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
It's number three on your worksheet. ;)
The minimum initial velocity that the ball must have for it to reach the top of the hill is 21 m/s. The correct option is D.
<h3>What is mechanical energy?</h3>
The mechanical energy is the sum of kinetic energy and the potential energy of an object at any instant of time.
M.E = KE +PE
A boy is trying to roll a bowling ball up a hill. The friction is ignored. The ball must have to reach the top of the hill with a velocity. The acceleration due to gravity, g = 9.8 m/s²
The conservation of energy principle states that total mechanical energy remains conserved in all situations where there is no external force acting on the system.
M.E bottom of hill = M.E on top of hill
Kinetic energy + Potential energy = Kinetic energy + Potential energy
1/2 mu² + 0 = 0 + mgh
At the top of hill, the velocity will become zero. So, final kinetic energy is zero.
Substituting the values, we have
1/2 x u² = 9.8 x 22.5
u = sqrt [2 x9.8 x 22.5 ]
u= 21 m/s
Thus, the minimum initial velocity that the ball must have for it to reach the top of the hill is 21 m/s.
Learn more about mechanical energy.
brainly.com/question/13552918
#SPJ1
It's hard to tell what's going on down there in the corner with the resistor and the ammeter. There seems to be as many as 3 or 4 wires in and out of the ammeter, which would be wrong. A real ammeter only has two ... one in and one out. (Same for a resistor.)
It's hard to say whether this circuit works, until we can clearly understand how everything is hooked up in that corner of the drawing.
