The bigger galaxy survives the smaller one gets eaten up
Answer:
5.5 N at 50.8° north of west.
Explanation:
To find the resultant of these forces, we have to resolve each force along the x- and y-direction, then find the components of the resultant force, and then calculate the resultant force.
The three forces are:
(east)
(west)
(at 60° north of west)
Taking east as positive x-direction and north as positive y-direction, the components of the forces along the 2 directions are:
Threfore, the components of the resultant force are:
Therefore, the magnitude of the resultant force is
And the direction is:
And since the x-component is negative, it means that this angle is measured as north of west.
In nature there are two categories of microorganisms as relating to health. Microorganisms that are considered harmful to humans are called pathogens and these cause disease. Examples include bacteria such as streptococcus which cause sore throat and salmonella which cause typhoid disease.
There are some microorganisms which are helpful to man and they live mostly on the skin of man or in his gut and are mostly bacteria. They are collectively called bacterial normal flora.
In man the normal bacterial flora of the skin include staphylococcus found on dry skin, cornybacteria found in moist skin sites and propionibacteria in the sebaceous sites (head, neck, trunk) of the body. Normal bacterial flora of the gut include Escherichia coli.
One of the major function of bacterial flora is actually to protect our bodies by competing for space with pathogens preventing them from gaining a foothold in our bodies.
<span>At first when the car is at the datum
point where is no elevation, the kinetic energy increases while the potential
energy is zero. As it travel the path and goes upward the kinetic enrgy
decreases while the potential energy increases. When it goes down again the
kinetic enrgy increases again while the potential energy decreases </span>
We have the following equation for height:
h (t) = (1/2) * (a) * t ^ 2 + vo * t + h0
Where,
a: acceleration
vo: initial speed
h0: initial height.
The value of the acceleration is:
a = -g = -9.8 m / s ^ 2
For t = 0 we have:
h (0) = (1/2) * (a) * 0 ^ 2 + vo * 0 + h0
h (0) = h0
h0 = 0 (reference system equal to zero when the ball is hit).
For t = 5.8 we have:
h (5.8) = (1/2) * (- 9.8) * (5.8) ^ 2 + vo * (5.8) + 0
(1/2) * (- 9.8) * (5.8) ^ 2 + vo * (5.8) + 0 = 0
vo = (1/2) * (9.8) * (5.8)
vo = 28.42
Substituting values we have:
h (t) = (1/2) * (a) * t ^ 2 + vo * t + h0
h (t) = (1/2) * (- 9.8) * t ^ 2 + 28.42 * t + 0
Rewriting:
h (t) = -4.9 * t ^ 2 + 28.42 * t
The maximum height occurs when:
h '(t) = -9.8 * t + 28.42
-9.8 * t + 28.42 = 0
t = 28.42 / 9.8
t = 2.9 seconds.
Answer:
The ball was at maximum elevation when:
t = 2.9 seconds.
