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#1 Re: Help Me ! » trig problem, law of cosines » 2016-08-04 02:39:48

thanks bob, unfortunately i dont understand the ruler/protracto method. the book i am using doesnt go into that sort of thing. thanks for your help.

#2 Help Me ! » trig problem, law of cosines » 2016-08-03 23:36:36

ninjaman
Replies: 5

hello,
i am having a problem with law of cosines. i am using the C^2 = A^2-2ABcosc+B^2
what do i do if i dont have a value for A?
i have tried to rearrange to find A but with no luck.
any help would be great.
i get to here usually,
2ABcosc = A^2+B^2-C^2,
i then work out 2Bcosc, as i would have those values but no A
sometimes i think square root both sides. this would remove the exponents on the right and square everything on the left but that does not work.
i am using a book called "trigonometry: essentials practice workbook with answers" by chris mcmullen. it uses only certain values that mean you do not require a calculator. the values you use are 0, 30, 45 and 90 degrees. sine, cosine and tangent for these values and only those values.
any help would be great thanks
simon

#3 Help Me ! » drawing phasor diagram » 2016-04-17 00:15:45

ninjaman
Replies: 0

hello

i have a parallel rlc circuit which seems to be inductive with a 1 amp magnitude(?)
the circuit is explained in a post below about rlc circuits in parallel with complex numbers.
i havent drawn a phasor diagram. the capacitor and resistor currents are really low, in the micro amps with the inductor at 1 amp which is the supply current. so i am guessing that the phasor diagram would be the voltage pointing to 3 oclock and the current pointing toward 12 oclock, the voltage would have a magnitude of 10(volts, the supply voltage) and the current would have a magnitude of 1 amp(the supply current).
would this be correct. i have read a few things on the internet but not to sure if i have grasped the idea.

thanks

simon

#4 Help Me ! » RLC parallel circuit, complex numbers, done it but not sure » 2016-04-15 04:16:18

ninjaman
Replies: 1

hello all,

been a while!

i have just started my HND electronics course and i have to do some circuit analysis.
i have a parallel RLC circuit. three components, R = 12k ohms, L = 10mH, C = 10nF
with a 10sin(1000t) supply
i have to find the total impedance in polar form
supply current in polar form
current through each component in polar form
q factor
bandwidth and
resonant frequency

i written out what i have done and attached it (hopefully)

thanks

simon

#5 Re: Help Me ! » laplace problem » 2015-05-13 01:57:03

no, my tutor has told me that he thinks it is wrong, though he hasnt checked his answers?

he told me he would check with other students and get back to me. i posted this question on another forum and they told me this is correct. im not sure as i am terrible at this. the question is almost exactly the same as a question posted on the internet.

http://www.intmath.com/laplace-transformation/10-applications.php

this is the web site, its the first question with the answer. the only difference is the capacitor, voltage and resistor values. the whole question is identical. i followed the example changing the values as i went. so i thought that it was accurate. i just wanted to check as i have no way of knowing if this is correct. i have changed the degree that i will be studying at uni because it has less maths. thats a change in career. thats how much i drag at maths.
i just wanted to know if this is correct answer with the correct work.

thanks
simon

ninjaman
Replies: 3

not sure here
any help!
thanks
simon

#7 Re: Help Me ! » not sure of the answer » 2015-04-30 21:30:19

i have just looked at simplification and cant seem to find anything that looks like the problem i have.

i did 1/400 on the calculator and pressed the S<>D button and got 0.0025 which i can rewrite as 25x10^-3 which is what i have to show.

i still don't understand how to do this.

