Answer:
The possible frequencies for the A string of the other violinist is 457 Hz and 467 Hz.
(3) and (4) is correct option.
Explanation:
Given that,
Beat frequency f = 5.0 Hz
Frequency f'= 462 Hz
We need to calculate the possible frequencies for the A string of the other violinist
Using formula of frequency
...(I)
...(II)
Where, f= beat frequency
f₁ = frequency
Put the value in both equations
Hence, The possible frequencies for the A string of the other violinist is 467 Hz and 457 Hz.
first we make a U turn and travel towards home in t = 20 s
so the distance of home from initial position is
Now after picking up the book we travel back with speed 20 m/s
so again after t = 20 s the displacement is given as
so the net displacement is given as
so it will be displaced by total displacement 200 m
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)