Recent content by Patricia Reid

Yes I think so because F=P*A which you need to change P into density which gives ρgz. Also the pressure is always changing with the depth you cannot use the fixed area so you need the change in height times the length. Thanks for all your help!

So F=ρA =ρgw∫dx integrate =ρgwx from 25 to zero I'm still doing something wrong though...

Ok thanks! I'm still a little confused though...So p(z)A is the force right? And if you were to integrate this: p(z)wh = p(z)wdz you would get p(z)wz from 25 to 0?

p=ρgz ?

Ok thanks! So pressure would be ρgh=(1000)(9.8)(25)= 2.45*105 Pa then you could use the equation p=F/A to find F but this would give you 6.125*108 N which is double than the actual answer of 3.1*108 N

Homework Statement A dam is made which is rectangular and flat in profile. It is a depth of 25m and a width of 100m and holds back fresh water which has a density of 1000 kg m3 . What is the total force that the water exerts on the dam? Homework Equations F=ρA |dF|=pdA d/dz p=-ρg The Attempt...

Ok thanks! Got it

I'm not really sure... it was on a sheet my professor handed out it's suppose to be a parametrization for the volume or something it should have actually have been written as V(T)≈(1+β(Δt))V0

Homework Statement An object has a density of 1250 kg /m3 at 10◦C and a coefficient of linear expansion of α = 2.5 × 10−5 1 /K . What is the object’s density when the temperature is 25◦C?Homework Equations V=βV0ΔT V(T)=(1+β(Δt))V0 The Attempt at a Solution How I got the answer is that I used...

So I ended up using the equation V=(1+β(Tf-Ti)) Vi and substituting v=m/ρ which gave me the right answer! Thanks for your help!

Pressure would increase with depth. So p=g*h*density =9.81*1000*0.119= 1163 PaThen for the next part the the force is equal to pressure times area which would equal 1163 Pa * (0.119m)2=16.4 N Then for finding the tension in the rope... The correct answer is 32.8 N what do you double the answer...

[QUO TE="uselesslemma, post: 5061484, member: 542831"]Did you try expressing the volume expansion formula in terms of ρ instead of V? You should find a result that is independent of the object's mass. Ok so you go... V=3αVΔT =3αΔTM/ρ =3(2.5*10-5)(298-283)M/1250 =9.0*10-7M Am I on the right track?

Ok thanks! So you could use the equation β=3α and then use the volume expansion formula. The equation that relates density to volume is ρ=M/V... but you don;t know what the mass is.

So 5 kg * m3/3000 kg = 0.001667 m3 Therefore the height must be 0.119 m but what does this tell you about pressure?

A 5.0 kg cube of metal has a density of 3000 kg m3 and is held by a vertical string while immersed in water which has a density of 1000 kg m3 . Assume that the cube is oriented so that all the faces are either vertical or horizontal. 1. What is the difference in pressure between the top of the...