Answer: y = 2.4×10^-6m or y= 2.4μm
Explanation: The formulae for the distance between the central bright fringe to any other fringe in pattern is given as
y = R×mλ/d
Where y = distance between nth fringe and Central bright spot fringe.
m = position of fringe = 4
λ = wavelength of light= 600nm = 600×10^-9 m
d = distance between slits = 1.50×10^-5m
R = distance between slit and screen = 2m
y = 2 × 4 × 600×10^-9/2
y = 4800×10^-9/2
y = 2400 × 10^-9
y = 2.4×10^-6m or y= 2.4μm
Answer:
minimum length of a surface crack is 18.3 mm
Explanation:
Given data
plane strain fracture toughness K = 82.4 MPa m1/2
stress σ = 345 MPa
Y = 1
to find out
the minimum length of a surface crack
solution
we will calculate length by this formula
length = 1/π ( K / σ Y)²
put all value
length = 1/π ( K / σ Y)²
length = 1/π ( 82.4
/ 345× 1)²
length = 18.3 mm
minimum length of a surface crack is 18.3 mm
The Factors are Temperature and kinetic energy, the temperature is because the particles are going to move fast which means the particles in a solid container or in solid pattern they will actually vibrate and they will expand , when temperature increase, more kinetic energy between the particles .
examples is that in the steel when you heat a steel the particles inside it will vibrate then the particles speed up because the vibration increases, therefore the temperature increases so a a thermal expansion occurs that the vibration of the particles will take up more space so the steel bar expands slightly in all Direction if the temperature Falls the reverse happens and the material or steel contracts which means get smaller .
another example is the thermometer, the thermometer has a liquid inside it which is Mercury or alcohol this liquid expands when the temperature rises, the tube is made narrow so that a small increase in volume of the liquid produces a large movement along the tube.
<span>If my memory serves me well, sensory receptors which would lead you to squint in bright light are called </span><span>C. photoreceptors</span>
Answer:
60 m
Explanation:
After 3 seconds of travel at 20 m/s, the projectile is 3·20 = 60 meters horizontally from the cannon.
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The vertical height after 3 seconds is 0.9 m, so the straight-line distance from cannon to target is √(60^2 +0.9^2) ≈ 60.007 meters.