To solve this problem we apply the thermodynamic equations of linear expansion in bodies.
Mathematically the change in the length of a body is subject to the mathematical expression
Where,
Initial Length
Thermal expansion coefficient
Change in temperature
Since we have values in different units we proceed to transform the temperature to degrees Celsius so
The coefficient of thermal expansion given is
The initial length would be,
Replacing we have to,
This means that the building will be 35.5cm taller
D, electron, the nucleus is not a single particle to begin with, the proton has a positive charge, a neutron has a neutral charge or no charge, and an electron has a negative charge
Answer:
50% of it .
Explanation:
50% of it is illuminated by the Sun.
A) a mouse, to an order of magnitude = 0.1 m ( a tenth of a meter ) That would be a big mouse but the alternatives are 1 meter or one hundredth of a meter... so go with 1/10th
<span>b) Easy = 1 meter </span>
<span>c) two choices 10m or 100 m . Go with 100 m </span>
<span>d) Stretch it out , trunk tip to tail tip - call it 10 m </span>
<span>e) Your choice 100 m or 1000 m..... These are estimates. So long as you are within one order of magnitude you can't really be given wrong. So I'd say 100m</span>
Answer:
spacing between the slits is 405.32043 ×
m
Explanation:
Given data
wavelength = 610 nm
angle = 2.95°
central bright fringe = 85%
to find out
spacing between the slits
solution
we know that spacing between slit is
I = 4
× cos²∅/2
so
I/4 = cos²∅/2
here I/4 is 85 % = 0.85
so
0.85 = cos²∅/2
cos∅/2 = √0.85
∅ = 2 ×
0.921954
∅ = 45.56°
∅ = 45.56° ×π/180 = 0.7949 rad
and we know that here
∅ = 2π d sinθ / wavelength
so
d = ∅× wavelength / ( 2π sinθ )
put all value
d = 0.795 × 610× / ( 2π sin2.95 )
d = 405.32043 × m
spacing between the slits is 405.32043 × m
