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jolli1 [7]
3 years ago
6

For a cylindrical capacitor, the capacitance does not depend on which of the following values?

Physics
1 answer:
IgorC [24]3 years ago
5 0

Answer:

Capacitance of cylindrical capacitor does not depends on the amount of charge on the conductors

Explanation:

Consider a cylindrical capacitor of length L, inner radius R₁ and outer radius R₂, permitivity ε₀ constant then capacitance of cylindrical capacitor is given by:

C=\frac{2\pi \epsilon_{o}L}{ln\frac{R_{2} }{R_{1}} }

From this equation it is clear that capacitance of cylindrical capacitor is independent of the amount of charge on the conductors where as directly  proportional permitivity constant and length of cylinder where as inversely proportional to natural log of ratio of  R₂ and R₁

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A 20-liter container contains 2.0 moles of oxygen at a pressure of 92 kpa. the average kinetic energy of translation of oxygen m
Vikentia [17]

From ideal gas law, PV=nRT

where P is the pressure, V is the volume of the container, n is number of moles, R is the gas constant and T is the temperature.

Hence, T=\frac{PV}{nR}=\frac{92*10^{3}*2.0*10^{-3}}{2*3.314}

T= 110.65 k

Kinetic Energy = \frac{3}{2}KT=\frac{3}{2}  (1.38*10^{-23})(110.65)

K.E=  2.2*10^{-21}J

<h3>What is a kinetic energy? </h3>

The energy an object has as a result of motion is known as kinetic energy.

A force must be applied to an object in order to accelerate it. We must put in effort in order to apply a force. After the work is finished, energy is transferred to the item, which then moves at a new, constant speed. Kinetic energy is the type of energy that is transferred and is dependent on the mass and speed attained.

Kinetic energy can be converted into other types of energy and transported between objects. A flying squirrel may run into a chipmunk that is standing still, for instance. Some of the squirrel's initial kinetic energy may have been transferred to the chipmunk or changed into another kind of energy after the collision.

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4 0
1 year ago
A truck is driving over a scale at a weight station. When the front wheels drive over the scale, the scale reads 5800 N. When th
aev [14]

Answer:

x_2=1.60m

Explanation:

From the Question We are told that

Initial Force F_1=5800N

Final Force F_2=6500N

Distance between the front and rear wheels \triangle x=3.20 m

Since

 \triangle x=3.20 m

Therefore

 x_1+x_2=3.20

 x_1=3.20-x_2

Generally the equation for The center of mass is at x_2 is mathematically

given by

 x_2 =\frac{(F_1x_1+F_2x_2)}{(F_1+F_2)}

 x_2=3.20F_1-\frac{x_2F_1+F_2x_2}{(F_1+F_2)}

 2*F_1*x_2 =3.20F_1

 x_2=1.60m

6 0
2 years ago
T=2pi square root 1/g solve for g.<br> Explanation would be really helpful.
Natalija [7]

I added individual steps for clarity. Note that g must be positive if the solution is to be real.

T=2\pi \sqrt{\frac{1}{g}}=2\pi g^{-\frac{1}{2}}\\g^{-\frac{1}{2}} = \frac{T}{2\pi}\\(g^{-\frac{1}{2}})^{-2} = (\frac{T}{2\pi})^{-2}\\g = \frac{4\pi^2}{T^2}\,\,\,, g>0}

Let me know if you have any questions.

7 0
3 years ago
.
Greeley [361]

Answer:

this is impossible for me

Explanation:

7 0
3 years ago
the average american receives 2.28 mSv dose equivalent from radon each year. Assuming you receive this dose, and it all comes fr
Dmitry [639]

Answer:

The approximate number of decays  this represent  is  N= 23*10^{10}  

Explanation:

 From the question we are told that

    The amount of Radiation received by an average american is I_a = 2.28 \ mSv

     The source of the radiation is S = 5.49 MeV \ alpha \ particle

 Generally

            1 \  J/kg = 1000 mSv

   Therefore  2.28 \ mSv = \frac{2.28}{1000} = 2.28 *10^{-3} J/kg

Also  1eV = 1.602 *10^{-19}J

  Therefore  2.28*10^{-3} \frac{J}{kg} = 2.28*10^{-3} \frac{J}{kg}  * \frac{1ev}{1.602*10^{-19} J} = 1.43*10^{16} ev/kg

           An Average american weighs 88.7 kg

      The total energy received is mathematically evaluated as

        1 kg ------> 1.423*10^{16}ev \\88.7kg  --------> x

Cross-multiplying and making x the subject

           x = 88.7 * 1.423*10^{16} eV

              x = 126.2*10^{16}eV

Therefore the total  energy  deposited is x = 126.2*10^{16}eV

The approximate number of decays  this represent  is mathematically evaluated as

            N = \frac{x}{S}

Where n is the approximate number of decay

   Substituting values

                             N = \frac{126 .2*10^{16}}{5.49*10^6}  

                                  N= 23*10^{10}  

                     

             

7 0
3 years ago
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