#8 Re: Help Me ! » not sure of the answer » 2015-04-30 07:46:44

thankyou for you help bobby, please could you explain how you got this. what is the process called so I can look it up.

thanks
simon

#9 Re: Help Me ! » not sure of the answer » 2015-04-30 02:00:37

hello

im not actually sure how to work it out. i have to rearrange it somehow. is there a name for that process.
the bottom part is supposed to be the top rearranged to be able to find I
i dont know how to do this. a little hint would be good

thanks
simon

i just had another look and i think that it is 1/50 instead, not 1/200
otherwise the rest is the same

#10 Re: Help Me ! » not sure of the answer » 2015-04-30 00:43:38

hello

thanks for that bobby

is this right?

thanks
simon

#11 Help Me ! » not sure of the answer » 2015-04-30 00:01:05

ninjaman
Replies: 9

hello
i have a problem  i have this

I =  5*10^-5/1+(2*10^-2) * s = ?

i dont know s, that shouldnt matter.

i think this equals (5*10^-3) 1/200+s

i have done this on latex but cant post it so this will probably look weird, its laplace transform.
i have to rearrange to find I
any help would be great

thanks

simon

#12 Re: Help Me ! » j notation, equivalent circuit resistance » 2014-11-01 03:17:24

hello

could someone please let me know if this is correct as I have been told that I have done this wrong though I didn't inform the person of how I did it. this is using division of conjugate complex numbers.

I have followed the method my tutor showed me and I think this is correct.

any help or a push in the write direction would be great!!!

thanks
simon

#13 Help Me ! » j notation, equivalent circuit resistance » 2014-10-31 22:44:03

ninjaman
Replies: 2

hello

I have a circuit with values
v1 = 20 + j0
r1 = 15 +j0
l1 = 0 +j10
c1 = 0 - j5

I have to use superposition and division of complex conjugate numbers.
the resistor, inductor and voltage source are in series. the capacitor is in parallel with the voltage source

I added the resistor and inductor, (15 + jo) + (0 + j10) = 15 + j10, find the conjugate of this 15 - j10
so then product over sum for the components that are in parallel. the inductor/resistor || capacitor

(0 - j5)*(15 - j10)
(15 + j10)*(15 - j10)

for the numerator I got

0 x 15 = 0
0 x -j10 = 0
-j5 x 15 = -j75
-j5 x -j10 = (j^2)50 = -50   minus 50 because of (j^2) = -1

so the result is -50 -j75, which im not sure about

then the second line

15 x 15 = 225
15 x -j10 =-j150
j10 x 15 = j150
j10 x -j10 = -(j^2)100,    (j^2) = -1 x -100 = 100

225 -j150 +j150 +100, the middle terms cancel leaving 225 + 100 = 325

so I have,     -50 -j75/325

then I divide each of the numerator by the denominator so,

-50/325 = -0.153846

-75/325 = -0.230769

I think I can use this to find polar notation r and theta(the zero with a line through)

im not sure if any of this is right though, any help would be great thanks

simon

#14 Re: Help Me ! » kinetic energy » 2014-05-20 09:18:09

hello

sorry I didn't realize shivams had posted something. I go to a college in Hampshire, not teeside.
please could you point out the other forum or thread as I don't want to have to post the same thing if its already been answered with advice
thanks
simon

#15 Re: Help Me ! » kinetic energy » 2014-05-20 06:47:09

thanks for that answer but im terrible with math.
the graph that the tutor showed me looked like a regular graph not a quadratic.

all the best
simon

#16 Re: Help Me ! » kinetic energy » 2014-05-20 06:10:42

I don't know, the question was given like that. the mass stays the same I imagine. I just think it may have been left in as a clue to finding the answer.

#17 Help Me ! » kinetic energy » 2014-05-20 02:59:55

ninjaman
Replies: 8

hello

I have these two questions

c)    Plot a graph of the kinetic energy of the mass against time. Explain your calculations and state formulae used.

d)    Plot a graph of the kinetic energy of the mass against distance. Explain your calculations and state formulae used.

