid are slowly overcoming the forces of attraction and spreading out due to the thermal energy they are absorbing. This makes the liquid less dense as it slowly changes into a gas after reaching its boiling point.
D) The particles start absorbing the energy form the surroundings as latent heat of evaporation. They need this energy to overcome the strong forces of attraction between particles to change into the gaseous state
E) The particles have spaced out due to the thermal energy absorbed, making the liquid lighter and it rises upwards.
