Angle, θ2 at which the light leaves mirror 2 is 56°
<u>Explanation:</u>
Given-
θ1 = 64°
So, α will also be 64°
According to the figure:
α + β = 90°
So,
β = 90° - α
= 90° - 64°
= 26°
β + γ + 120° = 180°
γ = 180° - 120° - β
γ = 180° - 120° - 26°
γ = 34°
γ + δ = 90°
δ = 90° - γ
δ = 90° - 34°
δ = 56°
According to the law of reflection,
angle of incidence = angle of reflection
θ2 = δ = 56°
Therefore, angle θ2 at which the light leaves mirror 2 is 56°
Answer:
p = 1.16 10⁻¹⁴ C m and ΔU = 2.7 10 -11 J
Explanation:
The dipole moment of a dipole is the product of charges by distance
p = 2 a q
With 2a the distance between the charges and the magnitude of the charges
p = 1.7 10⁻⁹ 6.8 10⁻⁶
p = 1.16 10⁻¹⁴ C m
The potential energie dipole is described by the expression
U = - p E cos θ
Where θ is the angle between the dipole and the electric field, the zero value of the potential energy is located for when the dipole is perpendicular to the electric field line
Orientation parallel to the field
θ = 0º
U = 1.16 10⁻¹⁴ 1160 cos 0
U1 = 1.35 10⁻¹¹ J
Antiparallel orientation
θ = 180º
cos 180 = -1
U2 = -1.35 10⁻¹¹ J
The difference in energy between these two configurations is the subtraction of the energies
ΔU = | U1 -U2 |
ΔU = 1.35 10-11 - (-1.35 10-11)
ΔU = 2.7 10 -11 J
Answer:
65
Explanation:
The resonant frequencies for a fixed string is given by the formula nv/(2L).
Where n is the multiple
.
v is speed in m/s
.
The difference between any two resonant frequencies is given by v/(2L)= fn+1 – fn
fundamental frequency means n=1
i.e fn+1 – fn = 390 -325
= 65
<h2>You input potential (stored) energy into the rubber band system when you stretched the rubber band back. Because it is an elastic system, this kind of potential energy is specifically called elastic potential energy.</h2><h2>Hope it helps..</h2>
Answer:
what help you need?????????
