The wavelength, which represents the size of the smallest detectable detail that uses ultraviolet light , is calculated as follows: 3×
/ 1.72×
or approximately 1.74×
m.
The distance between the two positive, two negative, or two minimal points on the waveform is known as the wavelength of the wave. The following formula expresses the relationship between the frequency and wavelength of light:
f = c / λ
where, f = frequency of light
c = speed of light
λ = wavelength of light
Given data = f = 1.72×Hz
Therefore, λ = 3× / 1.72×
λ = 1.74×m
The wavelength, which represents the size of the smallest detectable detail that uses ultraviolet light , is calculated as follows: 3× / 1.72× or approximately 1.74×m.
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Answer:
Yes there is if number of bulb is high the bulbs wouldn't glow much brighter
Explanation:
We assign the variables: T as tension and x the angle of the string
The <span>centripetal acceleration is expressed as v²/r=4.87²/0.9 and (0.163x4.87²)/0.9 = </span><span>T+0.163gcosx, giving T=(0.163x4.87²)/0.9 – 0.163x9.8cosx.
</span>
<span>(1)At the bottom of the circle x=π and T=(0.163x4.87²)/0.9 – .163*9.8cosπ=5.893N. </span>
<span>(2)Here x=π/2 and T=(0.163x4.87²)/0.9 – 0.163x9.8cosπ/2=4.295N. </span>
<span>(3)Here x=0 and T=(0.163x4.87²)/0.9 – 0.163x9.8cos0=2.698N. </span>
<span>(4)We have T=(0.163v²)/0.9 – 0.163x9.8cosx.
</span><span>This minimum v is obtained when T=0 </span><span>and v verifies (0.163xv²)/0.9 – 0.163x9.8=0, resulting to v=2.970 m/s.</span>
Answer:
<em>1.228 x </em>
<em> mm </em>
<em></em>
Explanation:
diameter of aluminium bar D = 40 mm
diameter of hole d = 30 mm
compressive Load F = 180 kN = 180 x 
N
modulus of elasticity E = 85 GN/m^2 = 85 x 
Pa
length of bar L = 600 mm
length of hole = 100 mm
true length of bar = 600 - 100 = 500 mm
area of the bar A = 
= 
= 1256.8 mm^2
area of hole a = 
= 
= 549.85 mm^2
Total contraction of the bar = 
total contraction = 
==> 
= <em>1.228 x </em><em> mm </em>
