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Question #836ca

Mark C.
Mark C.
Featured 4 months ago

Answer:

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

Explanation:

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 #alpha, beta# particle or a #gamma# ray enters the detector it ionises (liberates a low energy electron from) a particle of the low pressure gas (typically argon at about 0.1 atmospheres) filling the cavity.

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 #alpha# particles cannot penetrate the metal walls of the detector so a thin mica window is placed at the end of the tube. The difference in ionising power of the 3 types of radiation means that GM tubes aren’t equally sensitive to all of them, but this is a small problem given their sensitivity, range of operation and low cost.

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 #100 mus# and during this time no further detection of ionising radiation is possible - hence the term ‘dead time’.

Your diagram should help visualise it all:

https://www.bbc.co.uk/education/guides/zt9s2nb/revision/5

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!

hyperphysics.phy-astr.gsu.edu
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 #=(mu_0i^2)/(2pia)#
Force per unit length along the diagonal #=(mu_0i^2)/(2pisqrt2a)#
Now resultant vector of two forces per unit length along the sides #=sqrt(F_1^2+F_2^2)# along the diagonal.
Since moduli of #F_1 and F_2# are equal #:.sqrt(F_1^2+F_2^2)=sqrt2F#
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.

Question #c5671

Jane
Jane
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)

enter image source here 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.enter image source here

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.

                                       THANK YOU

Answer:

The boat's speed is #10.4# km/h and it takes 1.8 hrs to travel upstream.

Explanation:

We know that #s = d/t#, thus #t = d/s#.

Therefore,

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

We know that #t_"total" = 3 hrs#, and that #d = 15# km. If we let the speed in still water be #x#, then the speed against the current would be #x - 2# and with the current #x + 2#. Therefore:

#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 #y_1 = x^2 - 10x - 4# and #y_2 = 0#. Then use the intersect function to determine the intersection points.

enter image source here

A negative answer is impossible, so we deduce that the speed in still water of the boat is approximately #10.4 "km"/h#. When the boat is going upstream, its speed will be reduced by #2# km/h, to #8.4# km/h. Therefore,

#t = 15/8.4 = 1.8 hrs#

Hopefully this helps!

enter image source here

Let the radius of the great conical mound of height #h# built by workers be #r#.

Given that the weight of the finished mound is #M#. So the weight of finished mound per unit volume will be #m=M/(1/3pir^2h)#.

Now for the sake of our calculation let us consider the center of the circular base of the the mound #(O)# as origin with diameter of the mound lying along X-axis and height along Y-axis.

It is obvious that the rate of decrease of radius of the conical mound with height will be given by #r/h#.

Hence at an arbitrary height #y# its radius will be #(r-(ry)/h)=r(1-y/h)#

So the volume of an imaginary circular disk of infinitesimal thickness #dy# at this height #y# will be given by

#dv=pir^2(1-y/h)^2dy#

So weight of this thin disk will be

#=m*dv=M/(1/3pir^2h)*pir^2(1-y/h)^2dy#

#=(3M)/h*(1-y/h)^2dy#

So work done against gravitational pull for lifting this imaginary thin disk to a height of #y# will be given by

#dw="weight"(mdv)xx"height"(y)=(3M)/h*(1-y/h)^2dy*y#

The total work done #W# to finish the mound will be obtained by integrating this #dw# as follows where #y# varies from #0toh#

#W=(3M)/hint_0^h(1-y/h)^2ydy#

#=>W=(3M)/h[y^2/2-(2y^3)/(3h)+y^4/(4 h^2)]_0^h#

#=>W=(3M)/h[h^2/2-(2h^3)/(3h)+h^4/(4 h^2)]#

#=>W=(3M)/h*1/12[6h^2-8h^2+3h^2]#

#=>W=(3M)/h*1/12*h^2#

#=>W=1/4hM#

Proved

Alternative method

Considering the center of mass of the conical mound which is placed at #1/4h# we can calculate the work done in simpler way.

So work done to heap up uniform material found at ground level which is equivalent to lift the COM to a height of #h/4# from ground will be

#W=Mxxh/4=1/4hM#

#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 #alpha# and #kappa# corresponds to your substance.


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 #T# and #V#, which are given in the Helmholtz free energy Maxwell relation.

#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 #(3)# at constant temperature,

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

From #(2)#,

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

and also from #(2)#,

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

since the Helmholtz free energy is a state function and its cross-derivatives must be equal. Thus from #(3)# we get

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

or we thus go back to #(1)# to get:

#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 #alpha# and isothermal compressibility #kappa#,

#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 #DeltaU#, we first need to measure the temperature of the system, and reference #alpha# and #kappa#.

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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