# Further Mathematics: Number Systems Essay Example

Assignment title

Assignment 1

The number of moles () of an ideal gas are obtained using the equation
, where

is the pressure measured in Pascals (Pa),

is the volume measured in cubic metres (m3),

is the universal gas constant (8.31447JK-1mol-1),

is the temperature measured in Kelvin (K)

In a repeated experiment, the temperature of the gas was measured by a digital meter to the nearest Celsius. The following repeated measurements were obtained :

32C, 33C, 32C, 34C, 32C, 33C, 32C, 32C, 32C, 31C

1. What is the uncertainty due to the reading error of the digital meter

Working = {(32+33+32+34+32+33+32++32+32+31)/(10)}

Max = 32.35

Min = 32.295

Error = [(Max-Min)/working]*100

[(32.35-32.295)/32.3]*100

Error= 0.17027863

1. )
or
Calculate the uncertainty due to the random error of the repeated measurements (you can either use standard error =

1. Calculate the mean of the of the repeated measurements and round off to an appropriate number of significant figures

Working = {(32+33+32+34+32+33+32++32+32+31)/(10)}

The temperature of the gas must be in Kelvin before it is used in the equation. This is obtained by adding 273 to the Celsius temperature. The measured values of the other quantities are:

Pa

cm3

JK-1mol-1

1. with its uncertainty given to an appropriate number of significant figures.
Calculate the value of

N= [][750]

[][305.3]

1237.5*10^5

2538.407691

=48751.03413

1. Convert the binary 1101111010101101 into

1. 1101111010101101

1101= .64

67521.64

1101111010101101

Now assign each group its corresponding hex value

1101 = F, 1110 = 5, 1010 = 4, 1101 = 14

When put together, we get F54E16

1. Draw a truth table to determine the output states at Z with all possible combinational input states at A, B, C & D.

 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

Two complex numbers are such that and

1. . and the principal in rectangular (Cartesian) form. Now, findWrite down

a + bj = a(b + j) + b(a + j) = b + j + a + j = b – aj

= (a – j)(b + j) = c – j + aj – dj2 = c + bj – a(– 1) = c + aj

on remembering that j2 = – 1

1. into polar form i.e. in the form and Convert

The
is

If we substitute z = j into this we get

ej= 1 + j + + + … + + …

= 1 + j – – j+ + …

on using j2 = – 1 repeatedly. Gathering real and imaginary terms together then gives

ej= 1 – + + … + j

cos  = 1 – + + … and sin  =  – + …

to obtain the required relation:

ej = cos  + j sin 

Using this we have
as required.

1. in rectangular (Cartesian ) form. . Now, write and the principalFind

=
for
.

For
. = 0 we get

=  . = cos
+ z sin
.=
– z

For z = 1 we get

=  .= = z (Check: z1 = z2z = – z)

For z = 2 we get

=  .= 

= cos
+ zsin
.= –
– z

Find the m complex roots of the equation
using De Moivre’s theorem in rectangular (Cartesian) form. Where m is given by 3 + n modulo 5 and n is your dataset number.

[For example, for n=9 then n modulo 5 = 9 modulo 5 = 4. Therefore 3 + n modulo 5 = 3+4 = 7. And the question would be to find the 7 roots of
]

()5 = [3(cos

+ i sin
)]5

= 25(cos z + i sin
)5

= 32(cos

+ i sin
)

= 32(-3/2 + 1/2 i)

= -16√3 + 16 i.

 Task (LO 1.5) The complex impedance of a circuit containing a resistor of resistance R, inductor of inductance L and a capacitor of capacitance C is given by , where and is the frequency of the alternating power supply, measured in Hertz (Hz). to an appropriate number of significant figures the impedance L=240||(r+c) = = 500  600 = 30000 = = = 2 = 5002 = 2500 to an appropriate number of significant figures Arg (z) = Arg (z1) + Arg (z2) = + Arg = Arg (z1) – Arg (z2) Arg 1 = – Arg z1 = –