Recent content by mikep

how do you get that?

yeah when i entered this value in, i got it wrong

Can someone please help me with this question? Two point charges lie on the x axis. A charge of + 3.5 µC is at the origin, and a charge of -3.5 µC is at x = 10.0 cm. What is the net electric field at x = -2.0 cm? What is the net electric field at x = + 2.0 cm? for the first one i did k...

does anyone know how to do this problem? The total vapor pressure above a solution of C_6H_6(l) and C_7H_8(l) is 0.485 atm. What is the mole fraction of C_6H_6(l) in the liquid solution? P^o C6H6(l) = 0.526 atm P^o C7H8(l) = 0.188 atm the answer is 0.879 but I'm not sure how to get that...

A 2g bullet hits a 5kg wood block, which hangs from a 1.5m long string. This causes the block to swing through an arc of 5°. What was the speed of the bullet before it hit the block? I know that I need to use the momentum and circular motion equations. Can someone tell me what I need to do...

oh and i had it setup right too. thanks

are you sure cause i got the answer wrong so i thought i did something wrong. i think i made a wrong assumetion somewhere

In a hydraulic system the piston on the left has a diameter of 4.5 cm and a mass of 1.7 kg. The piston on the right has a diameter of 12 cm and a mass of 2.5 kg. If the density of the fluid is 750 kg/m3, what is the height difference h between the two pistons? can someone please help me with...

so the F_b = W_{hot} + W_{balloon} ? W_{balloon} = (291kg)(9.8N/kg) \rho_{hot} = 0.93kg/m^3

so Fb = 1atm + (d)(9.8N/kg)(809m^3) = (1.29kg/m^3)(9.8N/kg)(809m^3) is that right?

oh ok. so how would you find the density? i tried 291/809 = 0.36kg/m^3 but that wasn't correct

A hot air balloon plus cargo has a mass of 291 kg and a volume of 809 m3. The balloon is floating at a constant height of 6.25 m above the ground. The density of the outside air may be assumed to be 1.29 kg/m3. What is the density of the hot air in the balloon? can someone help me with this...

oh i get it. so te amount of heat left is 0.94kg and the final temperature would be 0°C because there is still some ice left

wouldn't it be 335034/670000 = 0.53kg ? because Q = m L = (2kg)(335000) = 670000

A large punch bowl holds 4.75 kg of lemonade (which is essentially water) at 20.0°C. A 2.00 kg ice cube at -10.2°C is placed in the lemonade. What is the final temperature of the system, and the amount of ice (if any) remaining? Ignore any heat exchange with the bowl or the surroundings. can...