Answer:
h> 2R
Explanation:
For this exercise let's use the conservation of energy relations
starting point. Before releasing the ball
Em₀ = U = m g h
Final point. In the highest part of the loop
Em_f = K + U = ½ m v² + ½ I w² + m g (2R)
where R is the radius of the curl, we are considering the ball as a point body.
I = m R²
v = w R
we substitute
Em_f = ½ m v² + ½ m R² (v/R) ² + 2 m g R
em_f = m v² + 2 m g R
Energy is conserved
Emo = Em_f
mgh = m v² + 2m g R
h = v² / g + 2R
The lowest velocity that the ball can have at the top of the loop is v> 0
h> 2R
It is true because <span>A pyramid of biomass is a representation of the amount of energy contained in biomass, at different trophic levels for a given point in time . The amount of energy available to one trophic level is limited by the amount stored by the level below. Because energy is lost in the transfer from one level to the next, there is successively less total energy as you move up trophic levels. Tree is a base as it provides food and energy.</span>
Answer:
-400km/hr
Explanation:
Velocity=displacement/time
=400/1
=400Km/hr
=-400km/hr (because west direction)
Answer:
Period of the signal.
Explanation:
So, this question is all about a concept in physics or astronomy which is called or known as Radiation Astronomy and Galactic Nuclei that are active. This concept talks most about Quasars; a powerful radiating object which derives its power from black holes.
When You take a look at Quasars, we get the to know that the more you think you can see, the more they move away from us.
Thus, when "You are observing the radiation from a distant active galaxy and you notice that the amplitude of the signal varies in strength regularly over a certain period. The maximum possible size for the source of this radiation can now be calculated from the "PERIOD OF THE SIGNAL.
NB: not the amplitude but the period.
The energy from the sun that reaches the corn is about two billionths.
