<span>Assuming pulley is frictionless. Let the tension be ‘T’. See equation below.</span>
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The pilot might be correct (I think), because, if the gravity of the planet is strong, then the planet’s gravity will pull the spaceship into its orbit, so the engines don’t need to be on for the ship to get pushed toward the planet.
<h2><em>A reference point is a place or object used for comparison to determine if something is in motion. An object is in motion if it changes position relative to a reference point. You assume that the reference point is stationary, or not moving.</em></h2>
Answer:
a) 1.082 × 10⁻¹⁹C ( e = 1.6 × 10⁻¹⁹C)
b) 3.466 × 10¹¹ N/C
Explanation:
a)
p(r) = -A exp ( - 2r/a₀)
Q = ₀∫^∞ ₀∫^π ₀∫^2xπ p(r)dV = -A ₀∫^∞ ₀∫^π ₀∫^2π exp ( - 2r/a₀)r² sinθdrdθd∅
Q = -4πA ₀∫^∞ exp ( - 2r/a₀)r²dr = -e
now using integration by parts;
A = e / πa₀³
p(r) = - (e / πa₀³) exp (-2r/a₀)
Now Net charge inside a sphere of radius a₀ i.e Qnet is;
= e - (e / πa₀³) ₀∫^a₀ ₀∫^π ₀∫^2π r² exp (-2r/a₀)dr
= e - e + 5e exp (-2) = 1.082 × 10⁻¹⁹C ( e = 1.6 × 10⁻¹⁹C)
b)
Using Gauss's law,
E × 4πa₀ ² = Qnet / ∈₀
E = 4πa₀ ² × Qnet × 1/a₀²
E = 3.466 × 10¹¹ N/C
r(t) models the water flow rate, so the total amount of water that has flowed out of the tank can be calculated by integrating r(t) with respect to time t on the interval t = [0, 35]min
∫r(t)dt, t = [0, 35]
= ∫(300-6t)dt, t = [0, 35]
= 300t-3t², t = [0, 35]
= 300(35) - 3(35)² - 300(0) + 3(0)²
= 6825 liters
