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: I am getting a result around the 54,800 area, so I am selecting the 55,000 HP answer.
Answer:
Essentially, your thumb is the main piece of your body that has saddle joints. The bones in your seats joint are in charge of moving forward and backward, side to side.
When all is said in done, the piece of the thumb joint that is subjected to extreme anxiety is that known as CMC joint or carpometacarpal joint. This joint is fundamentally shaped by the metacarpal bone and it explains with the trapezium bone of the wrist.
Explanation:
To solve this problem we must apply the concept related to the longitudinal effort and the effort of the hoop. The effort of the hoop is given as
Here,
P = Pressure
d = Diameter
t = Thickness
At the same time the longitudinal stress is given as,
The letters have the same meaning as before.
Then he hoop stress would be,
And the longitudinal stress would be
The Mohr's circle is attached in a image to find the maximum shear stress, which is given as
Therefore the maximum shear stress in the pressure vessel when it is subjected to this pressure is 600Psi
<span>First of all, the maximum speed occurs when the object passes through the
equilibrium position
The kinetic energy when the object has this max speed is
K= 1/2 * mass * (1.25 m/s)^2
The potential energy in the spring when the speed is equal to zero
U= 1/2 * k * xmax^2
The maximun force of the spring is
mass*acceleration = k*xmax
m * 6.89 m/s2 = k * xmax
xmax = 6.89* m / k
0.5 * m * 1.56 = 0.5 * k * xmax^2
</span>m * 1.56 = k * (<span>6.89* m / k )^2 </span>
<span>
1.56 m = 47.47 m^2 / k
m/k = 0.032862
period = 2 *pi*sqrt[m/k]
= 2 pi </span><span>sqrt [ </span><span>0.032862]
= 1.139 s
A fourth of the period elapses between the instants of max acceleration and maximum speed
= 1/4* period
= 1/4 * </span><span><span>1.139 s </span>
= 0.284s </span>
