(a) Suppose you charge a 2.5 F capacitor with two 1.5 volt batteries. How much charge was on each plate? 7.5 Correct: Your answe
r is correct. C (b) How many excess electrons were on the negative plate? electrons
1 answer:
Answer:
C = Q / V
Q = C * V = 2.5 F * 3 V = 7.5 Coulombs (assuming the batteries are in series)
N e = Q where N is the number of electrons and e the electronic charge
N = 7.5 / (1.60E-19) = 4.7E19 electrons
or 4.7 * 10^19 electrons
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Answer:
the answer might the number 2
Explanation:
Answer:
-56.9 m/s
Explanation:
Given:
Δy = -165 m
v₀ = 0 m/s
a = -9.8 m/s²
Find: v
v² = v₀² + 2aΔy
v² = (0 m/s)² + 2 (-9.8 m/s²) (-165 m)
v = -56.9 m/s
The answer to this is C could i please get the brainliest answer
" 30% " means " 0.30 "
30% ÷ 75 means (0.30 / 75) = <em>0.004</em>
Answer:
Explanation:
ME=mvh
m=0.0780kg
v=4.84ms⁻¹
h=5.36m
ME=0.0780*4.84*5.36
ME=2.02joules
