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KATRIN_1 [288]
1 year ago
6

(i) Rank the following five capacitors from greatest to smallest capacitance, noting any cases of equality. (a) a 20- μF capacit

or with a 4-V potential difference between its plates (b) a 30- μF capacitor with charges of magnitude 90μC on each plate (c) a capacitor with charges of magnitude 80 μC on its plates, differing by 2V in potential, (d) a 10- μF capacitor storing energy 125 μJ (e) a capacitor storing energy 250 μJ with a 10-J potential difference.
Physics
1 answer:
Nana76 [90]1 year ago
6 0

The Rank from greatest to smallest capacitance, noting any cases of equality is C>B>A>D>E.

<h3>What is capacitance?</h3>

The ratio of the quantity of electric charge held on a conductor to an electric potential difference is known as capacitance. Self capacitance and mutual capacitance are two concepts that are related to one another. Self capacitance is a property of all electrically chargeable objects. In this instance, the electrical potential difference between the object and the ground is measured. At a given potential difference, a material with a high self capacitance can carry more electric charge than one with a low capacitance. Understanding the concept of mutual capacitance is crucial for comprehending how the capacitor, one of the three basic linear electronic components, operates (along with resistors and inductors).

To learn more about capacitance ,visit:

brainly.com/question/14746225

#SPJ4

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A uniform electric field of magnitude 7.0 ✕ 104 N/C passes through the plane of a square sheet with sides 5.0 m long. Calculate
Vadim26 [7]

Answer:

1.52*10^6 Nm^2/C

Explanation:

Given that:

Electrical field E = 7.0 * 10^{-4}N/C

square side l = 5.0 m

Area A = 5.0 * 5.0

= 25.0 m²

Angle ( θ ) between area vector and E = (90° - 60°)

= 30°

The flux \phi_E can now be determined by using the expression

\phi_E = E*A*Cos \theta

\phi_E = 7.0 * 10^{-4}N/C *25.0m*Cos 30^0

\phi_E = 1515544.457 Nm^2/C

\phi_E = 1.52*10^6 Nm^2/C

5 0
3 years ago
Read 2 more answers
A distant galaxy is determined to be 150 million light years distant and moving away from us; using the Hubble law determine its
dlinn [17]

Your question kind of petered out there towards the end and you didn't specify
the terms, so I'll pick my own.

The "Hubble Constant" hasn't yet been pinned down precisely, so let's pick a
round number that's in the neighborhood of the last 20 years of measurements:

             <em>70 km per second per megaparsec</em>.

We'll also need to know that 1 parsec = about 3.262 light years.

So the speed of your receding galaxy is

         (Distance in LY) x (1 megaparsec / 3,262,000 LY) x (70 km/sec-mpsc) =

              (150 million) x  (1 / 3,262,000) x (70 km/sec) =

                                 <em>3,219 km/sec  </em>in the direction away from us (rounded)

4 0
3 years ago
A string of mass 60.0 g and length 2.0 m is fixed at both ends and with 500 N in tension. a. If a wave is sent along this string
Darya [45]

Answer:

a

The  speed of  wave is   v_1  = 129.1 \ m/s

b

The new speed of the two waves is v =  129.1 \ m/s

Explanation:

From the question we are told that

    The mass of the string is  m  =  60 \ g  =  60 *10^{-3} \ kg

    The length is  l  =  2.0 \ m

    The tension is  T  = 500 \ N

Now the velocity of the first wave is mathematically represented as

     v_1  = \sqrt{ \frac{T}{\mu} }

Where  \mu is the linear density which is mathematically represented as

      \mu  =  \frac{m}{l}

substituting values    

     \mu  =  \frac{ 60 *10^{-3}}{2.0 }

     \mu  =  0.03\ kg/m

So

   v_1  = \sqrt{ \frac{500}{0.03} }

   v_1  = 129.1 \ m/s

Now given that the Tension, mass and length are constant the velocity of the second wave will same as that of first wave (reference PHYS 1100 )

     

8 0
3 years ago
The car in the figure travels at a constant speed along the road shown. Draw vector showing its acceleration at the point C if t
yuradex [85]

Answer:

The vector form is as shown in the attachment

Explanation:

The figure as shown in the diagram, indicates that the car is moving along the road at a constant speed. Centripetal acceleration comes into play for an object moving in a circular motion at uniform speed. The centripetal acceleration is the acceleration experienced by an object while in uniform circular motion.

Mathematically from centripetal acceleration; a = v2/r

The equation shows that there is an inverse relationship between the acceleration and the radius of curvature as such the radius of curvature at the point A will be more than the radius of curvature at the point C, this shows that the centripetal acceleration at point C will be more than the centripetal acceleration at point A.

The attachment shows the figure and the representation in vectorial form.

6 0
3 years ago
Can someone help me with this question
geniusboy [140]

833.33 sec

5000m/6ms

Divide 5000 by 6 and you get your answer !!

6 0
3 years ago
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