I have used this formula for the first one,
Ek = ½ mv2 = ½ ma2t2
Ek is proportional to t2.

this for the second
Ek = ½ mv2 , but we can substitute, giving
Ek = mgy

in not too sure how to show the graphs, but is the above formula along the right lines?
also I have to explain the use of the formula. I think this is just explain why I used it and any transposing required. if anyone knows a better answer for this one please give some advice.

taken from here, http://colourpoint.co.uk/sample_files/physics_as.pdf

this is the first part of the question,

A horizontal force of 80 N acts on a mass of 6 kg resting on a horizontal surface.  The mass is initially at rest and covers a distance of 5 m in 0.92 s under the action of the force.  Assuming there are no energy losses due to air resistance and therefore that the acceleration is constant:

a)    Calculate the total energy expended in the acceleration, I got 11.81 acceleration

b)    Calculate the coefficient of kinetic friction between the mass and the surface.  Suggest materials from which the block and table might be made in order to give such results.  I got 0.155

thanks
simon

#18 Help Me ! » normal distribution, find percentage » 2014-05-05 08:48:05

ninjaman
Replies: 1

hello

I have a question

the lengths of pins produced by a machine follow a normal distribution with mean 2.54cm and standard deviation 0.04cm. a pin is rejected if the length is less than 2.44cm or more than 2.60cm

A/ find the percentage of pins that are accepted.

B/ if it is decided that 2.5% of the pins are to be rejected because they are too long and 2.5% because they are too short, determine the new range of acceptable lengths.

for the first part I have the formula, z = x-MU / sigma, I rearranged this

sigma is 0.04cm
MU is 2.54cm

so I used this z * sigma + MU = x, z * 2.58 = ? so here I thought I would take a value from a table and divide by 2.58 until I got an accurate z value. then use that z for the rest.
if anyone could confirm this method as im not to sure? the rest I think I can get.

thanks
simon

#19 Re: Help Me ! » evaluate integration » 2014-03-20 10:54:41

hello

I ended up with loge 1 * 1/3 e ^3*1 and, loge 0 * 1/3 e ^3*0
I also put those upper and lower limits in the x in the [] 2x/3 = 2*1/3 and 2*0/3
and e^3*1, e^3*0
any help

im fed up of this

thanks
simon:)

#20 Help Me ! » volumes of revolution » 2014-03-17 00:16:39

ninjaman
Replies: 1

hello

i have a question

s = integration sign

s e^sinx cosx dx

i have

sinx = u

du/dx = cosx

then

s e^u du

du/dx = cosx

du = cosx dx

s e^u du

e^u +c

e^sinx +c

not sure if this is correct, i havent yet read about the latex thing.
any help

thanks
simon:)

#21 Re: Help Me ! » evaluate integration » 2014-03-16 23:12:42

ive tried to add to this, i have read through what i have found and im still confused

#22 Help Me ! » mechanical engineering, » 2014-03-10 11:12:30

ninjaman
Replies: 0

hello

I have a question

a hollow steel shaft with a diameter ratio of 0.75 and a length of 4m is required to transmit 1Mw at 120rpm. the maximum shear stress is not to exceed 7 Mpa and the overall angle of twist is not to exceed 1.75degree. assume the modulus of rigidity for steel is 80Gpa.

a) determine the necessary outside diameter of the shaft so that both of the above limitations are satisfied

b) determine the actual maximum shear stress and the actual angle of twist.

first I tried finding the torque(T) using P(power) = Tw(w being 2piN( N is rotational speed in rps)) I think w is omega

I rearranged the formula to get T = P/w

so, 1000,000 / 2pi 2                   the second 2 is from 120(rpm) / 60(rps) to get seconds

= 79577.47

then, as I have degrees and would require radians I converted that

1.75 * pi / 180 = 0.0305 radians

so using a formula to find J ( the symbol for second moment of area)

the formula is J = T * L / G (theta sign), so

79577.47 * 4 / 80x10^9 * 0.0305 = 130454.87x10^-9 = J

which im not sure about. I have to find the outside diameter using D^4-d^4,  D is outside diameter and d is inside diameter.

