Recent content by wei1006

May I know why is the water level the same then?

Homework Statement A big ice cube is placed in a small glass of water. The level of water in the glass is marked. When the ice cube melts completely,the overall water level will remain the same. Explain. Homework Equations Upthrust= weight of fluid displaced The Attempt at a Solution Since...

Homework Statement A solution of sodium hydroxide(NaOH) contains 1.70g of hydroxide (OH-) ions per dm^3. 20.0 cm^3 of the solution reacts with 25.00cm^3 of 0.0400 mol/dm^3 phosphoric acid(H3PO4). How many moles of sodium hydroxide reacts with 1 mol of phosphoric acid? What will be the chemical...

Oh I see. Thank you! The resistance of the buzzer is extremely large(stated in part 2 of the question) and I guess that is related to why it is not regarded...

I see, thank you for the detailed explanation! I apologise for not adjusting the angle of the picture posted and not able to type out the question(it will only look messy and confusing when I type out the equation)as I have no access to the computer in school and was using a phone at that point...

Homework Statement Homework Equations (y-y1)/(x-x1)=mThe Attempt at a Solution I have attempted part i but I don't know how to do part ii. As point B is still part of the curve and the normal, do I still sub with the same normal eqn? :/ I have no idea how to start... Please help thanks[/B]

Homework Statement The following figure shows a potential divider circuit made up of a LDR and a variable resistor R. It is used to activate a buzzer circuit which sounds only at 2.6V and above. The resistance of the variable resistor varies from 100ohm to 10kohm while the resistance of the LDR...

Homework Statement As contact point Y moves to X, the lamps becomes brighter. Why is this so? Homework Equations R=p (length/Area) I=V/RThe Attempt at a Solution As Y moves closer to X, length of wire decreases, hence resistance decreases and brightness of lamp increases. I am not clear of...

1) Problem: given that x is an obtuse angle for which cos^2x/(1 + 5sin^2x) = 8/35, find the value of cosx/(1 - 5 sin x) without evaluating x. 2) Relevent equations: sin(-x) = - sin x cos(-x) = cos x sin(180° - x) = sin x cos(180° - x) = - cos x sin^2x + cos^2x = 1 3) Attempt: cos^2x/(1 +...

1) Question statement: Simplify 2sec^2x-2sec^2xsin^2x-sin^2x-cos^2x 2)Relevant equations: tan A=sinA/cos A 1+tan^2A=sec^A cot A=1/tanA cot A=cos A/sinA sin^2A+cos^2A=1 secA=1/cos A cosecA=1/sinA 1+cosec^2A= cot^2A sin2A=2sinAcosA cos2A=1-2sin^2A=cos^2A-sin^2A=2cos^A-1 tan2A=(2tanA)/1-tan^2A 3)...

Thank you for your help, have solved it

1) Question: Show that (sin3A-sinA)/(cosA+cos3A)=tanA 2) Relevant equations: tan A=sinA/cos A 1+tan^2A=sec^A cot A=1/tanA cot A=cos A/sinA sin^2A+cos^2A=1 secA=1/cos A cosecA=1/sinA 1+cosec^2A= cot^2A sin2A=2sinAcosA cos2A=1-2sin^2A=cos^2A-sin^2A=2cos^A-1 tan2A=(2tanA)/1-tan^2A 3)Attempt...

I am not sure how to continue...

1) Problem statement: Solve the trigonometric equation for the domain is. 0°<x<360° 5(sinx - cosx) = 4sinx - 3cosx 2) Relevent equations: secx=1/cosx cosec x=1/sinx cot x= 1/tanx tan x=sinx/cosx cot x=cosx/sinx sin^2 x + cos^2 x=1 1 + tan^2 x= sec^2 x 1+ cot^2 x = cosec^2 x Template of answer...

So hi, I am Wei and currently 15. I didn't have the time to really sit down and explore this forum when I joined in December so am introducing myself now. I am just a normal student who is highly imaginative and is passionate about science especially astronomy but never really study it in depth...