The sun is the mother star of the solar system, which only emits light to half of the planet, while the other part is always dark.
The sun emits light towards the earth, which dominates all life on earth. The movements of the Moon around the Earth and of the Earth around the Sun are complex. Movements of rotation around their own axes are superimposed on movements of orbital translation. The Earth and the Moon rotate around their own axes: This is rotation.
Answer:
Support at Cy = 1.3 x 10³ k-N
Support at Ay = 200 k-N
Explanation:
given:
fb = 300 k-N/m
fc = 100 k-N/m
D = 300 k-N
L ab = 6 m
L bc = 6 m
L cd = 6 m
To get the reaction A or C.
take summation of moment either A or C.
<em><u>Support Cy:</u></em>
∑ M at Ay = 0
(( x1 * F ) + ( D * Lab ) + ( D * L bc + D * L cd )
Cy = -------------------------------------------------------------------
( L ab + L bc )
Cy = 1.3 x 10³ k-N
<em><u>Support Ay:</u></em>
Since ∑ F = 0, A + C - F - D = 0
A = F + D - C
Ay = 200 k-N
“Charged objects have an imbalance of charge - either more negative electrons than positive protons or vice versa. And neutral objects have a balance of charge - equal numbers of protons and electrons. The principle stated earlier for atoms can be applied to objects. Objects with more electrons than protons are charged negatively; objects with fewer electrons than protons are charged positively.
In this discussion of electrically charged versus electrically neutral objects, the neutron has been neglected. Neutrons, being electrically neutral play no role in this unit. Their presence (or absence) will have no direct bearing upon whether an object is charged or uncharged. Their role in the atom is merely to provide stability to the nucleus.”
Hope this helps a bit.
!! (Credits to The Psychics Classroom) !!
Answer:
19.2*10^6 s
Explanation:
The equation for time dilation is:

Then, if it is observed to have a life of 6*10^6 s, and it travels at 0.95 c:

It has a lifetime of 19.2*10^6 s when observed from a frame of reference in which the particle is at rest.
The initial height of the first body is given by:

where
g is the gravitational acceleration
t is the time it takes for the body to reach the ground
Substituting t=1 s, we find

The second body takes takes t=2 s to reach the ground, so it was located at an initial height of

The second body started its fall 1 second before the first body, therefore when the second body started its fall, the first body was located at its initial height, i.e. at 4.9 m from the ground.