Answer:
78.315 m/s²
Explanation:
Amplitude, A = 4.70 cm = 0.047 m
Frequency, f = 6.50 Hz
Angular frequency, ω = 2 π f = 2 x 3.14 x 6.50 = 40.82 rad/s
Maximum acceleration, a = ω² A
a = 40.82 x 40.82 x 0.047
a = 78.315 m/s²
A day which in this case is longer than a year
What equation?
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The answer is 80 J of electrical energy
Answer:
Magnitude of the net force on q₁-
Fn₁=1403 N
Magnitude of the net force on q₂+
Fn₂= 810 N
Magnitude of the net force on q₃+
Fn₃= 810 N
Explanation:
Look at the attached graphic:
The charges of the same sign exert forces of repulsion and the charges of opposite sign exert forces of attraction.
Each of the charges experiences 2 forces and these forces are equal and we calculate them with Coulomb's law:
F= (k*q*q)/(d)²
F= (9*10⁹*3*10⁻⁶*3*10⁻⁶)(0.01)² =810N
Magnitude of the net force on q₁-
Fn₁x= 0
Fn₁y= 2*F*sin60 = 2*810*sin60° = 1403 N
Fn₁=1403 N
Magnitude of the net force on q₃+
Fn₃x= 810- 810 cos 60° = 405 N
Fn₃y= 810*sin 60° = 701.5 N

Fn₃ = 810 N
Magnitude of the net force on q₂+
Fn₂ = Fn₃ = 810 N