J = pi * (D^4 - d^4) / 32, this is the formula using diameter to find J, I have J and so would try to rearrange

I tried this,

D^4 - d^4 = pi * J / 32

any help would be great, I only know that diameter ratio is 0.75, the inside diameter is three quarters

If I did, d^4 = pi * J / 32, then d^4 answer would be three quarters of the D^4 answer. so if I took that and cut it into thirds that is what I would add to d^4 to get D^4???

any help would be great

thanks
simon
:):)

#23 Help Me ! » evaluate integration » 2014-03-03 00:49:55

ninjaman
Replies: 5

hello

s = integration sign (top number, bottom number)

i have this :

s (1, 0) (2xe^3x dx)

u = 2x

du/dx = x/2

dv/dx = e^3x

v = s 1/3 ^e3x

i really have no clue what im doing.
if someone could tell me what i do any of this it would really help

thanks
simon:)

i have 2x so to integrate it, i increase the power by 1 and divide by this figure. so 2x = 2x^1, add 1 = 2x^2/2 (correct?)

is 2xe^3x treated as one number, like 2*xe^3x. one answer the lecturer wrote was 2e^3x/3, = 2/3 s e^3x dx

i am looking on the internet at the moment for any answers and coming up with nothing.
does anyone know any good sites that explain with good, easy to follow examples.

thanks

#24 Re: Help Me ! » height » 2014-02-28 08:53:39

NEVERMIND, I GOT THE ANSWER THANKS ANYWAY ALL THE BEST!!!:D:D

hello

I have this question

a rectangular sheet of metal measuring 120mm by 80mm has equal squares of size x cut from each of the corners. the remaining flaps are then folded upwards to form an open tray. draw a neat sketch of the net and prove that the volume of the tray is given by:

v = 4x^3 - 400x^2 + 9600x

a) find the value of x such that the volume is a maximum, making sure you show how you prove your value is a maximum not a minimum. work out the maximum volume of the tray.

b) verify your result for question 6a using a second method. present your alternative method in writing, ensuring you compare both ease of use and accuracy of results of the two methods.

the second bit is whats getting me. I can use the results of length and width which are 88.612mm and 48.612mm respectively. I cant include height and have to find volume.

im guessing its something to do with the cut outs.
I have maximum volume of box when x = 15.594

length = 120 -2 * 15.694 = 88.612
width = 80 -2 * 15.694 = 48.612

so I have the starting width and length, the finishing width and length and two lots of something to be taken away.

I found something close, I have 15.69499126 = x, this came from the general quadratic equation

I did this,

120 - 2 * 15.69499126 = 88.61001748

88.61001748 + 2 = 90.61001748

120 - 90.61001748 = 29.38998252

divide this by 2 to get = 14.69499126

this last number is supposed to be x but it is a whole 1 out. 14 instead of 15. I think this is correct but there is a 1 missing somewhere.
I think if I put all this together I should be able to get the correct answer giving me a final volume and height.
unless I have done something really daft here and got a knot in my brain.....again!:D

NEVERMIND, I GOT THE ANSWER THANKS ANYWAY ALL THE BEST!!!

anyway, if anyone could point me in the right direction. I would really appreciate it

thanks
simon

#25 Help Me ! » elastic modulus » 2014-02-26 08:13:22

ninjaman
Replies: 0

hello

I have a question im stuck on.

a beam must withstand a bending moment of 285kNm. if the maximum stress must not exceed 200Mpa, determine the elastic modulus z and select an appropriate I section from the given table. (the table is in the book that I have)

im not sure what the math formula im supposed to use.

the elastic modulus is z = I / y

I don't know what I is or Y

I have M, the maximum bending moment Nm
and, sigma which is stress in pa

so I thought M/sigma = z.       285*10^3 * 200  = 1425

im not sure if the M in Mpa is mega(1000 000) or milli or what. when I use mega the answer is really small.

a push in the right direction please

many thanks

simon:)