Answer:
Explanation:
Part A) Using
light intensity I= P/A
A= Area= π (Radius)^2= π((0.67*10^-6m)/(2))^2= 1.12*10^-13 m^2
Radius= Diameter/2
P= power= 10*10^-3=0.01 W
light intensity I= 0.01/(1.12*10^-13)= 9*10^10 W/m^2
Part B) Using
I=c*ε*E^2/2
rearrange to solve for E= 
((I*2)/(c*ε))
c is the speed of light which is 3*10^8 m/s^2
ε=permittivity of free space or dielectric constant= 8.85* 10^-12 F⋅m−1
I= the already solved light intensity= 8.85*10^10 W/m^2
amplitude of the electric field E= (9*10^10 W/m^2)*(2) / (3*10^8 m/s^2)*(8.85* 10^-12 F⋅m−1)
---> E= (1.8*10^11) / (2.66*10^-3) = (6.8*10^13) = 8.25*10^6 V/m
Answer:
x = -1.20 m
y = -1.12 m
Explanation:
as we know that four masses and their position is given as
5.0 kg (0, 0)
2.9 kg (0, 3.2)
4 kg (2.5, 0)
8.3 kg (x, y)
As we know that the formula of center of gravity is given as
Similarly for y direction we have
Answer:
B. the stars to come back to the same positions in the sky.
Explanation:
In fact, the solar day is equivalent to more than a rotation, because when the point has turned completely, it is not, as it should, in the same position with respect to the Sun.
The reason for this is that while performing the rotation, the Earth simultaneously moved following its orbit around the Sun.
When the reference point completed its rotation, the Earth already moved almost 2,500,000 km., So that to see the Sun again it will be necessary to turn a little more.
Solar day is more than a rotation. The sidereal or sidereal day, commonly used by astronomers, is also based on the rotation of the Earth; but in this case a distant star is taken as a reference (sidereal comes from the Latin sidus which means "star").
The effective acceleration or deceleration due to gravity depends on the inclined angle of the track relative to ground; the steeper the slope is the greater the effective acceleration.
