Answer:
Explanation:
Given that
Beam current (i)=23.3µA
And the time to strike(t)=28s
Also, a fundamental charge e=1.602×10^-19C
Then, the charge quantity is given as,
q=it
Then, q=23.3×10^-6×28
q=6.524×10^-4C
Also, the number of electron N is given as
q=Ne
Therefore, N=q/e
So, N=6.524×10-4/1.602×10^-19
N=4.072×10^15
There are 4.072×10^15 electrons strike the tube screen every 28 s.
Answer:
Electric flux;
Φ = 30.095 × 10⁴ N.m²/C
Explanation:
We are given;
Charge on plate; q = 17 µC = 17 × 10^(-6) C
Area of the plates; A_p = 180 cm² = 180 × 10^(-4) m²
Angle between the normal of the area and electric field; θ = 4°
Radius;r = 3 cm = 3 × 10^(-2) m = 0.03 m
Permittivity of free space;ε_o = 8.85 × 10^(-12) C²/N.m²
The charge density on the plate is given by the formula;
σ = q/A_p
Thus;
σ = (17 × 10^(-6))/(180 × 10^(-4))
σ = 0.944 × 10^(-3) C/m²
Also, the electric field is given by the formula;
E = σ/ε_o
E = (0.944 × 10^(-3))/(8.85 × 10^(-12))
E = 1.067 × 10^(8) N/C
Now, the formula for electric flux for uniform electric field is given as;
Φ = EAcos θ
Where A = πr² = π × 0.03² = 9π × 10^(-4) m²
Thus;
Φ = 1.067 × 10^(8) × 9π × 10^(-4) × cos 4
Φ = 30.095 × 10⁴ N.m²/C
Force, F = ma
Where m = mass in kg, a = acceleration in m/s², Force, F is in N.
F = ma
2000 = m*2.2
2.2m = 2000
m = 2000/2.2
m ≈ 909.09
Mass is ≈ 909.09 kg.
Answer:
Angular frequency is 20 rad/s.
Explanation:
Given that,
A block is attached to an oscillating spring. The function below shows its position (cm) vs. time (s) is given by :
.....(1)
The general equation of oscillating particle is given by :
.......(2)
Compare equation (1) and (2) we get :
So, the angular frequency of the oscillation is 20 rad/s.
