Yes you're correct. For distance, SI is based on meters.
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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Answer: 0m/s²
Explanation:
Since the forces acting along the plane are frictional force(Ff) and moving force(Fm), we will take the sum of the forces along the plane
According newton's law of motion
Summation of forces along the plane = mass × acceleration
Frictional force is always acting upwards the plane since the body will always tends to slide downwards on an inclined plane and the moving acts down the plane
Ff = nR where
n is coefficient of friction = tan(theta)
R is normal reaction = Wcos(theta)
Fm = Wsin(theta)
Substituting in the formula of newton's first law we have;
Fm-Ff = ma
Wsin(theta) - nR = ma
Wsin(theta) - n(Wcos(theta)) = ma... 1
Given
W = 562N, theta = 30°, n = tan30°, m = 56.2kg
Substituting in eqn 1,
562sin30° - tan30°(562cos30°) = 56.2a
281 - 281 = 56.2a
0 = 56.2a
a = 0m/s²
This shows that the trunk is not accelerating
Answer:
44.6 N
Explanation:
Draw a free body diagram of the block. There are four forces on the block:
Weight force mg pulling down,
Normal force N pushing up,
Friction force Nμ pushing left,
and applied force F pulling right 30° above horizontal.
Sum of forces in the y direction:
∑F = ma
N + F sin 30° − mg = 0
N = mg − F sin 30°
Sum of forces in the x direction:
∑F = ma
F cos 30° − Nμ = 0
F cos 30° = Nμ
N = F cos 30° / μ
Substitute:
mg − F sin 30° = F cos 30° / μ
mg = F sin 30° + (F cos 30° / μ)
Plug in values:
mg = 20 N sin 30° + (20 N cos 30° / 0.5)
mg = 44.6 N
