A train travels 250 km westward from Carthage to Johnson City. The train arrives 2.5 hours after it left. What was the average v
1 answer:
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Answer:
Fn = 246.2 N
Explanation:
I assume you meant meters per second squared for acceleration of the elevator. All you have to after your picture/free-body diagram is made is add the forces together. You could do this by finding them separately like I did or use Fn = m(Fg + Fe). I just calculated magnitude and didn't worry about signs.
Answer:
a) q₁ = 15. 28 10⁻⁶C, b) q₂ = 5.64 10⁻⁶ C
Explanation:
For this exercise we use Newton's second law where force is Coulomb's electric force
Case 1. Distance (x₁ = 0.200 m) from the third sphere
F₁ = F₁₃ - F₂₃
F₁ = k q₁q₃ / x₁² - k q₂ q₃ / (0.4 - x₁)²
F₁ = k q₃ (q₁ / x₁² - q₂ / (0.4- x₁)²
Case2 Distance (x₂ = 0.6 m) from the third sphere
F₂ = F₁₃ + F₂₃
F₂ = k q₁q₃ / x₂² + k q₂q₃ / (0.4- x₂)²
F₂ = k q₃ (q₁ / x₂² + q₂ / (0.4-x₂)²
The distance is between the spheres, in the annex you can see the configuration of the charge and forces
Let's replace the values
F₁ = 8.99 10⁹ 3.00 10⁻⁶⁶ (q₁ / 0.2² - q₂ / (0.4-0.2)²
F₂ = 8.99 10⁹ 3.00 10⁻⁶ (q₁ / 0.6² + q₂ / (0.4-0.6)²
6.50 = 674. 25 10³ (q₁ –q₂)
3.50 = 26.97 10³ (q₁ / 0.36 + q₂ / 0.04)
We have a system of two equations with two unknowns, let's solve it. Let's clear q1 in the first and substitute in the second
q₁ = q₂ + 6.50 / 674 10³
3.50 / 26.97 10³ = (q₂ + 9.64 10⁻⁶) /0.36 + q₂ / 0.04
1.2978 10⁻⁴ = q₂ / 0.36 + q₂ / 0.04 + 26.77 10⁻⁶
q₂ (1 / 0.36 + 1 / 0.04) = 129.78 10⁻⁶ + 26.77 10⁻⁶
q₂ 27,777 = 156,557 10⁻⁶
q₂ = 156.557 10-6 /27.777
q₂ = 5.636 10⁻⁶ C
We look for q1 in the other equation
q₁ = q₂ + 6.50 / 674 10³
q₁ = 5.636 10⁻⁶ + 9.6439 10⁻⁶
q₁ = 15. 28 10⁻⁶C
