u are very correct@honey tnx brb

honey:
V(velocity of sound)= 2X/T,
i) in the first case, T=4, then 4V=2x, dividing through by 2,
X-2V=0---------------------(i)
ii) case 2, the distance from the wall was reduced by 200, time(T)= 3.4, therefore
(2X-200)=3.4V
2(x-100)=3.4V, dividing through by 2
x-1.7=100----------------------(ii), bringing the two simultaneopus equations together,

X-2V=0
x-1.7=100, subtracting eqn ii from i
v=333.33, x=666.67
a) speed of sound= 333.33ms^-1
b)distance from the wall in case 2 = (2x-200)/2= (x-100)= 666.67-100= 566.67m
i hope i am correct.

4 equ 1 V=2d/t whr t=4s we av 4d=2d .e 4d-2d=0.....1. 4 equ 2 distance bcoms d-200 nd t=3.4.v=2(d-200)/3.4 wich =3.4d-2d=-400...solvin simultaneousli u gt V=666.67 nd D=1333.3

a metal wire contains 5.0 *10^-22 electrons per cm3 and has across sectional area of 1.0mm2.if the electrons move along the wire with a meean drift velocity of 1.0mm/s.calaulate the current in amperes in the wire if the electronic charge is 1.6 *10-19c.
A)10A
b)8A
c)12A
d)15A

buzyguy: a metal wire contains 5.0 *10^22 electrons per cm3 and has across sectional area of 1.0mm2.if the electrons move along the wire with a meean drift velocity of 1.0mm/s.calaulate the current in amperes in the wire if the electronic charge is 1.6 *10-19c.
A)10A
b)8A
c)12A
d)15A

dQ/dt=C_e*de/dv*V*A, Where C_e= charge of an electron= (1.6x10^-19).... de/dv= rate of flow of electrons with respect to volume= 5.0x10^{22[/]ecm[sup]-3}....V=mean drift velocity= 1.00mm/s, since i want to deal in cm, then v=0.1cm/s.... A=area= 1.00mm/s= 0.01cm[su]2[/sup], now that we have gotten all, the parameters, we fix it in the equation..... dQ/dt=(5x10²²)*(0.1)*(0.01)*(1.6x10^-19), after multiplying, dQ/dt= I(current) = 8A...

honey: dQ/dt=C_e*de/dv*V*A, Where C_e= charge of an electron= (1.6x10^-19).... de/dv= rate of flow of electrons with respect to volume= 5.0x10^{22[/]ecm[sup]-3}....V=mean drift velocity= 1.00mm/s, since i want to deal in cm, then v=0.1cm/s.... A=area= 1.00mm/s= 0.01cm[su]2[/sup], now that we have gotten all, the parameters, we fix it in the equation..... dQ/dt=(5x10²²)*(0.1)*(0.01)*(1.6x10^-19), after multiplying, dQ/dt= I(current) = 8A...

gr8 job,bt u had to convert to m nt cm.ansa z 8 * 10-12A
bt nt in option anywy jst a trick.

I giv it to u.

buzyguy: gr8 job,bt u had to convert to m nt cm.ansa z 8 * 10-12A
bt nt in option anywy jst a trick.

I giv it to u.

Buzyguy bro.. Pls help us post oau postume questions for this year(Physics i mean)

a block of weight W on an inclined plane of angle 30deg is attached by a string passing over a pulley to another block of weight 50kg hanging vertically.the coefficient of static friction between the plane and the weight W is 0.3.if the system is just at the point of moving down the plane determine W.

topsonkay: Buzyguy bro.. Pls help us post oau postume questions for this year(Physics i mean)

ok no wahala

a lens of focal lenght f1=12cm is placed coaxially to the left of a second lens of focal lenght f2=18cm.both lenses are thin biconvex and are separated by 24cm.calculate the magnification of an object placed 15cm to the left of the first lens.

