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vovangra [49]
3 years ago
15

Suppose you have a dipole that's free to move in any way (including rotate - imagine it floating in space). And there's an objec

t with charge Q a distance r away. That distance r would be much larger than d, the distance between the charges of the dipole, so we draw the dipole small.

Physics
1 answer:
Setler [38]3 years ago
5 0

Complete Question

The complete question shown on the first uploaded image

Answer:

a)

The force on Q due to dipole is Attractive

b)

The charge Q exerts attractive force on the dipole

c)

Yes from the above parts, force depends on the sign of charge

d)

   F = kQq[\frac{d^{2}+2rd}{r^{2}(d+r)^{2}} ]

e)

The magnitude o force decrease by a factor of 8.0 times

Explanation:

The explanation is shown on the second uploaded image

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Answer:

Salt can dissolve in water making it an excellent soluble for water

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What is the most effective means of establishing awareness of hazards in commercial, industrial, and storage facilities with lar
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A solid nonconducting sphere of radius R has a charge Q uniformly distributed throughout its volume. A Gaussian surface of radiu
anyanavicka [17]

Answer:

1. E x 4πr² = ( Q x r³) / ( R³ x ε₀ )

Explanation:

According to the problem, Q is the charge on the non conducting sphere of radius R. Let ρ be the volume charge density of the non conducting sphere.

As shown in the figure, let r be the radius of the sphere inside the bigger non conducting sphere. Hence, the charge on the sphere of radius r is :

Q₁ = ∫ ρ dV

Here dV is the volume element of sphere of radius r.

Q₁ = ρ x 4π x ∫ r² dr

The limit of integration is from 0 to r as r is less than R.

Q₁ = (4π x ρ x r³ )/3

But volume charge density, ρ = \frac{3Q}{4\pi R^{3} }

So, Q_{1} = \frac{Qr^{3} }{R^{3} }

Applying Gauss law of electrostatics ;

∫ E ds = Q₁/ε₀

Here E is electric field inside the sphere and ds is surface element of sphere of radius r.

Substitute the value of Q₁ in the above equation. Hence,

E x 4πr² = ( Q x r³) / ( R³ x ε₀ )

7 0
4 years ago
Why do elements within a group have similar chemical properties
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4 0
3 years ago
The rate at which heat enters an air conditioned building is often roughly proportional to the difference in temperature between
erma4kov [3.2K]

Answer:

Considering first question

    Generally the coefficient of performance of the air condition  is mathematically represented as

   COP  =  \frac{T_i}{T_o - T_i}

Here T_i is the inside temperature

while  T_o is the outside temperature

What this coefficient of performance represent is the amount of heat the air condition can remove with 1 unit of electricity

So it implies that the air condition removes   \frac{T_i}{T_o - T_i} heat with 1 unit of electricity

Now from the question we are told that the rate at which heat enters an air conditioned building is often roughly proportional to the difference in temperature between inside and outside. This can be mathematically represented as

         Q \ \alpha \ (T_o - T_i)

=>        Q= k (T_o - T_i)

Here k is the constant of proportionality

So  

    since  1 unit of electricity  removes   \frac{T_i}{T_o - T_i}  amount of heat

   E  unit of electricity will remove  Q= k (T_o - T_i)

So

      E =  \frac{k(T_o - T_i)}{\frac{T_i}{ T_h - T_i} }

=>   E = \frac{k}{T_i} (T_o - T_i)^2

given that  \frac{k}{T_i} is constant

    =>  E \  \alpha  \  (T_o - T_i)^2

From this above equation we see that the  electricity required(cost of powering and operating the air conditioner) is approximately proportional to the square of the temperature difference.

 Considering the  second question

Assuming that  T_i   =  30 ^oC

 and      T_o  =  40 ^oC

Hence  

     E = K (T_o - T_i)^2

Here K stand for a constant

So  

        E = K (40 -  30)^2

=>      E = 100K

Now if  the  T_i   =  20 ^oC

Then

       E = K (40 -  20)^2

=>      E = 400 \ K

So  from this see that the electricity require (cost of powering and operating the air conditioner)when the inside temperature is low  is  much higher than the electricity required when the inside temperature is higher

Considering the  third question

Now in the case where the  heat that enters the building is at a rate proportional to the square-root of the temperature difference between inside and outside

We have that

       Q = k (T_o - T_i )^{\frac{1}{2} }

So

       E =  \frac{k (T_o - T_i )^{\frac{1}{2} }}{\frac{T_i}{T_o - T_i} }

=>   E =  \frac{k}{T_i} * (T_o - T_i) ^{\frac{3}{2} }

Assuming \frac{k}{T_i} is a constant

Then  

     E \ \alpha \ (T_o - T_i)^{\frac{3}{2} }

From this above equation we see that the  electricity required(cost of powering and operating the air conditioner) is approximately proportional to the square root  of the cube of the  temperature difference.

   

4 0
3 years ago
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