1. ### Distance in 3D

Homework Statement If distance |AB| = root 83, and i also know that |PA| is twice |PB|, then how do you find? |PA| or |PB| ? The whole question states: consider the points P such that the distance from P to A (-1,5,3) is twice the distance from P to B (6,2,-2). Show that the set of all...
2. ### What am i doing wrong (water thru a tube question)

well i didnt put brackets in right place. should be (20.97l/m /60) * 10^-3 = m^3) =3.495*10^-4 still doesnt solve prob though. any other help?? please help!??
3. ### What am i doing wrong (water thru a tube question)

Homework Statement A tube of radius 6.000 cm is connected to tube of radius 1.000 cm as shown above. (see attachment).. Water is forced through the tube at a rate of 20.97 liters/min. The pressure in the 6 cm tube is 1.015×105 Pa. The density of water is 1000 kg/m3. Assume that the water...
4. ### Ice displacing water question, -need help really bad

(1.1 x 6 x 6) x 917/1025 m = 35.428cm (the amount of water displaced). So, the ice sticks out of the water 1100-35.428 = 1064.572cm = 1.064m still wrong though ????
5. ### Ice displacing water question, -need help really bad

ice displacing water question, --need help really bad Homework Statement A rectangular block of ice 6 m on each side and 1.1 m thick floats in seawater. The density of the seawater is 1025 kg/m3. The density of ice is 917 kg/m3. How high does the top of the ice block float above the water...
6. ### Yes, this is a dumb question. sorry

Homework Statement how are litres related to meters???? (with respect to flow rates. so for eg, if something goes 20litres per minute, then how do you convert that to the fluids velocity?? (so i think im having probs cuz i cant seem to get the difference between a litre and a meter...
7. ### Hydralic lift

sorry for the redirect, but i am alot further along on this problem and posted different question (from the same problem) https://www.physicsforums.com/showthread.php?p=1572416#post1572416
8. ### Hydralic lift

Homework Statement A hydraulic lift has two connected pistons with cross-sectional areas 5 cm2 and 550 cm2. It is filled with oil of density 520 kg/m3. a) What mass must be placed on the small piston to support a car of mass 1500 kg at equal fluid levels? this was no prob. =13.61...
9. ### Hydralic lift

Homework Statement A hydraulic lift has two connected pistons with cross-sectional areas 5 cm^2 and 550 cm^2. It is filled with oil of density 520 kg/m3. a) What mass must be placed on the small piston to support a car of mass 1500 kg at equal fluid levels? Homework Equations...
10. ### Bernoulli problems

Homework Statement A long horizontal hose of diameter 3.8 cm is connected to a faucet. At the other end, there is a nozzle of diameter 1.8 cm. Water squirts from the nozzle at velocity 38 m/sec. Assume that the water has no viscosity or other form of energy dissipation. a) What is the...
11. ### Please explain equipotential surfaces/ contour maps

Ok thanks, that helps, but now i guess im having prob with the actual word. "EQUI" .... --what exactlly is equal in the contour map? i am assuming equi word root means "equal"
12. ### PD between 2 points

Homework Statement a current of 2A is flowing from a to b. what is the potential difference between the points? ______a______3ohm_____-20V+_____1ohm_______+6V-________b_____ hopefully my diagram makes sense. Homework Equations this is a really simple question and i cant answer...
13. ### Please explain equipotential surfaces/ contour maps

Homework Statement Can someone please define contour map for equipotential surfaces in really simple terms. I understand that the potential diff is from the neg side I dont understand what it means when the lines are closer together? I thought the lines are just divisions of the voltage...
14. ### Circuit with 2 batteries is confusing me

1.421V is wrong answer though, so can u please tell me what im doing wrong? thanks
15. ### Circuit with 2 batteries is confusing me

because picture says current is going cw, so 1st batt is -4V and 2nd batt is 6V = 2V so 2V enters Req. Req = 2ohms and total current in the circuit is .286A so V=IR = .286 X 2ohm = .579V 2V - .579V = 1.421V