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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

thanks for that answer but im terrible with math.

the graph that the tutor showed me looked like a regular graph not a quadratic.

thank you for your help

all the best

simon

**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

**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

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:)

**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:)

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

**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

:):)

**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

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

**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:)

**ninjaman**- Replies: 3

hello

I have to find height of a box, I have length and width but I don't know height. unfortunately im not gifted with these things. my brain stops ticking. anyone got any tips or hints

thanks

simon

**ninjaman**- Replies: 2

hello

i have 6t^2 -18t +12

i got (3t^2-3)(4t+2)

i did 6 * 12 = 72

2 36

3 24

4 18

6 12

8 9

middle term is 18 = 6 + 12

i put this in a box

3 3

6t^2 | 6 2

+12 | +12 4

not sure if this is correct?

any help

cheers

simon:)

no the bottom bit is (x-3)^1/2

i dont understand how that square root comes about

cheers

**ninjaman**- Replies: 7

hello,

i have this, y = (x+1)(2x+1)^3/(x-3)^1/2

i got

dy/dx = (x+1)(2x+1)^3/(x-3)^1/2 [1/x+1 + 3/2x+1 - 1/2(x-3)]

im not sure about this?

thanks

simon:)

nope!

nevermind!

I got it!

I HAVE GOT THIS MAN!!!!

how do you do the third with that, also I checked an online derivative calculator and it says that is wrong?

it gave the same answer with the power of 2, 48(2x-3)^2

I looked at the steps on this calculator and didn't understand them.

Im not sure how to go onto the third derivative.

would I use (2x-3) as U or 48(2x-3) as U?

**ninjaman**- Replies: 4

I have to do second and third order derivatives

(2x-3)^4

for the first one I got

8(2x-3)^3

for the second I got

48(2x-3)

the way I was shown is this,

2x-3=U du/dx = 2

y= u^4 dy/du = 4u^3

dy/du * du/dx = 4u^3 * 2 = 8u^3

since u = 2x-3

8(2x-3)^3

then I got lost

here I got stuck and not sure where I went wrong, any help

thanks

simon

d^2s/dt = -3sint - cost

d^3s/dt = -3cost - - sint = -3cost +sint

if this isn't right I don't know what is.

**ninjaman**- Replies: 3

hello

I have this question and given my answer for second and third derivatives. not sure if its right though

thanks a lot, I looked at the bit where you put in a radical(I think that's what its called, the sqrt sign) that threw me.

thanks also for the latex thing, that will help a lot when writing out maths stuff!

all the best

simon

**ninjaman**- Replies: 3

hello

I have a question im stuck with

y = (7x-4) 3/2 (3/2 is supposed to be 3 over 2, I don't know how to do that on here)

I got this,

u = 7x-4

du/dx = 7

y = u 3/2

dy/du = 3/2u -3/2 (the -3/2 is power, supposed to be above right of u)

du/dx * dy/du = 3/2u-3/2 * 7 = 21/2 u -3/2

since u = 7x-4, 21/2 (7x-4) -3/2

im thinking most of it is correct except for the -3/2, the fraction part has me concerned.

thanks

simon:)

**ninjaman**- Replies: 2

hello

I completed the answer for this question some time ago but I cant remember the start of it which I didn't write down. please could someone confirm what I have so far.

this is the question

form a ship at sea, the angles of elevation of the top and bottom of a vertical lighthouse standing on the edge of a vertical cliff are 31 degrees and 26 degrees respectively. if the lighthouse is 25m high, how high is the cliff?

H is height in meters (of cliff)

H + 25 / tan 31= H / tan 26, this is how I started the formula. should I add + 25 to the H + 25 / tan 26?

(H + 25) tan 26 = H tan 31

H + 25 tan 26 = H tan 31

H tan 26 + 25 tan 26 = H tan 31

25 tan 26 = H tan 31 - H tan 26

25 tan 26 = H (tan 31 - tan 26) this second bit works out to 0.1131280305, I divided the second bit by this to move it across the equals sign

25 tan 26 / 0.1131280305 = 107.8 (rounded up)

does this make any sense to anyone?

I cant remember how I started it. I have the rest written down.

Christmas soon!!!

all the best

simon:)

what would that make side a then?

I was told that the answer is gave was correct?