Answer:
Θ=0.01525 rad
or
Θ=0.87°
Explanation:
Given data
wavelength λ=2.5 µm =2.5×10⁻⁶m
Diameter d=0.20 mm =0.20×10⁻³m
To find
Angle Θ in radians and degree
Solution
Circular apertures have first dark fringe at
Θ=(1.22λ)/d
Substitute the given values
So
Θ=[1.22(2.5×10⁻⁶m)]/0.20×10⁻³m
Θ=0.01525 rad
or
Θ=0.87°
Down to the left, down the the right, criss cross!
The work output of a machine can be anything, depending on the friction
and other losses internal to the machine. But it can never be MORE than
the work <u>input</u>.
I suppose you might say that the work output can never be less than zero,
because there's no such thing as negative energy.
Explanation:
Formula to determine the critical crack is as follows.

= 1,
= 24.1
[/tex]\sigma_{y}[/tex] = 570
and, 
= 427.5
Hence, we will calculate the critical crack length as follows.
a = 
= 
= 
Therefore, largest size is as follows.
Largest size = 2a
= 
= 
Thus, we can conclude that the critical crack length for a through crack contained within the given plate is
.