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erma4kov [3.2K]
3 years ago
11

A 400 µF capacitor is constructed out of two parallel plates of very large area which are separated by 1 mm. A battery is used t

o charge the capacitor to a potential of 100 V . (a) If the capacitor is disconnected from the battery, how much work (in Joules) must be done to pull the plates apart to a separation of 2 mm?
Physics
1 answer:
Kay [80]3 years ago
6 0

Answer:

W=2 J

Explanation:

Given that

C= 400 µF

V= 100 V

d= 1 mm

Energy before disconnected from the battery

U= 1/2 CV²=Q²/(2C)

Energy after  disconnected from the battery

U'=1/2 C'V²=Q²/(2C')

The work done W

W= U' - U

W= Q²/(2C') - Q²/(2C)

W=\dfrac{Q^2}{2C}\left ( \dfrac{C}{C'} -1\right )

1/2 CV²=Q²/(2C)

W=\dfrac{CV^2}{2}\left ( \dfrac{C}{C'} -\right )

We know that

C=\dfrac{\varepsilon _oA}{d}

C'=\dfrac{\varepsilon _oA}{d'}

Given that

d'=2mm ,d= 1mm

C/C'= d'/d= 2

W=\dfrac{CV^2}{2}\left ( \dfrac{C}{C'} -\right )

By putting the values

W=\dfrac{400\times 10^{-6}\times 100^2}{2}( 2 -1 )

W=2 J

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Two containers (A and B) are in thermal contact with an environment at temperature T = 280 K. The two containers are connected b
zmey [24]

Answer:

The maximum amount of work is  W = 1563.289 \ J

Explanation:

From  the question we are told that

   The temperature of the environment is  T = 280\ K

    The volume of container A is  V_A = 2 m^3

    Initially the number of moles  is  n = 1.2 \ moles

     The volume of container B is V_B = 3.5 \ m^3

     

At equilibrium of the gas the maximum work that can be done on the turbine is mathematically represented as

             W =  P_A V_A  ln[ \frac{V_B}{V_A} ]

Now from the Ideal gas law

          P_A V_A =  nRT

So substituting for P_A V_A in the equation above

          W =  nRT ln [\frac{V_B}{V_A} ]

Where R is the gas constant with a values of  R =  8.314 \  J/mol

Substituting values we have that

            W = 1.2 * (8.314) * (280) * ln [\frac{3.5}{2} ]

          W = 1563.289 \ J

8 0
3 years ago
what equastion do you use to solve Riders in a carnival ride stand with their backs against the wall of a circular room of diame
Hitman42 [59]

Answer:

μsmín = 0.1

Explanation:

  • There are three external forces acting on the riders, two in the vertical direction that oppose each other, the force due to gravity (which we call weight) and the friction force.
  • This friction force has a maximum value, that can be written as follows:

       F_{frmax} = \mu_{s} *F_{n} (1)

       where  μs is the coefficient of static friction, and Fn is the normal force,

       perpendicular to the wall and aiming to the center of rotation.

  • This force is the only force acting in the horizontal direction, but, at the same time, is the force that keeps the riders rotating, which is the centripetal force.
  • This force has the following general expression:

       F_{c} =  m* \omega^{2} * r (2)

       where ω is the angular velocity of the riders, and r the distance to the

      center of rotation (the  radius of the circle), and m the mass of the

      riders.

      Since Fc is actually Fn, we can replace the right side of (2) in (1), as

      follows:

     F_{frmax} = m* \mu_{s} * \omega^{2} * r (3)

  • When the riders are on the verge of sliding down, this force must be equal to the weight Fg, so we can write the following equation:

       m* g = m* \mu_{smin} * \omega^{2} * r (4)

  • (The coefficient of static friction is the minimum possible, due to any value less than it would cause the riders to slide down)
  • Cancelling the masses on both sides of (4), we get:

       g = \mu_{smin} * \omega^{2} * r (5)

  • Prior to solve (5) we need to convert ω from rev/min to rad/sec, as follows:

      60 rev/min * \frac{2*\pi rad}{1 rev} *\frac{1min}{60 sec} =6.28 rad/sec (6)

  • Replacing by the givens in (5), we can solve for μsmín, as follows:

       \mu_{smin} = \frac{g}{\omega^{2} *r}  = \frac{9.8m/s2}{(6.28rad/sec)^{2} *2.5 m} =0.1 (7)

5 0
3 years ago
Two Polaroids are aligned so that the initially unpolarized light passing through them is a maximum. At what angle should one of
steposvetlana [31]

To solve this problem it is necessary to apply the law of Malus which describes the change in the Intensity of Light when it crosses a polarized surface.

Mathematically the expression is given as

I = I_0 cos^2\theta

Where,

I_0= Initial Intensity

I = Final Intensity after pass through the polarizer

\theta= Angle between the polarizer and the light

Since it is sought to reduce the intensity by half the relationship between the two intensities will be given as

\frac{I}{I_0} = \frac{1}{2}

Using the Malus Law we have,

I = I_0 cos^2\theta

cos^2\theta = \frac{I}{I_0}

cos^2\theta = \frac{1}{2}

\theta = cos^{-1}(\frac{1}{2})^2

\theta = 75.52\°

Angle with respect to maximum is 90-75.52 = 14.48\°

8 0
3 years ago
During which stage does the birth rate begin to decline?
sveticcg [70]

Answer:

During stage 3 - late expanding (of demogrpahic transition model)

Explanation:

During stage 3, birth rate begins to decline as infant mortality is lower and women have more access to education, family planning, and contraceptives.  Children are not needed as "free labor" as they might have been in earlier stages.

3 0
3 years ago
Read 2 more answers
An unknown fluid flows at a speed of 31 m/s. Suppose the fluid has a mass of 47 kg runs at this speed. What is the fluid’s kinet
Leya [2.2K]

Answer:

22583.5J

Explanation:

KE=1/2 mv^2

=1/2*47Kg*(31m/s^2)

=23.5Kg * 961m/s^2

=22583.5J

7 0
2 years ago
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