up off the ground a distance of 0.31 m before releasing her. Assume the partner’s velocity is zero at the beginning and the end
1 answer:
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1)
The electrostatic force between two charges is given by
(1)
where
k is the Coulomb's constant
q1, q2 are the two charges
r is the separation between the charges
When the two spheres are brought in contact with each other, the charge equally redistribute among the two spheres, such that each sphere will have a charge of
where Q is the total charge between the two spheres.
So we can actually rewrite the force as
And since we know that
r = 1.41 m (distance between the spheres)
(the sign is positive since the charges repel each other)
We can solve the equation for Q:
So, the final charge on the sphere on the right is
2)
Now we know the total charge initially on the two spheres. Moreover, at the beginning we know that
(we put a negative sign since the force is attractive, which means that the charges have opposite signs)
r = 1.41 m is the separation between the charges
And also,
So we can rewrite eq.(1) as
Solving for q1,
Since , we can substituting all numbers into the equation:
which gives two solutions:
Which correspond to the values of the two charges. Therefore, the initial charge q1 on the first sphere is
Answer:
The velocity of the student has after throwing the book is 0.0345 m/s.
Explanation:
Given that,
Mass of book =1.25 kg
Combined mass = 112 kg
Velocity of book = 3.61 m/s
Angle = 31°
We need to calculate the magnitude of the velocity of the student has after throwing the book
Using conservation of momentum along horizontal direction
Put the value into the formula
Hence, The velocity of the student has after throwing the book is 0.0345 m/s.