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belka [17]
3 years ago
9

Determine the equivalent capacitance between a and b for the group of capacitors in the drawing.

Physics
2 answers:
iragen [17]3 years ago
3 0

Answer:

There are different of arrangements for the capacitors to work inside a given system, as there are two different form of arrangements for the capacitors to be in, which are:

  1. The parallel connections of capacitor,
  2. The series connections of capacitor.

Explanation:

The parallel connection of capacitor is comprised of capacitors having the unequal amount of current, ΔI. And has a same potential difference across the number of capacitors. And the equivalent amount of capacitance,ΔC is calculated by the following equation:

  • ΔC(eq)=C₁ + C₂ + C₃ + C₄ + C(n),

For the series form of arrangements of the capacitors we have the different form of equation which can be explained by the total number of capacitors that are connected in series in combination as they are provided by the equal amount of current,ΔI across it, while the voltage or total amount of potential difference, Δv provided to the system is more different for each of the capacitor. And the equivalent amount of capacitance, ΔC(equivalent) can be determined by the following equation:

  • 1/C(eq)=1/C₁+1/C₂+1/C₃+1/C(n).

KonstantinChe [14]3 years ago
3 0

Answer:

The total effective capacitance is 2.033 μF

Explanation:

As the complete question is not given, the complete question is found online and is as attached herewith.

As with resistors start as far away from where you want total capacitance to be, so we have 3 capacitors in series, the 24 μF, the 12 μF, and the 8 μF

\dfrac{1}{C_{effective1}}=\dfrac{1}{C_{24}}+\dfrac{1}{C_{12}}+\dfrac{1}{C_{8}}=\dfrac{1}{24}+\dfrac{1}{12}+\dfrac{1}{8}=\dfrac{1}{4}\\C_{effective1}=4 \mu F

Next we have two 4 μF capacitors in parallel so we can combine them to

C_{effective2}=C_4+C_{effective1}\\C_{effective2}=4+4\\C_{effective2}=8 \mu F\\

Now we just have three capacitors in series so the total is

\dfrac{1}{C_{effective1}}=\dfrac{1}{C_{5}}+\dfrac{1}{C_{effective2}}+\dfrac{1}{C_{6}}=\dfrac{1}{5}+\dfrac{1}{8}+\dfrac{1}{6}=0.49166\\C_{effectivet}=1/0.49166=2.033 \mu F

So the total effective capacitance is 2.033 μF

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Answer:

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Figure 1. (a) The work done to lift the weight is stored in the mass-Earth system as gravitational potential energy. (b) As the weight moves downward, this gravitational potential energy is transferred to the cuckoo clock.

Let us calculate the work done in lifting an object of mass m through a height h, such as in Figure 1. If the object is lifted straight up at constant speed, then the force needed to lift it is equal to its weight mg. The work done on the mass is then W = Fd = mgh. We define this to be the gravitational potential energy (PEg) put into (or gained by) the object-Earth system. This energy is associated with the state of separation between two objects that attract each other by the gravitational force. For convenience, we refer to this as the PEg gained by the object, recognizing that this is energy stored in the gravitational field of Earth. Why do we use the word “system”? Potential energy is a property of a system rather than of a single object—due to its physical position. An object’s gravitational potential is due to its position relative to the surroundings within the Earth-object system. The force applied to the object is an external force, from outside the system. When it does positive work it increases the gravitational potential energy of the system. Because gravitational potential energy depends on relative position, we need a reference level at which to set the potential energy equal to 0. We usually choose this point to be Earth’s surface, but this point is arbitrary; what is important is the difference in gravitational potential energy, because this difference is what relates to the work done. The difference in gravitational potential energy of an object (in the Earth-object system) between two rungs of a ladder will be the same for the first two rungs as for the last two rungs.

Converting Between Potential Energy and Kinetic Energy

Gravitational potential energy may be converted to other forms of energy, such as kinetic energy. If we release the mass, gravitational force will do an amount of work equal to mgh on it, thereby increasing its kinetic energy by that same amount (by the work-energy theorem). We will find it more useful to consider just the conversion of PEg to KE without explicitly considering the intermediate step of work. (See Example 2.) This shortcut makes it is easier to solve problems using energy (if possible) rather than explicitly using forces.

More precisely, we define the change in gravitational potential energy ΔPEg to be ΔPEg = mgh, where, for simplicity, we denote the change in height by h rather than the usual Δh. Note that h is positive when the final height is greater than the initial height, and vice versa. For example, if a 0.500-kg mass hung from a cuckoo clock is raised 1.00 m, then its change in gravitational potential energy is

mgh=(0.500 kg)(9.80 m/s2)(1.00 m) =4.90 kg⋅m2/s2=4.90 Jmgh=(0.500 kg)(9.80 m/s2)(1.00 m) =4.90 kg⋅m2/s2=4.90 J

Note that the units of gravitational potential energy turn out to be joules, the same as for work and other forms of energy. As the clock runs, the mass is lowered. We can think of the mass as gradually giving up its 4.90 J of gravitational potential energy, without directly considering the force of gravity that does the

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3 years ago
A 55-kg skier starts from rest at the top of a ski jump, point A in Fig. 6–48, and travels down the ramp. If fric- tion and air
andre [41]

Explanation:

The first attachment contains the complete question while the second attachment is the solution.

6 0
4 years ago
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a truck is moving around a circular curve @ a uniform velocity of 13m/s. if the centifrical force on truck is 3,300N with truck
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<u>Answer</u>

81.94 m


<u>Explanation</u>

The centripetal force of an object moving in a circular path is given by:

F = mv²/r  Where m is the mass of the object, v is the constant velocity and r is the radius of the curve.

F = mv²/r

3,300 = (1600×13²)/r

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That is not yet possible. that technology has not yet been invented.
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Engineers and science fiction writers have proposed designing space stations in the shape of a rotating wheel or ring, which wou
lyudmila [28]

Answer:

w = 1.976 rpm

Explanation:

For simulate the gravity we will use the centripetal aceleration a_c, so:

a_c = w^2r

where w is the angular aceleration and r the radius.

We know by the question that:

r = 60.5m

a_c = 2.6m/s2

So, Replacing the data, and solving for w, we get:

2.6m/s = w^2(60.5m)

W = 0.207 rad/s

Finally we change the angular velocity from rad/s to rpm as:

W = 0.207 rad/s = 0.207*60/(2\pi)= 1.976 rpm

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3 years ago
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