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Featured 4 months ago

This isn’t simple, but I’ll try and then give a diagram and reference ... off we go!

A G.M. tube is a detector of ionising radiation that works on the principle of ‘avalanches’ of electrons accelerated by a high electric field. The field exists between the neutral case and a central ‘spike’ at a p.d. (voltage) of several hundred volts (typically +400-600V.) When an

The electron accelerates rapidly in the strong electrical field and collides with other atoms, liberating further electrons from them. The effect repeats until a burst of current arrives at the central spike. This is the ‘avalanche’ of electrons.

There is another difficulty in detecting all 3 types of radiation, the

So that the detector can “reset” it is necessary to have a quenching agent present to ‘mop up’ the remaining electrons after a detection (normally a halogen, such as bromine.) This takes up to

Your diagram should help visualise it all:

And the reference (gives *loads* more details): https://en.m.wikipedia.org/wiki/Geiger%E2%80%93M%C3%BCller_tube hmmm that didn’t work (Socratic doesn’t like the formatting of Muller!) just stick Geiger Muller tube into Wikipedia!

Featured 4 months ago

Referring to the figure above and force per unit length between two infinite parallel current carrying wires is given as

#F/(DeltaL)=(mu_0I_1I_2)/(2pir)#

We also know that if currents are parallel to each other these attract and repel if currents are anti-parallel.

In the given question currents are parallel. Each current carrying wire experiences three attractive forces. Two forces are directed along the sides of the square and third force is directed along the diagonal. Total force is resultant of three vectors.

Force per unit length along the side

Force per unit length along the diagonal

Now resultant vector of two forces per unit length along the sides

Since moduli of

We get sum of all three force per unit length as

#F/(DeltaL)=(sqrt2mu_0i^2)/(2pia)+(mu_0i^2)/(2pisqrt2a)# , directed along the diagonal.

#=>F/(DeltaL)=(mu_0i^2)/(2pia)(sqrt2+1/sqrt2)#

Inserting given values we get

#F/(DeltaL)=(4pixx10^-7xx(18.7)^2)/(2pixx0.245)xx3/sqrt2#

#=>F/(DeltaL)=6.06xx10^-4Nm^-1# , rounded to two decimal places.

Featured 3 months ago

Let's explain the molecular theory of friction to have a close look into the incidence.

what happens is that whatever smooth we consider any object in this world,it is only true upto a certain marco level.We know that all substances are made up of very smaller particles,so during formation of any thing deformity coexists at their surfaces,and having a micro level vision makes it appear as this.(see the diagram)

So,you can follow these ridges present at the interface of two substances in contact.Now a depreesion between two ridges of any substance will be filled by the ridge of the other,so that when you try to move one w.r.t the other,there is relative hindrances by the edge of the depressions on the ridge of the substance you want to move.

Now the harder you apply force to move it,the ridge of one substance,and the ridge of the other,lying next to one depression,undergoes **COLD WELDING** i.e a process where without application of any external heat,but only due to application of massive force,the ridges are broken down to molecules and rearrange and solidify from molten state.As a result these new bonds formed between two heterogenous substances now exert even greater opposing force for creating relative motion at the interface.

Upto this static frictional force was acting,now when you exert a more amount of force which is capable of overcome these effects,the body starts moving,but now kinetic frictional force acts only,where hindrances is only due to ridge-ridge effect,not cold welding as rapidly moving bodies don't give enough time for cold welding to happen,which explains the following graph,why kinetic frictional force has a bit lower magnitude than static frictional force.

This theory as well helps to explain why frictional force acting is directly proportional to normal force,not the surface area of contact,which seems to be something odd.But I am not going to explain that right now,hoping you to solve it and if you can't then only please ask me.

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THANK YOU
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Featured 3 months ago

The boat's speed is

We know that

Therefore,

#d/("speed against current") + d/("speed with current") = t_"total"#

We know that

#3 = 15/(x - 2) + 15/(x + 2)#

#3(x^2 - 4) = 15(x + 2) + 15(x - 2)#

#3x^2 - 12 = 15x + 30 + 15x - 30#

#3x^2 - 30x - 12 = 0#

#x^2 - 10x - 4 = 0#

This cannot be factored, thus we use the quadratic formula or a graphical approach. Use your graphing calculator to enter

A negative answer is impossible, so we deduce that the speed in still water of the boat is approximately

#t = 15/8.4 = 1.8 hrs#

Hopefully this helps!

