You stretch a spring ball system 0.552 m away from its equilibrium point and watch it oscillate. You find that the system's angu
1 answer:
Answer:
The maximum speed of ball is 0.784 m/s.
Explanation:
Given that,
Amplitude = 0.552 m
Angular frequency = 1.42 rad/s
We need to calculate the maximum speed of ball
Using formula of speed
Where, A = amplitude
= angular frequency
Put the value into the formula
Hence, The maximum speed of ball is 0.784 m/s.
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Answer:
44 N
Explanation:
The electrostatic forces between two charges is given by:
where
k is the Coulomb's constant
q1 and q2 are the two charges
r is their separation
We notice that the force is directly proportional to the charges.
In this problem, initially we have a force of
F = 22 N
on a q2 = 4.0 C, exerted by a charge q1.
If the charge is doubled,
q2 = 8.0 C
This means that the force will also double, so it will be
Answer:
0.22 b
Explanation:
Quadrupole moment of the nucleon is,
And also,
And,
Now,
For Bismuth and A is 209.
Therefore, the expected value of quadrupole is 0.22 b which is quite related with experimental value which is 0.37 b