U are from bu4 la dont lie.. I know ur form 1 class is 1 daisi

Guys, mind helping out on a few questions here?

1st pic: Question 9
2nd pic: Question 3 and 4

2b, 3c, 4 ( try) b right? Nt sure about 9.

Q9 requires you to calculate the pressure at the base of the tank due to water.

And you seemed to know the formula Pressure = Force (or Weight) / Area.

You have calculated the mass of water (m) is 1050 kg. Can you find the weight of water?

Mind explaning it, I don't know the answers too

What I did:

Your approach is correct, but lacking experience.

Try multiply the mass of water with g = 9.8 m/s².

Owh, I get it now.

Mind helping out on the other 2 questions?

wat . You also didn't meant me harm why apologise

I said

I never said I am actually from BU3. He like me to be from Bu3 then so be it. I don't mind Btw you are?

That 2nd question in 2nd pic answer label wrong or memang mcm tu? A -> C then B?

I believe question 4 is B. didn't see the numbering properly lol

This post has been edited by 5p3ak: Oct 6 2013, 01:06 PM

Q3

Purpose:
Find the tension in the rope that holds the hot air balloon in equilibrium (static).

Understanding:
The balloon floats because of the buoyant force (from the surrounding air) exerted on it. Without the rope, the balloon will slowly lift up high into the air, because the pressure difference between the top and bottom of the balloon results in a net force that tends to accelerate an object upwards. That's the buoyancy effect.

Solution:
The tension in the rope = the net force (lift) = Buoyancy force (↑) − weight of balloon due to gravity (↓)

Buoyancy force = Weight of air displaced by the balloon

Can you do Q4 using Archimedes' principle?

for Q9 is it a very direct question where Pressure = Depth x Density x 9.8

P = 0.7 x 1000 x 9.8 = 6860 Pa?

This post has been edited by kangaroo96: Oct 6 2013, 11:02 AM

The pressure formula P = hρg (for fluid) can be derived from P = W / A.

weight, W = mg

mass, m = ρV

volume, V = hA

In fact, if the question asks for the total pressure below the base of the water tank due to the compression load on the concrete foundation, then the formula P = hρg is not sufficient.

Question 9, the answer should be A, 1m X 1.5m X 0.7m and you get 1.05 m^3. Using the density you get 1050kg. Multiply by 9.8 and you get 10290 N acting on the base. Pressure = Force / Area . Sub 10290 into F and 1.5 into area and you get your pressure 6860 Pa.

or we could just use the P=hpg method?

For the second picture, Q3 should be C, as the force pushing it up is 15000 N while the the weight is 800 N, thus it has a resultant force of 7000 N which is canceled out by the tension of the string/rope whatever at 7 X10^3 N. For Q4, Archimedes' principle.

This post has been edited by Heartnsoul: Oct 6 2013, 11:58 AM

I'm sorry , I was meant to type D . Sorry for the confusion .

MaximR, what do you think of Screen's second picture, question 4. I assumed the answer was B, then I remembered about the mass of the objects. Do you think it is A or B? Have not work on my Physic in a while and the add. math I have been doing is really making me blur at the moment

Edit : Could not find a similar question in both of my reference books

This post has been edited by Heartnsoul: Oct 6 2013, 12:13 PM

Tension = Upthrust - Weight of object = Weight of water displaced - Weight of object

Since the objects will float when the tension is removed , Upthrust > Weight of object

So , based on my reasoning , since the mass of 3 > 2 > 1 , weight of 3 > 2 > 1 ,

T3 > T2 > T1

[ Ans : B ]

Critical_Fallacy

Is my reasoning correct ?

This post has been edited by maximR: Oct 6 2013, 01:00 PM

Although your answer B is correct, it is worthwhile to mention that, the weight of water displaced by a fully immersed polystyrene DOES NOT equal to the weight of the polystyrene, unless it floats freely on the water surface.

The weight of the polystyrene is negligible (0 N) because it has very low density.

Thus, we have

Tension = Weight of displaced water − 0.

Assuming that all Polystyrene (P) have the same cross section area, so volume of P3 > P2 > P1.

Since the parameters of water density and standard gravity, g are unchanged, then tension T3 > T2 > T1.

This post has been edited by Critical_Fallacy: Oct 6 2013, 01:02 PM

Edited my post above , forgotten the fact that Tension = Buoyant force - Weight of object ( net force on object )

Thank you . These questions serve as good practices ; I realise I will need to re-visit certain concepts .

Critical_Fallacy

I have an old past year paper book --- so I have no choice but to upload those original ones in B.M .
Can you help me with this ? [ Q 39 ]



Thanks !

Just want to be more specific. Because we are calculating the weight of the displaced water, W = mg,

where m, the mass of the displaced water is given by m = ρV, we have W = ρVg.

Therefore, it is easier to reason with the Volume of the displaced water because we can justify the volume of an object by vision alone. For mass, we need to measure since it is a fundamental physical property.

Noted with thanks .