All Answers 358501 Answers

Melody, your answer is certainly simpler!

.

k = (1/L)*ΔL/ΔT  

ΔT = (1/L)*ΔL/k

Brass ΔT = (1/2.583)*(D-2.583)/(19*10^-6)

Steel ΔT = (1/2.585)*(D-2.585)/(11*10^-6)

By dividing the Brass by the Steel we get:

1 = (2.585/2.583)*(D-2.583)/(D-2.585)*11/19

This gives D ≈ 2.588 cm

Plug this back in to one of the equations above to get ΔT ≈ 96.89degC

32°C + 96.89°C = 128.89°C ≈ 129°C

.

Hi 

These questions are not difficult.  It is just the wording that you have to get your head around.

You do not need to use most of the information that you are given you just need to spot which bits are relevant.

The first one gives you a speed and a distance and you just need to work out how long it will take.

you can use the formula     speed=distance / time

Question 2 is even easier.  I do not want to give you the answer.  Think about it.  Ignore the irrelevant info and hone in on the little bit of info that you actually need to use.  It will be so much better if you work it out for yourself.   

A great answer anon.  

Alan, I also answered this question a little while ago.  My answer is a bit simpler.   

Annual compounding

$$FV=PV(1+i)^n$$

$$\\1500=1000(1+0.05)^n\\
3=2(1.05)^n\\
1.5=1.05^n\\
log1.5=log1.05^n\\
log1.5=nlog1.05\\
n=\frac{log1.5}{log1.05}\\$$

$${\frac{{log}_{10}\left({\mathtt{1.5}}\right)}{{log}_{10}\left({\mathtt{1.05}}\right)}} = {\mathtt{8.310\: \!386\: \!222\: \!520\: \!560\: \!2}}$$    

8.31 years

Daily compounding

$$\\1500=1000(1+0.05/365)^{365n}\\
1.5=(1+0.05/365)^{365n}\\
log1.5=log(1+0.05/365)^{365n}\\
log1.5=365n*log(1+0.05/365)\\
365n=\frac{log1.5}{log(1+0.05/365)}\\
n=\frac{log1.5}{365log(1+0.05/365)}\\$$

$${\frac{{log}_{10}\left({\mathtt{1.5}}\right)}{\left({\mathtt{365}}{\mathtt{\,\times\,}}{log}_{10}\left({\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{0.05}}}{{\mathtt{365}}}}\right)\right)}} = {\mathtt{8.109\: \!857\: \!581\: \!140\: \!344\: \!3}}$$

8.11 years  (2dp)

The difference if 0.2 of a year = 2.4 months

Well if you look around for more than 2.4 months you will start losong money.

BUT If the interest is only added yearly with the first one then you might have to wait for the full year 9 years before you can withdraw the $1500.

With the daily one I would assume that the interest is added to the account daily and you can withdraw the $1500 in 8.11 years.

Now  the daily one is looking quite a lot better.  

The coldest point will be at the minimum (the lowest point) of the surface defined by the function of x and y.  The gradient of this surface in the x-direction is zero there, as is the gradient in the y-direction. These conditions can be used to find the values of x and y, which, when plugged back into the expression fgor temperature will give you the minimum value.

Just complete the square on both x and y.

The minimum value together with the corresponding values of x and y can then be read off.

5% compounded daily would be better

tan(2θ) = 2*tan(θ)/(1- tan²(θ))

Taking the limit as θ→pi/2 we have 2*tan(pi/2)/(1- tan²(pi/2)) = tan(pi) = 0.

(Geno, I think you got a sign wrong: should be 1 + tan² = sec² in your fourth line)

.

Young's modulus (E) is Stress/Strain

Stress = force/unit area = mass*g/(pi*radius²)

Strain = change in length/original length = δL/L

 Hence:

δL = L*mass*g/(pi*radius²*E)

$${\mathtt{deltaL}} = {\frac{{\mathtt{0.38}}{\mathtt{\,\times\,}}{\mathtt{147}}{\mathtt{\,\times\,}}{\mathtt{9.8}}}{\left({\mathtt{\pi}}{\mathtt{\,\times\,}}{{\mathtt{0.021}}}^{{\mathtt{2}}}{\mathtt{\,\times\,}}{\mathtt{9}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{{\mathtt{9}}}\right)}} \Rightarrow {\mathtt{deltaL}} = {\mathtt{0.000\: \!043\: \!903\: \!185\: \!783\: \!3}}$$

Note that I've turned length measurements into metres.

δL ≈ 4.4*10^-5m  = 0.044 mm

.

The bulk modulus is (ΔV/V)*ΔP

1.8*10⁹ = 6*10^-3*ΔP

ΔP= 3*10¹¹ N/m²

10⁵ N/m² = 1bar (≈ 1atm) so ΔP ≈ 3*10⁶ atm

.

What graph?

