iAmiAdam
New Member
Yeah, I meant in College, I don't tend to type properly on forums anways.
If you could PM them to me Marinyork that'd be great, I'm sure no one else wants to see me learning haha.
Rearranging equations.
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Yeah, I meant in College, I don't tend to type properly on forums anways.
If you could PM them to me Marinyork that'd be great, I'm sure no one else wants to see me learning haha.
twentysix by twentyfive
Clinging on tightly
- Location
- Over the Hill
No No - we all want to see them. Then we can have a race
Seriously keep practising as MarinYork says. The more you do the easier it will get
Hello iAmiAdam here's a few for you. I'll add in some of the solutions at some time later as I have the feeling you may not have been taught some of this stuff and the temptation to take a peek may be overwhelming. There's probably a typo in there somewhere.
1. a[sup]3[/sup] + 8b[sup]2[/sup] =8(c +d) Find in terms of a
[sup] [/sup]
2. x[sup]3[/sup] + 2x[sup]2[/sup] - 8x = (t + u)(x-2) rearrange for t
[sup]
[/sup]
3. b[sup]2/3[/sup] = [sup]3[/sup]√(4a) [sup]
[/sup]
4. 64[sup](2/3)[/sup] =
5. c[sup]2[/sup]=a[sup]2[/sup]+ b[sup]2[/sup] -2ab cosC
Find CosC
6. sinθ = O/H find in terms of θ where the inverse of sin is sin[sup]-1[/sup]
7. D=ST find T.
8. 7x[sup]2[/sup] + 8x = -2 and find x (hint rearrange and use the quadratic equation formula)
9. 1/ (1/x) find x.
10. 1/f = (1/u) + (1/v) Find in terms of v
11. F= (Q[sub]1[/sub]Q[sub]2[/sub])/(4πR[sup]2[/sup] ) Find R
12. (p[sub]1[/sub]V[sub]1[/sub])/(T[sub]1[/sub])[sub] [/sub] = (p[sub]2[/sub]V[sub]2[/sub])/T[sub]2[/sub] find in terms of T[sub]2[/sub]
13. 1/R = (1/R[sub]1[/sub]) + (1/R[sub]2[/sub]) rearrange for R
14. (a/b) / (c/d) express a,b,c and d as a single simpler fraction
15. (2/3) / (4/5) express as a single fraction without using a calculator
16. P=VI, V=IR, Find P in terms of V and R without I.
17. A question on planetary motion for you
Given F=(mv[sup]2[/sup] )/r
F=(GMm)/r[sup]2[/sup] and T=(2πr)/v
(a) Find an equation in terms of T, M and r (without v but can include constants). I'm not going to tell you what the symbols mean as an exercise in the power of rearranging equations.
(b) An approximate equation grabbed from this to roughly find the orbit time is where t stands for time
t[sup]2[/sup] / t[sup]2[/sup][sub]earth [/sub]= r[sup]3[/sup] / r[sup]3[/sup][sub]earth[/sub][sup] [/sup]
given r = 2.3x10[sup]11[/sup] m
r[sub]earth[/sub] = 1.5x10[sup]11[/sup] m
rearrange and find by some means the orbit time of Mars hint
