A ball is spun around in circular motion such that it completes 50 rotations in 25 s.
1 answer:
Answer:
(A) The period of its rotation is 0.5 s (2) The frequency of its rotation is 2 Hz.
Explanation:
Given that,
a ball is spun around in circular motion such that it completes 50 rotations in 25 s.
(1). Let T be the period of its rotation. It can be calculated as follows :
(2). Let f be the frequency of its rotation. It can be defined as the number of rotations per unit time. So,
Hence, this is the required solution.
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Answer:
a) F₃₁ = 63.0 μN
b) F₃₂ = - 14.0 μN
c) q₂ = - 5.0 nC
Explanation:
a)
- Assuming that the three charges can be taken as point charges, the forces between them must obey Coulomb's Law, and can be found independent each other, applying the superposition principle.
- So, we can find the force that q₁ exerts along the x-axis on q₃, as follows:
b)
- Since total force exerted by q₁ and q₂ on q₃ is 49.0 μN, we can find the force exerted only by q₂ (which is along the x-axis only too) just by difference, as follows:
c)
- Finally, in order to find the value of q₂, as we know the value and sign of F₃₂, we can apply again the Coulomb's Law, solving for q₂, as follows: