The effective spring constant of the system is 39.6 N/m
Explanation:
The frequency of oscillation of a spring-mass system is given by
where
k is the spring constant of the system
m is the mass
In this problem, we have:
f = 29 Hz is the frequency of vibration of the eyeball system
m = 7.5 g = 0.0075 kg is the mass
We can therefore re-arrange the equation to find the effective spring constant of the system. We find:
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Answer:
0.6
Explanation:
The volume of a sphere =
Therefore
r of the disc =
Using conservation of angular momentum;
The of the sphere =
of the disc =
= 0.6
Given that,
s = √(2t + 1)
Time, t = 4 s
Acceleration , a = ??
Since,
Acceleration = velocity / time
Velocity = distance/ time
Acceleration = distance/ time²
s/t² = √(2t+1)/t²
putting t = 4 sec, we have
a = √(2*4+1)/4²
a = √(5)/16
a= 0.139 ft/s²
Therefore, acceleration of the given particle will be 0.139 feet/ second².
Answer: Force F will be one-sixteenth of the new force when the charges are doubled and distance halved
Explanation:
Let the charges be q1 and q2 and the distance between the charges be 'd'
Mathematical representation of coulombs law will be;
F1=kq1q2/d²...(1)
Where k is the electrostatic constant.
If q1 and q2 is doubled and the distance halved, we will have;
F2 = k(2q1)(2q2)/(d/2)²
F2 = 4kq1q2/(d²/4)
F2 = 16kq1q2/d²...(2)
Dividing equation 1 by 2
F1/F2 = kq1q2/d² ÷ 16kq1q2/d²
F1/F2 = kq1q2/d² × d²/16kq1q2
F1/F2 = 1/16
F1 = 1/16F2
This shows that the force F will be one-sixteenth of the new force when the charges are doubled and distance halved