Answer:
α = 141.5° (counterclockwise)
Explanation:
If
q₁ = +q
q₂ = -q
q₃ < 0
b = 2*a
We apply Coulomb's Law as follows
F₁₃ = K*q₁*q₃ / d₁₃² = + K*q*q₃ / (2*a)² = + K*q*q₃ / (4*a²)
F₂₃ = K*q₂*q₃ / d₂₃² = - K*q*q₃ / (5*a²)
(d₂₃² = a² + (2a)² = 5*a²)
Then
∅ = tan⁻¹(2a/a) = tan⁻¹(2) = 63.435°
we apply
F₃x = - F₂₃*Cos ∅ = - (K*q*q₃ / (5*a²))* Cos 63.435°
⇒ F₃x = - 0.0894*K*q*q₃ / a²
F₃y = - F₂₃*Sin ∅ + F₁₃
⇒ F₃y = - (K*q*q₃ / (5*a²))* Sin 63.435° + (K*q*q₃ / (4*a²))
⇒ F₃y = 0.0711*K*q*q₃ / a²
Now, we use the formula
α = tan⁻¹(F₃y / F₃x)
⇒ α = tan⁻¹((0.0711*K*q*q₃ / a²) / (- 0.0894*K*q*q₃ / a²)) = - 38.5°
The real angle is
α = 180° - 38.5° = 141.5° (counterclockwise)
Answer:
Neither A or B
Explanation:
The 37.3mv is not the signal voltage
sensor ground circuit does not has excessive resistance.
ANSWER i believe it's A. wax, which is made up of atoms that share electrons equally
Answer:
I₃/Io % = 0.8.59
Explanation:
A polarizer is a complaint sheet for light in the polarization direction and blocks the perpendicular one. When we use two polarizers the transmission between them is described by Malus's law
I = I₀ cos² θ
Let's apply the previous exposures in our case, the light is indicatively not polarized, so the first polarized lets half of the light pass
I₁ = ½ I₀
The light transmitted by the second polarizer
I₂ = I₁ cos² θ
I₂ = (½ I₀) cos2 28
The transmission by the polarizing third is
I₃ = I₂ cos² θ₃
The angle of the third polarizer with respect to the second is
θ₃ = 90-28
θ₃ = 62º
I₃ = (½ I₀ cos² 28 cos² 62)
Let's calculate
I₃ = Io ½ 0.7796 0.2204
I₃ = Io 0.0859
I₃/Io= 0.0859 100
I₃/Io % = 0.8.59
Answer:
Explanation:
Given:
Initial velocity, u = 0 m/s (at rest)
Final velocity, v = 22 m/s
Time, t = 9 s
Diameter, d = 58 cm
Radius, r = 0.29 m
Using equation of motion,
v = u + at
a = (22 - 0)/9
= 2.44 m/s^2
v^2 = u^2 + 2a × S
S = (22^2 - 0^2)/2 × 2.44
= 99.02 m
S = r × theta
Theta = 99.02/0.29
= 341.44 °
1 rev = 360°
341.44°,
= 341.44/360
= 0.948 rev
= 0.95 rev
B.
Final angular speed, wf = v/r
= 22/0.29
= 75.86 rad/s