Lowered its infancy motel the weight by introducing increase prenatal and neonatal care for mothers have a social and cultural b
1 answer:
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Answer:
B. 7.07 m/s
Explanation:
The velocity of the stone when it leaves the circular path is its tangential velocity, , which is given by
where is the angular speed and
is the radius of the circular path.
is given by
where is the frequency of revolution.
Thus
Using values from the question,
<em>Note the conversion of 75 cm to 0.75 m</em>
Answer:
n=6.56×10¹⁵Hz
Explanation:
Given Data
Mass=9.1×10⁻³¹ kg
Radius distance=5.3×10⁻¹¹m
Electric Force=8.2×10⁻⁸N
To find
Revolutions per second
Solution
Let F be the force of attraction
let n be the number of revolutions per sec made by the electron around the nucleus then the centripetal force is given by
F=mω²r......................where ω=2π n
F=m4π²n²r...............eq(i)
as the values given where
Mass=9.1×10⁻³¹ kg
Radius distance=5.3×10⁻¹¹m
Electric Force=8.2×10⁻⁸N
we have to find n from eq(i)
n²=F/(m4π²r)