The equivalent capacitance (
) of an electrical circuit containing four capacitors which are connected in parallel is equal to: A. 21 F.
<h3>The types of circuit.</h3>
Basically, the components of an electrical circuit can be connected or arranged in two forms and these are;
<h3>What is a parallel circuit?</h3>
A parallel circuit can be defined as an electrical circuit with the same potential difference (voltage) across its terminals. This ultimately implies that, the equivalent capacitance () of two (2) capacitors which are connected in parallel is equal to the sum of the individual (each) capacitances.
Mathematically, the equivalent capacitance () of an electrical circuit containing four capacitors which are connected in parallel is given by this formula:
Ceq = C₁ + C₂ + C₃ + C₄
Substituting the given parameters into the formula, we have;
Ceq = 10 F + 3 F + 7 F + 1 F
Equivalent capacitance, Ceq = 21 F.
Read more equivalent capacitance here: brainly.com/question/27548736
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The answer would be center of mass, B
Answer:
595391.482946 m/s
Explanation:
E = Energy = 1.85 keV
I = Current = 5.15 mA
e = Charge of electron = 
t = Time taken = 1 second
m = Mass of proton = 
Velocity of proton is given by
The speed of the proton is 595391.482946 m/s
Current is given by
Number of protons is
The number of protons is
1) Frequency: 
the energy of the photon absorbed must be equal to the ionization enegy of the atom, which is
The energy of a photon is given by
where 
is the Planck's constant. By using the energy written above and by re-arranging thsi formula, we can calculate the frequency of the photon:
2) Wavelength: 91.2 nm
The wavelength of the photon can be found from its frequency, by using the following relationship:
where 
is the speed of light and f is the frequency. Substituting the frequency, we find
Answer:
angle minimum θ = 41.3º
Explanation:
For this exercise let's use Newton's second law in the condition of static equilibrium
N - W = 0
N = W
The rotational equilibrium condition, where we place the axis of rotation on the wall
We assume that counterclockwise rotations are positive
fr (l sin θ) - N (l cos θ) + W (l/2 cos θ) = 0
the friction force formula is
fr = μ N
fr = μ W
we substitute
μ m g l sin θ - m g l cos θ + mg l /2 cos θ = 0
μ sin θ - cos θ + ½ cos θ= 0
μ sin θ - ½ cos θ = 0
sin θ / cos θ = 1/2 μ
tan θ = 1/2 μ
θ = tan⁻¹ (1 / 2μ)
θ = tan⁻¹ (1 (2 0.57))
θ = 41.3º
