Answer:
Distance is directly proportional to the velocity
Explanation:
In 1929, Edwin Hubble's wrote an article that talked about relationship between the distance and recession speed/velocity of galaxies which led to what is known as the Hubble Law. This law states that galaxies are moving away from the earth at velocities proportional to their distances.
Thus is written as;
v = H_o•d
Where;
v is velocity
d is distance
H_o is Hubble's constant rate of cosmic expansion.
He came to this conclusion by generating a graph known as Hubble's classic graph which was a graph of observed velocity vs distance for nearby galaxies.
Deltas are complex depositional landforms that develop at the mouths of rivers. They are composed of sediment that is deposited as a river enters a standing body of water and loses forward momentum. Famous deltas include the Mississippi delta in Louisiana and the Nile delta in Egypt.
While falling, both the sheet of paper and the paper ball experience air resistance. But the surface area of the sheet is much more than that of the spherical ball. And air resistance varies directly with surface area. Hence the sheet experiences more air resistance than the ball and it falls more slowly than the paper ball.
Hope that helps!
Answer:
D. The oxygen side is partially negative because electrons are pulled toward the oxygen side.
Explanation:
The water molecule is polar by the virtue of covalent bonds and the hydrogen bonds within and between its molecule.
The oxygen side is partially negative because the electrons are pulled toward the oxygen side.
Between oxygen and hydrogen that makes up the water molecule, oxygen is more electronegative.
An electronegative atom has more affinity for electrons. Since the electrons in the molecule of water is shared between hydrogen and oxygen, the more electronegative specie which is water draws the electron more to itself.
This leaves a net negative charge on the oxygen atom.
Answer:
x=4.06m
Explanation:
A body that moves with constant acceleration means that it moves in "a uniformly accelerated movement", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.
When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.
Vf=Vo+a.t (1)\\\\
{Vf^{2}-Vo^2}/{2.a} =X(2)\\\\
X=Xo+ VoT+0.5at^{2} (3)\\
Where
Vf = final speed
Vo = Initial speed
T = time
A = acceleration
X = displacement
In conclusion to solve any problem related to a body that moves with constant acceleration we use the 3 above equations and use algebra to solve
for this problem
Vf=7.6m/s
t=1.07
Vo=0
we can use the ecuation number one to find the acceleration
a=(Vf-Vo)/t
a=(7.6-0)/1.07=7.1m/s^2
then we can use the ecuation number 2 to find the distance
{Vf^{2}-Vo^2}/{2.a} =X
(7.6^2-0^2)/(2x7.1)=4.06m
