Author Topic: Olly_s15 Question Thread (Read 4025 times) Tweet Share

olly_s15 · « **Reply #30 on:** May 16, 2010, 11:11:22 pm »

just looking at it, i think they meant to enclose the denominator in a square root which would provide an circular function anti derivative hence obtaining something like -pi/2.

olly_s15 · « **Reply #31 on:** June 15, 2010, 10:16:07 pm »

ok having trouble with making y the subject while solving for this DE

$y'(x)=8-\frac{y}{5}$

kamil9876 · « **Reply #32 on:** June 15, 2010, 11:01:37 pm »

olly_s15 · « **Reply #33 on:** June 16, 2010, 09:20:55 pm »

Quote from: kamil9876 on June 15, 2010, 11:01:37 pm

$\frac{dy}{dx}=8- \frac{y}{5}$

$\frac{dx}{dy}=\frac{5}{40-y}=\frac{-5}{y-40}$

not really what I meant but it actually helped eliminate the problem i encountered - thanks

new question:

A solid sphere of radius 6 cm has a cylindrical hole of radius 1 cm bored through it's center. What is the volume of water in the bowl?

olly_s15 · « **Reply #34 on:** July 12, 2010, 10:40:42 pm »

A large container of coffee holds 10 liters of boiling water in which coffee is dissolved so that the concentration of the coffee is 40g/L. If hot water is added at a rate of 0.5L/min while the coffee solution is run off at the same rate, find the concentration of coffee in the container after 5 minutes. Assume the mixture remains uniform.

Answer at back of book:31.15g/L

my answer is different :S

Mao · « **Reply #35 on:** July 12, 2010, 11:30:11 pm »

The book is correct.

The DE is $\frac{dQ}{dt} = \frac{-Q\times 0.5}{10}$

olly_s15 · « **Reply #36 on:** July 13, 2010, 09:30:27 pm »

Quote from: Mao on July 12, 2010, 11:30:11 pm

The book is correct.

The DE is $\frac{dQ}{dt} = \frac{-Q\times 0.5}{10}$

ok cheers for that i'll re-attempt

olly_s15 · « **Reply #37 on:** August 01, 2010, 08:23:12 pm »

An object moves with a position vector $\tilde{r}(t)=(t^3-t)\tilde{i}+2t^{-1}\tilde{j}$ . At what time is the object travelling parallel to the unit vector $\tilde{j}$

fady_22 · « **Reply #38 on:** August 01, 2010, 08:58:59 pm »

Quote from: olly_s15 on August 01, 2010, 08:23:12 pm

An object moves with a position vector $\tilde{r}(t)=(t^3-t)\tilde{i}+2t^{-1}\tilde{j}$ . At what time is the object travelling parallel to the unit vector $\tilde{j}$

$\tilde{r}(t)=(t^3-t)\tilde{i}+2t^{-1}\tilde{j}=k\tilde{j}$

Therefore: $(t^3-t)=0$
t=1, 0. (reject negative solution)

t also cannot equal 0 as the j component will be undefined.

Therefore, it occurs at t=1.

olly_s15 · « **Reply #39 on:** August 01, 2010, 09:38:07 pm »

Quote from: fady_22 on August 01, 2010, 08:58:59 pm

Quote from: olly_s15 on August 01, 2010, 08:23:12 pm
An object moves with a position vector $\tilde{r}(t)=(t^3-t)\tilde{i}+2t^{-1}\tilde{j}$ . At what time is the object travelling parallel to the unit vector $\tilde{j}$
$\tilde{r}(t)=(t^3-t)\tilde{i}+2t^{-1}\tilde{j}=k\tilde{j}$

Therefore: $(t^3-t)=0$
t=1, 0. (reject negative solution)

t also cannot equal 0 as the j component will be undefined.

Therefore, it occurs at t=1.

oh okay i see thanks for that

olly_s15 · « **Reply #40 on:** August 31, 2010, 10:22:29 pm »

New one!

A 1.0-kg mass is just in equilibrium on a 12degree inclined plane. The angle of inclination is increased to 24degrees. What is the size of an applied force on the mass parallel to the plane so that is accelerates upwards at 1.0m/s^2?

Mao · « **Reply #41 on:** August 31, 2010, 10:36:37 pm »

Tricky.

Assumption: There is static friction force. Assume the coefficient for static friction is the same as the coefficient for kinetic friction. [In real life, this is NOT true]

Thus, initially at equilibrium, friction act upwards wile gravity acts downwards.

$\mu N = mg \sin 12$

Now, accelerating upwards at 1 m/s² means friction acts downwards, as well as gravity.

$F - (\mu N + mg \sin 24) = ma \implies F = ma + mg \sin 12 + mg \sin 24$

EDIT: I can't read <.<

olly_s15 · « **Reply #42 on:** August 31, 2010, 10:48:52 pm »

nvm i worked out the correct answer

letting A denote applied force, F denote friction
$A-(F+mgsin24)=ma \implies A= ma + \mu N +mgsin24 \implies A= 1 + (tan12*9.8cos24)+9.8sin24$

Search the forums now!

We have moved!

Author Topic: Olly_s15 Question Thread (Read 4025 times) Tweet Share

olly_s15

Re: Olly_s15 Question Thread

olly_s15

Re: Olly_s15 Question Thread

kamil9876

Re: Olly_s15 Question Thread

olly_s15

Re: Olly_s15 Question Thread

olly_s15

Re: Olly_s15 Question Thread

Mao

Re: Olly_s15 Question Thread

olly_s15

Re: Olly_s15 Question Thread

olly_s15

Re: Olly_s15 Question Thread

fady_22

Re: Olly_s15 Question Thread

olly_s15

Re: Olly_s15 Question Thread

olly_s15

Re: Olly_s15 Question Thread

Mao

Re: Olly_s15 Question Thread

olly_s15

Re: Olly_s15 Question Thread

Recent Posts

User:
Password: