Answer:
The percentage of the mechanical energy of the oscillator lost in each cycle is 6.72%
Explanation:
Mechanical energy (Potential energy, PE) of the oscillator is calculated as;
PE = ¹/₂KA²
During the first oscillation;
PE₁ = ¹/₂KA₁²
During the second oscillation;
A₂ = A₁ - 0.0342A₁ = 0.9658A₁
PE₂ = ¹/₂KA₂²
PE₂ = ¹/₂K (0.9658A₁)²
PE₂ = (0.9658²)¹/₂KA₁²
PE₂ = (0.9328)¹/₂KA₁²
PE₂ = 0.9328PE₁
Percentage of the mechanical energy of the oscillator lost in each cycle;
Therefore, the percentage of the mechanical energy of the oscillator lost in each cycle is 6.72%
Constellations are based off of many Greek and Roman fables. Many of their gods and beliefs are pictured in the stars, which is where we get most of our constellations. Hope this helps!
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
D. there would be less energy transferred to the higher trophic levels
We have that the tension in rope 2 is mathematically given as
the tension in rope 2 is
From the question we are told
A crate hangs from a rope that is attached to a metal ring.
The metal ring is suspended by a <em>second </em>rope that is attached <u>overhead </u>at two points, as shown.
What is the angle if the tension in rope 1 is 0.640 times the tension in rope 2?
<h3> Angle of the tension</h3>
Generally the equation for the <em>Tension</em> is mathematically given as
Therefore
the tension in rope 2 is
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