Answer:
Explanation:
We shall apply Doppler's effect of sound .
speaker is the source , Jason is the observer . Source is moving at 10 m /s , observer is moving at 6 m/s .
apparent frequency =
V is velocity of sound , v₀ is velocity of observer and v_s is velocity of source and f_o is real frequency of source .
Here V = 340 m/s , v₀ is 6 m/s , v_s is 10 m/s . f_o = f
apparent frequency =
=
So m = 346 , n = 330 .
Answer:
a) -2.516 × 10⁻⁴ V
b) -1.33 × 10⁻³ V
Explanation:
The electric field inside the sphere can be expressed as:
The potential at a distance can be represented as:
V(r) - V(0) =
V(r) - V(0) = ₀
V(r) = ₀
Given that:
q = +3.83 fc = 3.83 × 10⁻¹⁵ C
r = 0.56 cm
= 0.56 × 10⁻² m
R = 1.29 cm
= 1.29 × 10⁻² m
E₀ = 8.85 × 10⁻¹² F/m
Substituting our values; we have:
= -2.15 × 10⁻⁴ V
The difference between the radial distance and center can be expressed as:
V(r) - V(0) =
V(r) - V(0) =
V(r) =
V(r) =
V(r)
V(r) = -0.00133
V(r) = - 1.33 × 10⁻³ V
Answer:
F = 0.1575 N
Explanation:
When the third sphere touches the first sphere, the charge is distributed between both spheres, then now the first sphere has only half of his original charge.
In this moment then
Sphere one has a charge = Q/2
Sphere three has a charge = Q/2
Now when the third sphere touches the second sphere again the charge is distributed in a manner that both sphere has the same charge.
How the total charge is Q = Q/2 + Q = 3/2Q, when the spheres are separated each one has 3/4Q
Sphere two has a charge = 3/4Q
Sphere three has a charge = 3/4Q
The electrostatic force that acts on sphere 2 due to sphere 1 is:
F =
F=
how = 0.42
Then
F =
F = 0.1575 N