Here is a diagram

$$\\sin23=\frac{F_x}{64N}\\\\
F_x=64sin23\;\;\;\;N\\\\\\
cos23=\frac{F_y}{64N}\\\\
F_y=64cos23\;\;\;\;N\\\\\\$$

(a) Net downward vertical force on block = mass*g - F*sin(θ)   or 54*9.8 - F*sin(25°) N

Resistive force = μ*(mass*g - F*sin(θ)) = 0.1*(54*9.8 - F*sin(25°) ) N

Hence, to get the block moving we must have F*cos(θ) = μ*(mass*g - F*sin(θ))

F = μ*mass*g/(cos(θ) + μ*sin(θ)) 

$${\mathtt{F}} = {\frac{{\mathtt{0.1}}{\mathtt{\,\times\,}}{\mathtt{54}}{\mathtt{\,\times\,}}{\mathtt{9.8}}}{\left(\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{25}}^\circ\right)}{\mathtt{\,\small\textbf+\,}}{\mathtt{0.1}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{25}}^\circ\right)}\right)}} \Rightarrow {\mathtt{F}} = {\mathtt{55.789\: \!263\: \!395\: \!076\: \!610\: \!5}}$$

or F ≈ 55.8 N

(b) Net force horizontal force on the block once it is moving = F*cos(θ) - μ_k*(mass*g - F*sin(θ))

where μ_k is kinetic coefficient of friction.  This force equals mass*a, where a is acceleration, so:

mass*a = F*cos(θ) - μ_k*(mass*g - F*sin(θ))

a = F*(cos(θ) + μ_k*sin(θ))/mass - μ_k*g

$${\mathtt{a}} = {\frac{{\mathtt{55.789\: \!263\: \!395\: \!1}}{\mathtt{\,\times\,}}\left(\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{25}}^\circ\right)}{\mathtt{\,\small\textbf+\,}}{\mathtt{0.03}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{25}}^\circ\right)}\right)}{{\mathtt{54}}}}{\mathtt{\,-\,}}{\mathtt{0.03}}{\mathtt{\,\times\,}}{\mathtt{9.8}} \Rightarrow {\mathtt{a}} = {\mathtt{0.655\: \!436\: \!494\: \!326\: \!533\: \!9}}$$

or a ≈ 0.655 m/s²

.

The resistive force is given by μ, where μ is coefficient of friction and M is normal force.  For constant velocity there must be no net force acting on the piston, so μ*M = applied force

(a)   0.03*M = 550.     M = 550/0.03   $${\mathtt{M}} = {\frac{{\mathtt{550}}}{{\mathtt{0.03}}}} \Rightarrow {\mathtt{M}} = {\mathtt{18\,333.333\: \!333\: \!333\: \!333\: \!333\: \!3}}$$ or M ≈ 18333N

(b) I'll leave you to do this.

.

$$\\T=x^2+y^2-4x+2y\\\\
T+4+1=(x^2-4x+4)+(y^2+2y+1)\\\\
T+5=(x-2)^2+(y+1)^2\\\\
T=(x-2)^2+(y+1)^2\;-5\\\\$$

Now anything squared has to be  greater or equal to zero so T has to be greater to or equal to -5 degrees. 

This will occur when x=2 and y=-1

Do it on the calculator on the home page  

There are slightly different formulas depending on your exact question.

just type

atan(1.83/4.67)=

straight into the calc on the home page.  if you type it in rather than just using the buttons provided you will need to press the = button on the actual calculator to get it to finish.

Make sure you choose degrees (or radians) from the toggle button on the bottome left.  

-7(3-2i)=-21+14i

To find the amount of money that you must pay per period to pay back a loan, use this formula:

A  =  [P·r(1 + r)^n] / [ (1 + r)^n - 1]

where:  A = payback amount per period

             P = initial loan amount

             r = interest rate per period           (example 9% per year rate paid monthly:  0.09/12)

             n = total number of payments      (example: monthly payments for 8 years:  12·8)

Or...

ex 1

$${\frac{{\mathtt{6.8}}}{{\mathtt{9}}}} = {\frac{{\mathtt{34}}}{{\mathtt{45}}}} = {\mathtt{0.755\: \!555\: \!555\: \!555\: \!555\: \!6}}$$(ignore the decimal without a Fraction)

ex 2

$${\frac{{\mathtt{13}}}{{\mathtt{3.2}}}} = {\frac{{\mathtt{65}}}{{\mathtt{16}}}} = {\mathtt{4.062\: \!5}}$$

Hope I helped!!!!

I think to press enter to see what happens.

example 1

$${\mathtt{0.8}} = {\frac{{\mathtt{4}}}{{\mathtt{5}}}}$$

Example 2

$${\mathtt{0.85}} = {\frac{{\mathtt{17}}}{{\mathtt{20}}}}$$

Example 3

$${\mathtt{0.934}} = {\frac{{\mathtt{467}}}{{\mathtt{500}}}}$$

Example 4

$${\mathtt{0.154\: \!1}} = {\frac{{\mathtt{1\,541}}}{{\mathtt{10\,000}}}}$$

I think this is the final example

Example 5

$${\mathtt{0.879\: \!1}} = {\frac{{\mathtt{8\,791}}}{{\mathtt{10\,000}}}}$$

hope I helped!!!!

6 will be a solution, if, when you enter 6 for x into the equation, the left side will equal the right side.

Left side:  6^2 - 4  =  36 - 4  =  32

Right side:  5(6)  =  30

They are not equal, so 6 is not a solution.