Start with the conservation of energy. The spring potential energy and the mass' kinetic energy will fluctuate over time, but their sum will stay constant. The maximum spring potential energy equals the maximum kinetic energy.
0.5mv² = 0.5kx²
m is the mass, v is the maximum velocity, k is the spring constant, and x is the maximum displacement along the spring.
Given values:
m = 4.1kg
v = 0.78m/s
Calculate the maximum kinetic energy.
Max KE = 0.5mv² = 1.247J
Set this equal to the maximum spring potential energy.
Max spring PE = 0.5kx² = 1.247J
<em>x² = 2.494/k</em>
<em />
The spring force is F = kx
Max F = kx = 13N
x = 13/k
<em>x² = 169/k²</em>
Set both values of x² equal to each other and solve for k the spring constant:
2.494/k = 169/k²
2.494k = 169
k = 67.8N/m
Use k to find x:
Max F = kx = 13N
67.8x = 13
x = 0.192m
The frequency of the system is given by:
f = (1/(2π))√(k/m)
f is the frequency, k is the spring constant, m is the mass.
f = (1/(2π))√(67.8/4.1)
f = 0.647Hz
The answer is evolution. When a specifies evolves over time they change and adapt to their environment.
Answer:
As an additional security measure in an AA&E storage facilities, securing drainage structures must be considered if their cross section is greater than 96 inches and any dimension is greater than 6 inches, they must be barred and welded at the intersections to prevent any human from crawling into the area.
Explanation:
U.S. national security relys on ensuring DoD sensitive or classified assets such as classified material, arms, ammunition, and explosives (AA&E) and nuclear weapons, in properly safeguard facilites and storage containers, voiding DoD assets loss or compromise and additional to supplemental intrusion detection systems and guard
patrols, security lighting, communications, fences and clear zones, drainage structures, key control, and other security measures required to protect AA&E.
Answer:
Wavelength at which the light reflected by the film is brightest = 567.5 nm
Explanation:
We are given;
index of refraction n2 = 1.33
Thickness;(t) = 320 nm
Now the wavelength at which the light reflected by the film is brightest is gotten from the formula for path difference in critical interference as;
Path difference = (m + ½)(λ/n)
Where;
path difference = 2 x thickness = 2(320) = 640 nm
λ = Wavelength at which the light reflected by the film is brightest
n is Refractive index
m is an integer = 0,1,2,3...
Thus; at m = 0;
We have;
640 = (0 + ½)(λ/1.33)
640 = (λ/2.66)
λ = 640 x 2.66
λ = 1702.4 nm
at m = 1;
We have;
640 = (1 + ½)(λ/1.33)
640 = (3/2)(λ/1.33)
λ = 640 x 1.33 x 2/3
λ = 567.5 nm
at m = 2;
We have;
640 = (2 + ½)(λ/1.33)
640 = (5/2)(λ/1.33)
λ = 640 x 2 x 1.33/5
λ = 340.5 nm
Since we are told that the wavelength is between 400 – 690 nm.
Thus, the wavelength at which the light reflected by the film is brightest is the higher value gotten that is between 400nm and 690nm.
Thus, Wavelength at which the light reflected by the film is brightest = 567.5 nm
Answer:
Jackson 2: Smart 3: Ahmed
Explanation:
square + circle = egg
oops, wrong guy