AS Physics Mechanics Help?

Original post by serculis

A pump is positioned directly on top of a reservoir. It ejects water through a nozzle of area 6cm² at a velocity of 10m/s. Calculate the power.

I really don't understand how to go about this... can someone help me understand please and why you would do what you're doing?

https://uk.answers.yahoo.com/question/index?qid=20150323113259AAQ4StD
^^^^^^^^^^^^^^^^
another question about mechanics I'm struggling with :frown:

Hello and welcome to the TSR physics sub-forum. :smile:

What are your thoughts on how to go about calculating the first problem?

Work backwards as well as forwards if it helps.

i.e. what is the definition of power and what are the units of measurement?

Do you know how to calculate the kinetic energy of a moving mass?

Do you now what the density of water is?

Do you know the relationship between density and mass?

Reply 3

OP

Original post by lerjj

Do you have a diagram for the first one? I'm not quite sure what the question is (I have an idea but don't want to confuse you if my guess is wrong).

For the latter, that method is confusing I think. A more systemat approach would be:
Equilibrium implies that forces and moments are both balanced

\therefore T_1 + T_2 = 40N

2.5T_1 = 1.5T_2

you know how to solve simultaneous equations I guess?

Hmmmm sounds like an interesting method.... I do remember, but could you show your working out of the tension using simultaneous equations please?

Reply 4

Original post by serculis

Hmmmm sounds like an interesting method.... I do remember, but could you show your working out of the tension using simultaneous equations please?

Why don't you have a go first?

Reply 5

OP

Actually I looked up simultaneous equations to jog my memory a bit, but I don't see how

fits into something like
x + 5y = 6
2x -3 = 5

for example

Can you explain please? Once I can put it into the right format I'll be able to simultaneously solve it.

Reply 6

Original post by serculis

Actually I looked up simultaneous equations to jog my memory a bit, but I don't see how

fits into something like
x + 5y = 6
2x -3 = 5

for example

Can you explain please? Once I can put it into the right format I'll be able to simultaneously solve it.

Algebra: Subtract 1.5T₂ from both sides of the equation to maintain balance:

2.5T₁ - 1.5T₂ = 1.5T₂ - 1.5T₂

i.e.

2.5T₁ - 1.5T₂ = 0

You now have two equations in the same form which can be solved simultaneously.

What answers do you get?

Reply 7

OP

Original post by uberteknik

Algebra: Subtract 1.5T₂ from both sides of the equation to maintain balance:

2.5T₁ - 1.5T₂ = 1.5T₂ - 1.5T₂

i.e.

2.5T₁ - 1.5T₂ = 0

You now have two equations in the same form which can be solved simultaneously.

What answers do you get?

Sorry I really don't understand how that fits in and makes it solvable! :frown:

Can you show how you work it out simultaneously please?

Reply 8

Original post by serculis

Sorry I really don't understand how that fits in and makes it solvable! :frown:

Can you show how you work it out simultaneously please?

You are given two expressions (both in the correct form as stated and fully explained by llerj in post 2) which are in the form as you requested and stated you could solve on your own.

I quote them again (with units omitted for clarity):

T_1 + T_2 = 40

2.5T_1 = 1.5T_2

These must be solved simultaneously to yield the values of T₁ and T₂ with the same method you would use to solve for x and y.

Spoiler

the value of T₁ can now be substituted into the first expression to yield T₂

Spoiler

Reply 9

OP

Thank you very much! That's much easier to understand now but there were some parts I didn't think of including like expanding brackets...

One rule I don't understand is:

Why does the numerator and denominator switch, and then you specifically multiply 40 by the numerator only?

I also found another way to do it. Treat T₂as a pivot and apply the W₁D_{1 =}W₂D₂ equation. I used this method on another question though > http://pbrd.co/1boO057 :biggrin:

Reply 10