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)
This heats up when an electric current passes through it, and produces light as a result. The resistance of a lamp increases as the temperature of its filament increases. The current flowing through a filament lamp is not directly proportional to the voltage across it.
151.9j
Explanation:
PE=1/2kx^2
PE=1/2(980)(.50)= 245j
PE=(1/2)(980)(.81)= 396.9j
396.9- 245= 151.9j
Answer:
Energy gained by the second particle = 12Uo
Explanation:
Given Data;
Resistant force = 12F
Initial kinetic energy = Uo
Calculating the kinetic energy gained, we have;
u = f *r
where f= resistant force = 20F
r = initial kinetic energy = Uo
Therefore,
U = 12 * uo
= 12 Uo
Therefore, energy gained by the second particle = 12Uo