18. 2= e[sup]s[/sup] find s where e≈ 2.718
19. 2=log[sub]10[/sub] 100 or 10[sup]2[/sup]=100
without a calculator find log[sub]10[/sub] (1000)
20. Using the rules of indices find y where e[sup]x[/sup]e[sup]-x[/sup] = y
[sup] [/sup]
[sup]
[/sup]
1. a[sup]3[/sup] + 8b[sup]2[/sup] =8(c +d) Find in terms of a
a[sup]3[/sup] = 8(c+d) - 8b[sup]2 [/sup]
a[sup]3[/sup] = 8(c+d - b[sup]2[/sup])[sup] [/sup]
a= [sup]3[/sup]√(8(c+d - 8b[sup]2[/sup])[sup] [/sup]
a=[sup]3[/sup]√(8) [sup]3[/sup]√(c+d - b[sup]2[/sup] )[sup] [/sup]
a=2 [sup]3[/sup]√(c+d - b[sup]2[/sup] )
a[sup]3[/sup] = 8(c+d - b[sup]2[/sup])[sup] [/sup]
a= [sup]3[/sup]√(8(c+d - 8b[sup]2[/sup])[sup] [/sup]
a=[sup]3[/sup]√(8) [sup]3[/sup]√(c+d - b[sup]2[/sup] )[sup] [/sup]
a=2 [sup]3[/sup]√(c+d - b[sup]2[/sup] )
2. x[sup]3[/sup] + 2x[sup]2[/sup] - 8x = (t + u)(x-2) rearrange for t
(x+4)(x-2)x= (t + u)(x-2)
(x+4)x = t+ u
t= x[sup]2[/sup] + 4x + u (left in this format as or another depending on whether u can be substituted in
(x+4)x = t+ u
t= x[sup]2[/sup] + 4x + u (left in this format as or another depending on whether u can be substituted in
[/sup]
3. b[sup]2/3[/sup] = [sup]3[/sup]√(4a) [sup]
[/sup]
[sup]3[/sup]√(b[sup]2[/sup])=[sup]3[/sup]√(4a)
([sup] 3[/sup]√(b[sup]2[/sup]) )[sup]3[/sup] =([sup]3[/sup]√(4a))[sup]3 [/sup]
b[sup]2[/sup] = (4a)
b= ±2√(a)
In indices b[sup]2/3[/sup]= (4a)[sup](1/3) [/sup]
(b[sup]2/3[/sup] )[sup]3[/sup] = ((4a)[sup](1/3)[/sup] )[sup]3 [/sup]
b[sup]2[/sup] = 4a
b= (4a)[sup](1/2)[/sup] check with +v value. (2√(a) )[sup]2[/sup] = 2√(a)2√(a) = 2x2√(a)√(a) = 4a
([sup] 3[/sup]√(b[sup]2[/sup]) )[sup]3[/sup] =([sup]3[/sup]√(4a))[sup]3 [/sup]
b[sup]2[/sup] = (4a)
b= ±2√(a)
In indices b[sup]2/3[/sup]= (4a)[sup](1/3) [/sup]
(b[sup]2/3[/sup] )[sup]3[/sup] = ((4a)[sup](1/3)[/sup] )[sup]3 [/sup]
b[sup]2[/sup] = 4a
b= (4a)[sup](1/2)[/sup] check with +v value. (2√(a) )[sup]2[/sup] = 2√(a)2√(a) = 2x2√(a)√(a) = 4a
4. 64[sup](2/3)[/sup] =
[sup]3[/sup]√(64[sup]2[/sup]) get rid of the cube root first = 4[sup]2[sub] [/sub][/sup]=16 note that 64x64 = 4096 and [sup]3[/sup]√4096 = 16 but one way may be easier
5. c[sup]2[/sup]=a[sup]2[/sup]+ b[sup]2[/sup] -2ab cosC
Find CosC
CosC = (a[sup]2[/sup] + b[sup]2[/sup] - c[sup]2[/sup] )/ (2ab)
6. sinθ = O/H find in terms of θ where the inverse of sin is sin[sup]-1[/sup]
sin[sup]-1[/sup]sinθ = sin[sup]-1[/sup] (O/H) ⇔ θ = sin[sup]-1[/sup] (O/H)
7. D=ST find T.
T = D/S
8. 7x[sup]2[/sup] + 8x = -2 and find x (hint rearrange and use the quadratic equation formula)
9. 1/ (1/x) find x.
