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