I would say D would be the answer. but is this even a real like homework question
Responda:
1) E = 6 × 10 ^ 6NC ^ -1 2) Q = 6 × 10 ^ -5
Explicação:
Dado o seguinte:
Carga (q) = 3uC = 3 × 10 ^ -6C
Força elétrica (Fe) = 18N
Intensidade do campo elétrico (E) =?
1)
Lembre-se:
Força elétrica (Fe) = carga (q) * Intensidade do campo elétrico (E)
Fe = qE; E = Fe / q
E = 18N / (3 × 10 ^ -6C)
E = 6N / 10 ^ -6C
E = 6 × 10 ^ 6NC ^ -1
2)
Lembre-se:
E = kQ / r ^ 2
E = intensidade do campo elétrico
Q = carga de origem
r = distância de espera = 30cm = 30/100 = 0,3m
K = 9,0 × 10 ^ 9
6 × 10 ^ 6 = (9,0 × 10 ^ 9 * Q) / 0,3 ^ 2
9,0 × 10 ^ 9 * Q = 6 × 10 ^ 6 * 0,09
Q = 0,54 × 10 ^ 6 / 9,0 × 10 ^ 9
Q = 0,06 × 10 ^ (6-9)
Q = 0,06 × 10 ^ -3
Q = 6 × 10 ^ -5 = 60 × 10 ^ -6 = 60μC
Answer:
10
Explanation:
This is tough. The last number 0.2 has only one significant figure. So while the sum of all the numbers is 12.3, you must only leave one sig figure. Rounding to the tenths gives 10.
D is the answer. Bc the pointy
The mass needed at peg 1 is a 5g mass.
The 15g should hang at peg 5.
The reason is force x distance clockwise is equal to force x distance anti-clockwise
