Answer: 6.36
Explanation:
Given
Radius of grindstone, r = 4 m
Initial angular speed of grindstone, w(i) = 8 rad/s
Final angular speed of the grindstone, w(f) = 12 rad/s
Time used, t = 4 s
Angular acceleration of the grinder,
α = Δw / t
α = w(f) - w(i) / t
α = (12 - 8) / 4
α = 4/4 = 1 rad/s²
Number of complete revolution in 4s =
Δθ = w(i).t + 1/2.α.t²
Δθ = 8 * 4 + 1/2 * 1 * 4²
Δθ = 32 + 1/2 * 16
Δθ = 32 + 8
Δθ = 40 rad/s
40 rad/s = 40/2π rpm = 6.36 rpm
Therefore, the grindstone does 6.36 revolutions during the 4 s interval
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)
Answer:
anyone know this or should i get my brother
the answer is CaO because that's what my homework says is correct