Further Mathematics: Number Systems Essay Example

Assignment title

Assignment 1

Task (LO 1.1)

The number of moles (Further Mathematics: Number Systems) of an ideal gas are obtained using the equation
Further Mathematics: Number Systems 1, where

Further Mathematics: Number Systems 2is the pressure measured in Pascals (Pa),

Further Mathematics: Number Systems 3is the volume measured in cubic metres (m3),

Further Mathematics: Number Systems 4is the universal gas constant (8.31447JK-1mol-1),

Further Mathematics: Number Systems 5is 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. ) Further Mathematics: Number Systems 7
    or Further Mathematics: Number Systems 6
    Calculate the uncertainty due to the random error of the repeated measurements (you can either use standard error =

Further Mathematics: Number Systems 8

  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 equationFurther Mathematics: Number Systems 9. This is obtained by adding 273 to the Celsius temperature. The measured values of the other quantities are:

Further Mathematics: Number Systems 10Pa

Further Mathematics: Number Systems 11cm3

Further Mathematics: Number Systems 12JK-1mol-1

  1. with its uncertainty given to an appropriate number of significant figures. Further Mathematics: Number Systems 13
    Calculate the value of

N= [Further Mathematics: Number Systems 14][750]

Further Mathematics: Number Systems 15[Further Mathematics: Number Systems 16][305.3]

Further Mathematics: Number Systems 171237.5*10^5

2538.407691

=48751.03413

Task (LO 1.2)

  1. Convert the binary 1101111010101101 into

  1. 1101111010101101

1101= .64

67521.64

  1. Hexadecimal

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.

Further Mathematics: Number Systems 18

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

Task (LO 1.3)

Two complex numbers are such that Further Mathematics: Number Systems 19 and Further Mathematics: Number Systems 20

  1. .Further Mathematics: Number Systems 23 and the principalFurther Mathematics: Number Systems 22 in rectangular (Cartesian) form. Now, findFurther Mathematics: Number Systems 21Write down

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

Further Mathematics: Number Systems 24= (a – j)(b + j) = c – j + aj – dj2 = c + bj – a(– 1) = c + aj

on remembering that j2 = – 1

  1. Further Mathematics: Number Systems 27 into polar form i.e. in the form Further Mathematics: Number Systems 26 and Further Mathematics: Number Systems 25Convert

The
Further Mathematics: Number Systems 28is

Further Mathematics: Number Systems 29

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

(remember  is in radians)

cos  = 1 – + + … and sin  =  – + …

to obtain the required relation:

ej = cos  + j sin 

Using this we have
as required.

  1. in rectangular (Cartesian ) form.Further Mathematics: Number Systems 32 . Now, write Further Mathematics: Number Systems 31 and the principalFurther Mathematics: Number Systems 30Find

=
Further Mathematics: Number Systems 33for
Further Mathematics: Number Systems 34 .

For
Further Mathematics: Number Systems 35 . = 0 we get

= Further Mathematics: Number Systems 36 . = cos
Further Mathematics: Number Systems 37+ z sin
Further Mathematics: Number Systems 38 .=
Further Mathematics: Number Systems 39– z

For z = 1 we get

= Further Mathematics: Number Systems 40 .= Further Mathematics: Number Systems 41= z (Check: z1 = z2z = – z)

For z = 2 we get

= Further Mathematics: Number Systems 42 .= Further Mathematics: Number Systems 43

= cos
Further Mathematics: Number Systems 44+ zsin
Further Mathematics: Number Systems 45 .= –
Further Mathematics: Number Systems 46– z

Task (LO 1.4)

Find the m complex roots of the equation
Further Mathematics: Number Systems 47 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
Further Mathematics: Number Systems 48]

(Further Mathematics: Number Systems 49)5 = [3(cos
Further Mathematics: Number Systems 50
+ i sin
Further Mathematics: Number Systems 51)]5

= 25(cos z + i sin
Further Mathematics: Number Systems 52)5

= 32(cos
Further Mathematics: Number Systems 53
+ i sin
Further Mathematics: Number Systems 54)

= 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
Further Mathematics: Number Systems 55 , where
Further Mathematics: Number Systems 56and
Further Mathematics: Number Systems 57is the frequency of the alternating power supply, measured in Hertz (Hz).

Further Mathematics: Number Systems 58

  1. to an appropriate number of significant figures Further Mathematics: Number Systems 59
    the impedance

L=240||(r+c)

= = 500  600 = 30000

= =
Further Mathematics: Number Systems 60

= 2 = 5002 = 2500

  1. Further Mathematics: Number Systems 61to an appropriate number of significant figures

Arg (z) = Arg (z1) + Arg (z2) =Further Mathematics: Number Systems 62 +
Further Mathematics: Number Systems 63

Arg =
Arg (z1) – Arg (z2)

Arg 1 = – Arg z1 = Further Mathematics: Number Systems 64Further Mathematics: Number Systems 65

Save Your Time for
More Important Things
Let us write or edit the article on your topic