Point charge q1 of 30 nC is separated by 50 cm from point charge q2 of -45 nC. As shown in the diagram, point a is located 30 cm
from q1 and 20 cm q2. What are the magnitude and direction of the electric field at point a? Let the electrostatic constant k= 8.99 x 10^9 N x m^2/C^2
1 answer:
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Answer:
Angular velocity is same as frequency of oscillation in this case.
ω = x
Explanation:
- write the equation F(r) = -K with angular momentum <em>L</em>
- Get the necessary centripetal acceleration with radius r₀ and make r₀ the subject.
- Write the energy of the orbit in relative to r = 0, and solve for "E".
- Find the second derivative of effective potential to calculate the frequency of small radial oscillations. This is the effective spring constant.
- Solve for effective potential
- ω = x
The spring constant is 181.0 N/m
Explanation:
We can solve the problem by applying the law of conservation of energy. In fact, the elastic potential energy initially stored in the compressed spring is completely converted into gravitational potential energy of the dart when the dart is at its maximum height. Therefore, we can write:
where the term on the left represents the elastic potential energy of the spring while the term on the right is the gravitational potential energy of the dart at maximum height, and where
k is the spring constant of the spring
x = 2.08 cm = 0.0208 m is the compression of the spring
m = 12.3 g = 0.00123 kg is the mass of the dart
is the acceleration due to gravity
h = 3.25 m is the maximum height of the dart
Solving for k, we find:
Learn more about potential energy:
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Explanation:
The general equation of an AC current is given by :
Where
I₀ is the peak value of current
is angular frequency
So,
We know that,
So, the frequency is 50 Hz and the maximum rms value of current is 14.14 A.