Answer:
temperature, atmosphere pressure,wind, topography, humidity etc
The answer is most likely A
Based on the length of the Ethernet cable and the mass, the tension in the cable can be found to be 80 N.
<h3>How much tension is in the cable?</h3>
The tension in the cable can be found as:
= 4 x mass x length x frequency
Solving for the frequency is:
= 1 / (0.800 / 4)
= 1 / 0.20
= 5.0 Hz
The tension is therefore:
= 4 x 0.20 x 4.00 x 5
= 80N
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Answer:
Recall the Diffraction grating formula for constructive interference of a light
y = nDλ/w Eqn 1
Where;
w = width of slit = 1/15000in =6.67x10⁻⁵in =
6.67x10⁻⁵ x 0.0254m = 1.69x10⁻⁶m
D = distance to screen
λ = wavelength of light
n = order number = 1
Given
y1 = ? from 1st order max to the central
D = 2.66 m
λ = 633 x 10-9 m
and n = 1
y₁ = 0.994m
Distance (m) from the central maximum (n = 0) is the first-order maximum (n = 1) = 0.994m
Q b. How far (m) from the central maximum (m = 0) is the second-order maximum (m = 2) observed?
w = width of slit = 1/15000in =6.67x10⁻⁵in =
6.67x10⁻⁵ x 0.0254m = 1.69x10⁻⁶m
D = distance to screen
λ = wavelength of light
n = order number = 1
Given
y1 = ? from 1st order max to the central
D = 2.66 m
λ = 633 x 10⁻⁹ m
and n = 2
y₂ = 0.994m
Distance (m) from the central maximum (n = 0) is the first-order maximum (n = 2) =1.99m
Number of barrels are 3.0. Each barrel contains 42 gallons of oil. Thus, total volume of oil will be 42×3=126 gallons.
Converting gallons into m^{3}
1 gallon=0.00378 m^{3}
Thus, 126 gallons=0.4769 m^{3}
Thickness of oil film is 2.5\times 10^{2} nm, converting it into meters as follows:
1 nm=10^{-9} m
Thus,
2.5\times 10^{2} nm=1.5\times 10^{-7}m
Now, volume V of oil is related to area A and thickness T as follows:
V=A×T
rearranging,
A=\frac{V}{T}=\frac{0.4769 m^{3}}{2\times 10^{-7}m}=2.38\times 10^{6}m^{2}
Thus, square meters of oil will be 2.38\times 10^{6}m^{2}