Featured 1 month ago

Let the **radius** of the great conical mound of height

Given that the weight of the finished mound is

Now for the sake of our calculation let us consider the center of the circular base of the the mound

It is obvious that the rate of decrease of radius of the conical mound with height will be given by

Hence at an arbitrary height

So the volume of an imaginary circular disk of infinitesimal thickness

So weight of this thin disk will be

So work done against gravitational pull for lifting this imaginary thin disk to a height of

The total work done

Proved

**Alternative method**

Considering the center of mass of the conical mound which is placed at

So work done to heap up uniform material found at ground level which is equivalent to lift the COM to a height of

Featured 1 month ago

#DeltaU = int_(V_1)^(V_2) ((delU)/(delV))_TdV = int_(V_1)^(V_2) -P + T((delP)/(delT))_VdV#

Now decide what gas law to use, or what

Well, from the total differential at constant temperature,

#dU = cancel(((delU)/(delT))_VdT)^(0) + ((delU)/(delV))_TdV# ,

so by definition of integrals and derivatives,

#DeltaU = int_(V_1)^(V_2) ((delU)/(delV))_TdV# #" "bb((1))#

The natural variables are

#dA = -SdT - PdV# #" "bb((2))#

This is also related by the isothermal Helmholtz relation (similar to the isothermal Gibbs' relation):

#dA = dU - TdS# #" "bb((3))#

Differentiating

#((delA)/(delV))_T = ((delU)/(delV))_T - T((delS)/(delV))_T#

From

#((delA)/(delV))_T = -P#

and also from

#((delS)/(delV))_T = ((delP)/(delT))_V#

since the Helmholtz free energy is a state function and its cross-derivatives must be equal. Thus from

#-P = ((delU)/(delV))_T - T((delP)/(delT))_V#

or we thus go back to

#barul|stackrel(" ")(" "DeltaU = int_(V_1)^(V_2) ((delU)/(delV))_TdV = int_(V_1)^(V_2)-P + T((delP)/(delT))_VdV" ")|#

*And what remains is to distinguish between the last term for gases, liquids and solids...*

**GASES**

Use whatever gas law you want to find it. If for whatever reason your gas is ideal, then

#((delP)/(delT))_V = (nR)/V#

and that just means

#((delU)/(delV))_T = -P + (nRT)/V#

#= -P + P = 0# which says that

ideal gases have changes in internal energy as a function of only temperature.One would get

#color(blue)(DeltaU = int_(V_1)^(V_2) 0 dV = 0)# .Not very interesting.

Of course, if your gas is **not** ideal, this isn't necessarily true.

**LIQUIDS AND SOLIDS**

These data are tabulated as **coefficients of volumetric thermal expansion** **isothermal compressibility**

#alpha = 1/V((delV)/(delT))_P#

#kappa = -1/V((delV)/(delP))_T#

#alpha/kappa = [ . . . ] = ((delP)/(delT))_V# at VARIOUS temperatures for VARIOUS condensed phases. Some examples at

#20^@ "C"# :

#alpha_(H_2O) = 2.07 xx 10^(-4) "K"^(-1)# #alpha_(Au) = 4.2 xx 10^(-5) "K"^(-1)# (because that's REAL useful, right?)#alpha_(EtOH) = 7.50 xx 10^(-4) "K"^(-1)# -
#alpha_(Pb) = 8.7 xx 10^(-5) "K"^(-1)# -
#kappa_(H_2O) = 4.60 xx 10^(-5) "bar"^(-1)# #kappa_(Au) = 5.77 xx 10^(-7) "bar"^(-1)# #kappa_(EtOH) = 1.10 xx 10^(-4) "bar"^(-1)# #kappa_(Pb) = 2.33 xx 10^(-6) "bar"^(-1)#

In that case,

#((delU)/(delV))_T = -P + (Talpha)/kappa#

Thus,

#color(blue)(DeltaU) = int_(V_1)^(V_2) -P + (Talpha)/kappadV#

#= color(blue)((alphaTDeltaV)/kappa - int_(V_1)^(V_2) P(V)dV)#

Next, the pressure will vary a lot if the volume varies a little for a solid or liquid...

Hence, to find

Then, we'll need to measure how the volume changes (to high precision!) with the change in pressure to empirically find the pressure as a function of the volume for the given solid or liquid.

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