for whole numbers 1/something is the inverse of the number itself. The inverse of the inverse cancels out and is just itself. Another way of looking at it is take 1/x and turn it upside down and multiply by it. So 1/(1/x) = 1*(x/1) obviously the 1s cancel so we're left with x
10. 1/f = (1/u) + (1/v) Find in terms of v
1/v = 1/f - 1/u
Taking the RHS of the equation we can write this as (u - f)/(uf)
u/(uf) is equal to 1/f and f/(uf) is equal to 1/u
So using this 1/v = (u-f)/(uf)
v = (uf)/(u-f) this is the same as writing 1/ {(1/f)- (1/u)}
Taking the RHS of the equation we can write this as (u - f)/(uf)
u/(uf) is equal to 1/f and f/(uf) is equal to 1/u
So using this 1/v = (u-f)/(uf)
v = (uf)/(u-f) this is the same as writing 1/ {(1/f)- (1/u)}
12. (p[sub]1[/sub]V[sub]1[/sub])/(T[sub]1[/sub])[sub] [/sub] = (p[sub]2[/sub]V[sub]2[/sub])/T[sub]2[/sub] find in terms of T[sub]2[/sub]
13. 1/R = (1/R[sub]1[/sub]) + (1/R[sub]2[/sub]) rearrange for R
Similarly as before
the RHS of the equation can be rewritten as (R[sub]2[/sub] + R[sub]1[/sub] )/ (R[sub]1[/sub]R[sub]2[/sub] )
and then invert this to find R
so R = (R[sub]1[/sub]R[sub]2[/sub])/(R[sub]2[/sub]+R[sub]1[/sub])
the RHS of the equation can be rewritten as (R[sub]2[/sub] + R[sub]1[/sub] )/ (R[sub]1[/sub]R[sub]2[/sub] )
and then invert this to find R
so R = (R[sub]1[/sub]R[sub]2[/sub])/(R[sub]2[/sub]+R[sub]1[/sub])
14. (a/b) / (c/d) express a,b,c and d as a single simpler fraction
When dealing with fractions like this the split denominator is turned upside down and placed next to the top so (a/b)*(d/c) = (ad)/(bc)
15. (2/3) / (4/5) express as a single fraction without using a calculator
As before 4/5 is placed upside down and multiplied with 2/3 so we have (2/3)*(5/4) = (2x5)/(3x4) = 10/12 which cancels to 5/6 try on a calculator if you don't believe it. This trick saves a lot of jabbing into calculators. Alternatively you could multiply up to make common factors on the bottom half of both subfractions here 15 so you get (10/15) / (12/15) the 15s actually cancel for reasons outlined above so we're left with 10/12. If you cannot still see this think about multiplying both top and bottom by (5/4), we can do this as it's equivalent to multiplying by 1. So we get {(2/3)*(5/4)} /{(4/5)*(5/4)} this leaves just the top half.
16. P=VI, V=IR, Find P in terms of V and R without I.
Write I = V/R and sub in P= VI
to get P = V*(V/R) = V[sup]2[/sup]/R
to get P = V*(V/R) = V[sup]2[/sup]/R
17. A question on planetary motion for you
Given F=(mv[sup]2[/sup] )/r
F=(GMm)/r[sup]2[/sup] and T=(2πr)/v
(a) Find an equation in terms of T, M and r (without v but can include constants). I'm not going to tell you what the symbols mean as an exercise in the power of rearranging equations.
Set F =(GMm)/r[sup]2[/sup] and F=(mv[sup]2[/sup] )/r equal to each other. GM/r = v[sup]2[/sup] . Now v[sup]2[/sup] = (2πr)[sup]2[/sup]/T[sup]2 [/sup]
Sub in v[sup]2[/sup] into the first bit. GM/r = (2πr)[sup]2[/sup]/T[sup]2[/sup] and rearrange to get T[sup]2[/sup] = 4π[sup]2[/sup]r[sup]3[/sup]/GM. G is a constant, M is effectively a constant as is 4π[sup]2[/sup] leaving T[sup]2[/sup] ∝ r[sup]3[/sup]
Sub in v[sup]2[/sup] into the first bit. GM/r = (2πr)[sup]2[/sup]/T[sup]2[/sup] and rearrange to get T[sup]2[/sup] = 4π[sup]2[/sup]r[sup]3[/sup]/GM. G is a constant, M is effectively a constant as is 4π[sup]2[/sup] leaving T[sup]2[/sup] ∝ r[sup]3[/sup]
(b) An approximate equation grabbed from this to roughly find the orbit time is where t stands for time
t[sup]2[/sup] / t[sup]2[/sup][sub]earth [/sub]= r[sup]3[/sup] / r[sup]3[/sup][sub]earth[/sub][sup] [/sup]
given r = 2.3x10[sup]11[/sup] m
r[sub]earth[/sub] = 1.5x10[sup]11[/sup] m
rearrange and find by some means the orbit time of Mars hint
Set t[sub]earth[/sub] = 1 year or 365 days or 365.25 days
t[sup]2[/sup] (for mars) = (t[sup]2[/sup]earth r[sup]3[/sup] )/ r[sup]3[/sup][sub]earth[/sub][sup]
[/sup]t[sup]2[/sup] (for mars)=1 year[sup]2[/sup] (2.3x10[sup]11[/sup] m)[sup]3[/sup] /(1.5x10[sup]11[/sup])[sup]3[/sup]
The (1x10[sup]11[/sup] )[sup]3[/sup] s cancel from top and bottom leaving t[sup]2[/sup] (for mars) = (2.3/1.5)[sup]3[/sup]
Either put in (2.3/1.5)[sup]3/2[/sup] into the calculator leaving t= 1.9 years (approximately)
(2.3/1.5)[sup]3/2[/sup] x 365.25 days = 693 days which is not bad Mars has an orbital period close to 687 days but has the 2nd most eccentric orbit. So not too bad.
