Answer:
Looks like you have:
a = -.324 cos 2.5 t
In this case ω^2 A = .324
ω = 2.5
f = ω / (2 * pi) = 2.5 / 6.28 = .40 / sec
Answer:
<h2>Magnitude of the second charge is

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Explanation:
According to columbs law;
F = 
F is the attractive or repulsive force between the charges = 12N
q1 and q2 are the charges
let q1 = - 8.0 x 10^-6 C
q2=?
r is the distance between the charges = 0.050m
k is the coulumbs constant =9*10⁹ kg⋅m³⋅s⁻⁴⋅A⁻²
On substituting the given values
12 = 9*10⁹*( - 8.0 x 10^-6)q2/0.050²
Cross multiplying

Acceleration = (Vf - Vi)/t
Since Vf= 60m/s
Vi= 15m/s
T= 15s
=> a= (60m/s - 15m/s)/15s
= 3
So the acceleration is 3m/s^2
<span>a. KE in electron volts is 1020 eV.
b. KE in Joules is e(1020) = (1.6022E-19)(1020) = 1.634E-16
c. KE = (1/2)mv^2, so v = sqrt[2*KE/m] = 18.94E6 m/s
note: m is the mass of an electron = 9.109e-31 kg
I hope my answer has come to your help. Thank you for posting your question here in Brainly.
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