The answer is D, human.
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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
Answer:
63
Explanation:
You first have to add all the numbers together.
22+72+79+72+70 = 315
You divide the total by the amount of numbers (5)
315/5 = 63
The mean is 63
The coefficient of friction between the road and the car's tire is determined as 0.78.
<h3>Acceleration of the car</h3>
The acceleration of the car is calculated as follows;
v² = u² - 2as
0 = u² - 2as
a = u²/2s
where;
- u is the initial velocity = 97 km/h = 26.94 m/s
a = (26.94)²/(2 x 47)
a = 7.72 m/s²
<h3>Coefficient of friction</h3>
μ = a/g
μ = (7.72)/9.8
μ = 0.78
Learn more about coefficient of friction here: brainly.com/question/14121363
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