Ω = 2.81
A = 0.232
k = 29.8
x = A cos(ωt + Ф)
at t = 0:
x = A = A cos(ωt + Ф) = A cos(Ф)
Ф = 0
at t = 1.42, with Ф = 0:
x = A cos(ωt)
U = 1/2 k x² = 1/2 k [A cos(ωt)]²
Answer: An exercise, activity, or workout that targets a specific training need Progression - Gradually increasing your skill level and ability over time Overload -Working your body harder than usual to improve how it functions Reversibility - Losing what you have gained by taking time off Tedium - getting bored by doing the same thing
Explanation:
Answer:
Epx= - 21.4N/C
Epy= 19.84N/C
Explanation:
Electric field theory
The electric field at a point P due to a point charge is calculated as follows:
E= k*q/r²
E= Electric field in N/C
q = charge in Newtons (N)
k= electric constant in N*m²/C²
r= distance from load q to point P in meters (m)
Equivalences
1nC= 10⁻⁹C
known data
q₁=-2.9nC=-2.9 *10⁻⁹C
q₂=5nC=5 *10⁻⁹C
r₁=0.840m
Calculation of the electric field at point P due to q1
Ep₁x=0
Calculation of the electric field at point P due to q2
Calculation of the electric field at point P(0,0) due to q1 and q2
Epx= Ep₁x+ Ep₂x==0 - 21.4N/C =- 21.4N/C
Epy= Ep₁y+ Ep₂y=36.95 N/C-17.11N =19.84N/C
Displacement from the center line for minimum intensity is 1.35 mm , width of the slit is 0.75 so Wavelength of the light is 506.25.
<h3>How to find Wavelength of the light?</h3>
When a wave is bent by an obstruction whose dimensions are similar to the wavelength, diffraction is observed. We can disregard the effects of extremes because the Fraunhofer diffraction is the most straightforward scenario and the obstacle is a long, narrow slit.
This is a straightforward situation in which we can apply the
Fraunhofer single slit diffraction equation:
y = mλD/a
Where:
y = Displacement from the center line for minimum intensity = 1.35 mm
λ = wavelength of the light.
D = distance
a = width of the slit = 0.75
m = order number = 1
Solving for λ
λ = y + a/ mD
Changing the information that the issue has provided:
λ = 1.35 * 10^-3 + 0.75 * 10^-3 / 1*2
=5.0625 *10^-7 = 506.25
so
Wavelength of the light 506.25.
To learn more about Wavelength of the light refer to:
brainly.com/question/15413360
#SPJ4
