Answer:
Magnitude of the force between the charges is F = 1.92×10^20N
Explanation:
Given the magnitude of force according to coulombs law
F =K[(q1*q2)/r2]
Where q1 and q2 are the charges
r is the distance between the charges
K is the coulombs constant
Substituting the given values, we have;
F = 8.98×10^9 × 1.5×10^6 × 3.2×10^4/1.5²
F = 43.1×10^19/2.25
F = 19.16×10^19N
F = 1.92×10^20N
A) We want to find the work function of the potassium. Apply this equation:
E = 1243/λ - Φ
E = energy of photoelectron, λ = incoming light wavelength, Φ = potassium work function
Given values:
E = 2.93eV, λ = 240nm
Plug in and solve for Φ:
2.93 = 1243/240 - Φ
Φ = 2.25eV
B) We want to find the threshold wavelength, i.e. find the wavelength such that the energy E of the photoelectrons is 0eV. Plug in E = 0eV and Φ = 2.25eV and solve for the threshold wavelength λ:
E = 1243/λ - Φ
0 = 1243/λ - Φ
0 = 1243/λ - 2.25
λ = 552nm
C) We want to find the frequency associated with the threshold wavelength. Apply this equation:
c = fλ
c = speed of light in a vacuum, f = frequency, λ = wavelength
Given values:
c = 3×10⁸m/s, λ = 5.52×10⁻⁷m
Plug in and solve for f:
3×10⁸ = f(5.52×10⁻⁷)
f = 5.43×10¹⁴Hz
Answer:
One bulb could go out and the strand will stay on.
Explanation:
In series circuit, there is only one path provided for the current to flow. So, all the lights are required to be in working condition, for the others to work. And if anyone light bulb goes out, the circuit will become incomplete and the rest of the strand will also go out. Because there is only one path for current flow which is now broken.
On the other hand, in parallel circuits, each light bulb has a separate connection with the source. Current path to each bulb is independent of the others. Therefore, if one bulb goes out, the rest of the strand will stay on.
So, the correct option is:
<u>One bulb could go out and the strand will stay on.</u>
Answer:
C = 2.9 10⁻⁵ F = 29 μF
Explanation:
In this exercise we must use that the voltage is
V = i X
i = V/X
where X is the impedance of the system
in this case they ask us to treat the system as an RLC circuit in this case therefore the impedance is
X = 
tells us to take inductance L = 0.
The angular velocity is
w = 2π f
the current is required to be half the current at high frequency.
Let's analyze the situation at high frequency (high angular velocity) the capacitive impedance is very small
→0 when w → ∞
therefore in this frequency regime
X₀ = 
the very small fraction for which we can despise it
X₀ = R
to halve the current at f = 200 H, from equation 1 we obtain
X = 2X₀
let's write the two equations of inductance
X₀ = R w → ∞
X= 2X₀ = 
w = 2π 200
we solve the system
2R = \sqrt{R^2 +( \frac{1}{wC} )^2 }
4 R² = R² + 1 / (wC) ²
1 / (wC) ² = 3 R²
w C = 
C = 
let's calculate
C = 
C = 2.9 10⁻⁵ F
C = 29 μF
I would do a windmill project!
Hope it helps!:)