a ray of light traveling in water hits a water/glass interface at an angle of incidence 30deg.calculate the angle of refraction in the glass in degrees given that the refractive indices of water and glass are 1.33 and 1.50 respectively
a)19.5
b)22.1
c)26.3
d)34.3

buzyguy: ok no wahala

be back 4 d oau questns later

A Simple astronomical telescope in normal adjustment has an objective of focal length 100cm and an eyepiece of focal length 5cm. 1.Where is the final image formed A. 20cm B.1/25cm C. 5cm D. 25cm
2. What is the angular magnification.. A.20 B.25 C.1/20 D.1/25

buzyguy: a ray of light traveling in water hits a water/glass interface at an angle of incidence 30deg.calculate the angle of refraction in the glass in degrees given that the refractive indices of water and glass are 1.33 and 1.50 respectively
a)19.5
b)22.1
c)26.3
d)34.3

water-glass=aNg/aNw water-glass=1.5/1.33 ;=1.13 R.i=angle of incidence/angle of refraction 1.13=sin30/sinr.. Sinr=0.5/1.13... R=sin^-1(0.4424) R=26.3deg.. Option C

buzyguy: a block of weight W on an inclined plane of angle 30deg is attached by a string passing over a pulley to another block of weight 50kg hanging vertically.the coefficient of static friction between the plane and the weight W is 0.3.if the system is just at the point of moving down the plane determine W.

the whole pulley system can be considered as a pulley having the block W on the inclined plane, and the mass 50kg hanging vertically on the high end of the pulley.... so the following can be deduced, since the block is about to move down, the minimum forward force =mg.sin30, and also the frictional forces are [1]the tension of the rope =F_T and [1] static friction =0.3mg.... now £_ax=mg.sin30-0.3mg-F_T (since mg=W), =0.5W-0.2W-F_T=ma ....£_by= F_T-50g=50a ...... adding the two equations, 0.2W-50g=a(50+m), now assuming the block is yet to move, a=0 (taking g=10ms^-2), then the equation becomes 0.2W=500 , dividing through by 2, W=1000N m=100Kg.

honey: the whole pulley system can be considered as a pulley having the block W on the inclined plane, and the mass 50kg hanging vertically on the high end of the pulley.... so the following can be deduced, since the block is about to move down, the minimum forward force =mg.sin30, and also the frictional forces are [1]the tension of the rope =F_T and [1] static friction =0.3mg.... now £_ax=mg.sin30-0.3mg-F_T (since mg=W), =0.5W-0.2W-F_T=ma ....£_by= F_T-50g=50a ...... adding the two equations, 0.2W-50g=a(50+m), now assuming the block is yet to move, a=0 (taking g=10ms^-2), then the equation becomes 0.2W=500 , dividing through by 2, W=1000N m=100Kg.

a little bit tricky,u cld see dat d tension in rope was nt givn datz negligible which means Wsin30=fr force so far its moving downward lik u did above.tthen Wsin30=uN
WHERE N IS Wcos30 normal resolution
so simply,Wsin30=0.3*Wcos30
therefore,Wtan30=0.3
W=0.3/tan30
cn u see dat d 50kg is useless

[quote author=topsonkay] water-glass=aNg/aNw water-glass=1.5/1.33 ;=1.13 R.i=angle of incidence/angle of refraction 1.13=sin30/sinr.. Sinr=0.5/1.13... R=sin^-1(0.4424) R=26.3deg.. Option C[/quoteu]u got ittt bro

at what altitude aboove the earth's surface would the acceleration due to gravity be 4.9m/s2?assume that the mean radius of the earth is 6.4*10^6m and acceleration due to gravity on the earth's surface is 9.8m/s2

at what temperature will the celsius and fahrenheit temperature scales record the same reading?

steel rails 8.0m long are laid end-to-end in winter when the temperature is -10°C
a)how much space should be left between them to allow for expansion in the summer when the temperature could reach 50°c?
B)if tthe rails had been butted against each other at -10°c,what stress would each have to withstand at 50°c in order to prevent buckling?
(for steel,take ¤=1.2 * 10-5/k and E=21 * 10^10N/m2)

three moles of a gas expand isothermally.if the initial temperature of the gas is 300k and its final volume is twice the initial volume,calculate the work done by the gas.

buzyguy: at what temperature will the celsius and fahrenheit temperature scales record the same reading?

-40

topsonkay: -40

gud vry correct!

buzyguy: gud vry correct!

Can i show workings for the purpose of others?

buzyguy: three moles of a gas expand isothermally.if the initial temperature of the gas is 300k and its final volume is twice the initial volume,calculate the work done by the gas.

 fnk we nid the molar gas constant "R" in solvin ƌis question.

buzyguy: at what altitude aboove the earth's surface would the acceleration due to gravity be 4.9m/s2?assume that the mean radius of the earth is 6.4*10^6m and acceleration due to gravity on the earth's surface is 9.8m/s2

g=GM/R2..since G is constant lets equate.9.8r^2/m^2=4.9(h+r)^2/m^2 whr h =d altitude..since m id equal,it can cut out so we av 9.8r^2=4.9(h+3)^2 removin bra we av 9.8r^2=4.9h^2+9.8hr+4.9r^2..clllectin like terms,we av 9.8r^2-4.9r^2=4.9h^2+9.8hr..wich equals 4.9r^2=H^2+2hr..completin ƌ square we av 4.9r^2+r^2=(h+r)^2....suareroot of (5.9r^2)-r=H...our H=9145.5......not so sure sha

topsonkay: Can i show workings for the purpose of others?

yes boss to mak it complete

buzyguy: three moles of a gas expand isothermally.if the initial temperature of the gas is 300k and its final volume is twice the initial volume,calculate the work done by the gas.

work done by gas in an isothermal process {W=P₁V₁*In(V₂/V₁)}, from the ideal gas equation, PV=nRT, and the final volume is twice the initial volume, then W=nRT*In2 , replacing the parameters in the equation, W=3*8.314*300*0.693= 5300J approx, i hope i am correct.

honey: work done by gas in an isothermal process {W=P₁V₁*In(V₂/V₁)}, from the ideal gas equation, PV=nRT, and the final volume is twice the initial volume, then W=nRT*In2 , replacing the parameters in the equation, W=3*8.314*300*0.693= 5300J approx, i hope i am correct.

absolutely correct! Bt it shld be 5186.5j chck ur calculator

lavosier:
g=GM/R2..since G is constant lets equate.9.8r^2/m^2=4.9(h+r)^2/m^2 whr h =d altitude..since m id equal,it can cut out so we av 9.8r^2=4.9(h+3)^2 removin bra we av 9.8r^2=4.9h^2+9.8hr+4.9r^2..clllectin like terms,we av 9.8r^2-4.9r^2=4.9h^2+9.8hr..wich equals 4.9r^2=H^2+2hr..completin ƌ square we av 4.9r^2+r^2=(h+r)^2....suareroot of (5.9r^2)-r=H...our H=9145.5......not so sure sha

u tried

fg=(GMem)/r2=mg
=>r2=GMe/g=GMe/4.9 eq(1)

on the earth's surface,g=GMe/Re2
=>GMe=gRe2=(9.(6.4*10^6)^2
=4.01*10^14

eq.1 then gives r2=(4.01*10^14)/4.9,or r=9.05*10^6m from the earth's centre
now,(9.05-6.4)*10^6=2.65*10^6m from the earth's surfac

My generals are here...that's great...let me take a seat and watch as my ogas they perform their magic here

Wow. This guys r on point.point.

a stone is projected upwards with a velocity of 20m/s.Two seconds later a second stone is similarly projected with the same velocity.When the two stones meet,d second one is rising at a velocity of 10m/s.neglect air resistance .calculate the length of time the two stone is in motion b4 they met.?..