t[sup]2[/sup] (for mars) = (t[sup]2[/sup]earth r[sup]3[/sup] )/ r[sup]3[/sup][sub]earth[/sub][sup]
[/sup]t[sup]2[/sup] (for mars)=1 year[sup]2[/sup] (2.3x10[sup]11[/sup] m)[sup]3[/sup] /(1.5x10[sup]11[/sup])[sup]3[/sup]
The (1x10[sup]11[/sup] )[sup]3[/sup] s cancel from top and bottom leaving t[sup]2[/sup] (for mars) = (2.3/1.5)[sup]3[/sup]
Either put in (2.3/1.5)[sup]3/2[/sup] into the calculator leaving t= 1.9 years (approximately)
(2.3/1.5)[sup]3/2[/sup] x 365.25 days = 693 days which is not bad Mars has an orbital period close to 687 days but has the 2nd most eccentric orbit. So not too bad.
18. 2= e[sup]s[/sup] find s where e≈ 2.718
What number do we need to raise e to to get 2? You could try some numbers in your calculator. Clearly e[sup]1[/sup] > 2 and e[sup]0[/sup] = 1 so we need a number between 0 and 1
If you used trial and error on the calculator you'd find e[sup](1/2)[/sup] is too small and eventually find that we need to put in a number around 0.693. You can change 2 for totally different numbers and different values come out. The different values of the function are plotted on a graph. We call it the natural logarithm and you can think of it as an inverse of e. On a calculator it is denoted as the 'ln' button but can sometimes be written log[sub]e[/sub] in text.
If you used trial and error on the calculator you'd find e[sup](1/2)[/sup] is too small and eventually find that we need to put in a number around 0.693. You can change 2 for totally different numbers and different values come out. The different values of the function are plotted on a graph. We call it the natural logarithm and you can think of it as an inverse of e. On a calculator it is denoted as the 'ln' button but can sometimes be written log[sub]e[/sub] in text.
19. 2=log[sub]10[/sub] 100 or 10[sup]2[/sup]=100
without a calculator find log[sub]10[/sub] (1000)
20. Using the rules of indices find y where e[sup]x[/sup]e[sup]-x[/sup] = y
Using the rules of indices e[sup]x[/sup]e[sup]-x[/sup]= e[sup](x-x)[/sup] = e[sup]0[/sup] and e[sup]0[/sup] =1 so y=1 e[sup]-x[/sup] could also have been written on the denominator as e[sup]x[/sup] and cancellation to 1 occurs - just the rules of indices all over again
[sup]
[/sup]
iAmiAdam
New Member
Wow, a lot of that looks foreign to me. I'll take a look and do what I can of it.
Thanks for that though, you've been a massive help for me. My chemistry teacher helped me out during his free the other day and explained all the maths I'll need in both, the only thing I struggled with is the equations, I got standard form, sig figs etc. so he's explained a lot of it too, so I'll do these as a test.
Thanks for that though, you've been a massive help for me. My chemistry teacher helped me out during his free the other day and explained all the maths I'll need in both, the only thing I struggled with is the equations, I got standard form, sig figs etc. so he's explained a lot of it too, so I'll do these as a test.
iAmiAdam
New Member
Just done the first one, but it's different to your answer, so I don't think it's right.
Standard form is very important, vitally important in chemistry and physics. It's all about manipulating the indices properly and learning the SI prefices, milli, micro, nano, etc.
Equations a lot of them come with just three things F=ma, V=IR and so on but you'll need square and cube roots.
Already spotted a horrendous typo that's carried over
Glad you had a go at the 1/f = 1/u + 1/v and got half way straight away, that's very good. I'll put up some workings. I've corrected the few typos and errors in there I made and put up solutions to most of the questions leaving a few of the more difficult ones till later. You might want to look particularly at the answers for questions 9, 10, 13, 14 and 15.
Equations a lot of them come with just three things F=ma, V=IR and so on but you'll need square and cube roots.
Already spotted a horrendous typo that's carried over
Glad you had a go at the 1/f = 1/u + 1/v and got half way straight away, that's very good. I'll put up some workings. I've corrected the few typos and errors in there I made and put up solutions to most of the questions leaving a few of the more difficult ones till later. You might want to look particularly at the answers for questions 9, 10, 13, 14 and 15.