Ch 11.2 Help
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question 4
4) A wooden block with a volume of 7.316 x 7.316 x 7.316 cm3 floats on water. What is the volume of the wood above the water level if the density of the wood is 0.6901 g/cm3 (see sheet 20) ? Indicate with a negative (positive) sign whether this fraction is zero (greater than zero) if the density of the wood is greater than the density of water.
Answer is NEGATIVE.
i) First convert the Volume of the object to m^3. The volume given is (7.316)^3=392 cm^3. Now, 1 m^3>1000000 cm^3, so the volume of the object is 0.000392 m^3.
ii) We know that the density of the fluid (water) is 1000 kg/m^3
iii) Convert the density of the object to SI units: 1 g/cm^3 has 1000 kg/m^3. 0.6901 g/cm^3 * 1000=690.1 kg/m^3
iv) The formula that we need to use is Vsub=(rho_object/rho_fluid)*Vobject. Now we are calculating the volume of the submerged part. So Vsub=0.0002705 m^3
v) The question is asking for the ABOVE volume, so we subtract the submerged volume from the total volume of the object that we have in the first step. So, Vabove=0.0003920.0002705= 0.0001215 m^3. And this is our final answer.
Hope this helps.
Answer is NEGATIVE.
i) First convert the Volume of the object to m^3. The volume given is (7.316)^3=392 cm^3. Now, 1 m^3>1000000 cm^3, so the volume of the object is 0.000392 m^3.
ii) We know that the density of the fluid (water) is 1000 kg/m^3
iii) Convert the density of the object to SI units: 1 g/cm^3 has 1000 kg/m^3. 0.6901 g/cm^3 * 1000=690.1 kg/m^3
iv) The formula that we need to use is Vsub=(rho_object/rho_fluid)*Vobject. Now we are calculating the volume of the submerged part. So Vsub=0.0002705 m^3
v) The question is asking for the ABOVE volume, so we subtract the submerged volume from the total volume of the object that we have in the first step. So, Vabove=0.0003920.0002705= 0.0001215 m^3. And this is our final answer.
Hope this helps.
guest145 Guest
Re: Ch 11.2 Help
4)
1. Calculate the volume in SI units.
2 volume  density*volume will give the answer.
Just use the given density. After all the unit conversions and such you wind up with the same thing.
1. Calculate the volume in SI units.
2 volume  density*volume will give the answer.
Just use the given density. After all the unit conversions and such you wind up with the same thing.
asdf Guest
Question 3
For question three just take your cm^3 value and divide by 1000. Then multiply by 9.81. Should work.
yo Guest
Re: Ch 11.2 Help
5) A hydraulic lift has a circular piston of 41.25 cm diameter at the point where the heavy load is to be lifted (see sheet 15)and a small diameter piston with a diameter of 3.549 cm which activates the lift. What is the ratio of the load over input force ? Indicate with a negative (positive) sign whether the liquid on the input side has to be pushed along a larger (same) distance as the distance the loadpiston travels during the lift.
.4125^2/.03549^2 gives me 135 and i put it as negative but i got it wrong...what did i do wrong?
.4125^2/.03549^2 gives me 135 and i put it as negative but i got it wrong...what did i do wrong?
steveo Guest
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