Some clouds do not rain and others do because the clouds that do not rain do not have enough water in them. White clouds are ones that do not have enough water in them.
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True because sometimes a breathalyzer isn't as accurate as we think and if there is multiple sources of blood they would want to rule out who's blood is who's.
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:
X Component is 183.85N
Explanation:
The x component of the force on the block due to the rope;
X = F cos @ where if is the force, @ is the angle mad with the block.
X = F cos @
X = 240 cos 40
Cos 40= 0.7660, so
X = 240 × 0.7660
X component= 183.85N// rounded to two decimal places.
Answer:
The second law of a vibrating string states that for a transverse vibration in a stretched string, the frequency is directly proportional to the square root of the string's tension, when the vibrating string's mass per unit length and the vibrating length are kept constant
The law can be expressed mathematically as follows;
The second law of the vibrating string can be verified directly, however, the third law of the vibrating string states that frequency is inversely proportional to the square root of the mass per unit length cannot be directly verified due to the lack of continuous variation in both the frequency, 'f', and the mass, 'm', simultaneously
Therefore, the law is verified indirectly, by rearranging the above equation as follows;
From which it can be shown that the following relation holds with the limits of error in the experiment
m₁·l₁² = m₂·l₂² = m₃·l₃² = m₄·l₄² = m₅·l₅²
Explanation:
