Physics help

[Saladin]

Can someone please tell me how to solve this question:


The question in in the Edexcel AS student book page 126

In a resistance experiment, a gold ring was connected into a circuit as a resistor. The connection touched diametrically opposite points on the circular ring. Its diameter is 2 cm, and the metal's cross section is 3 mm by 0.5 mm. The voltemeter connected accross the ring measured 8.3 mV, whislt the current through it measured 18 A.

a) Calculate the resistivity of gold
b) Describe a practical difficulty in this experiment

[astarmathsandphysics]

Gimme half am hour to get up

[loggerhead]

heres my effort, not to sure if its right.

R = P * L / A

first we need to work out R, L and A

R = V / I
R = 0.0083 /18
R = 0.000461

L = half circumference
L = 1/2 * pi * 0.02
L = 0.0314

A = 2 * cross sectional area
A = 2 * 0.003 * 0.0005
A = 0.000003

therefore 0.000461 = P * 0.0314 / 0.000003
P = 0.000461 * 0.000003 / 0.0314
P = 4.4 * 10^-8 ohm m

hope thats right.

not to sure about part b) possibly to do with such small measurements that theres a high %age error, or that the cross sectional area is inconstant

[astarmathsandphysics]

It might be difficult to get contacts on exactly opposite sides of the ring. Loggerhead I think your answer is cirrect.

[Saladin]

I did not understand how u calculated L though....

[astarmathsandphysics]

Circumference of the rind is pi*d. see the diagram.

[Saladin]

Thanx a lot astar, the illustration really helped.

[astarmathsandphysics]

Io is one of Jupiter’s moons. Some of the electrons released from the volcanic surface of
Io have an average velocity of 2.9 × 107 m s–1 towards Jupiter. The distance between
Jupiter and Io is 4.2 × 105 km.

(b) In this way a current of 3.0 × 106 A is created between Io and Jupiter. Calculate the
     number of electrons that arrive at Jupiter every second.
It will be I/e=3*10^6/1.6*10^-19=1.875*10^25

[Saladin]

here it is

15 b

[astarmathsandphysics]

The voltage across the 5ohm resitor, v=IR=0.6*5=3ohms so the voltage across the 30 ohm resistor and R is 9-3=6V for both since they are in parallel.
Use V=IR for the30 ohm resitor
V=IR
6=I*30 so I =6/30=0.2A

Thethe other 0.6-0.2=0.4A goes through R so I₂=0.4A

[Saladin]

its 15 not 13 its 15b,,,

[Saladin]

srry this is it

[astarmathsandphysics]

resistance of B is greater than A. For parallel circuits useP=V^2/R
The voltages are the same so B will have less power so is dimmer

[Saladin]

15 The current I in a length of aluminium of cross-sectional area A is given by the formula
I = nevA
where e is the charge on an electron.
(a) State the meanings of n and v.
(2)
n: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
v: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
(b) Show that the units on the left hand side of the equation are consistent with those on
the right hand side.

[astarmathsandphysics]

i=nave
A=Q/s =m^-3*m^2*m/s*Q
Q/s =Q/s