Recent content by kara123

thankyou all for the help figured it out I must have been really tired while doing that algebra because I'm not sure what I was thinking the answer is P=VI

yes i ment E

e=QxV t=Q/I p=(QxV)/(Q/I) =V/I The expression I came up with for a) is the potential difference divided by current to get power but I have no idea if that is even right if someone could just prompt me in the right direction that would be greatly appreciated

Thanks for all the help!

9.8 N/kg

i meant to put the acceleration as 2.25 m/s2 but my displacament is still 18.8m

okay nevermind i see what your saying since finding the acceleration your dividing by the mass anyways so using any different mass will still provide the same acceleration

okay so the only force acting on the sled horizontally is the force of friction so to find the acceleration it would just be -270.48N/120 kg which would equal -2.5m/s2 once subbing that into the equation vf2=vi2+2aavd to solve for displacement I would get an answer of 18.8 so 19m

well to find the force of friction you have to use the mass so yes if the mass were different the answer would be different

-began by finding Fnet=Fg-Ff =1176-270.48 =905.52 acceleration = 905.52/120 kg =7.546 m/s2 sub numbers into vf2=vi2+2aavd -got an answer of -5.6m so clearly it,s wrong because displacement...

got it thank you!

okay so the answer would be 799.68 J of work is done by pushing a 15kg box across a horizontal floor at a constant velocity

so since work of friction equals -799.68 J then work applied must equal 799.68 J or excel it?

would it be zero since acceleration is zero

-i had begun by finding the normal force =147 N -then found the force of friction=99.96 N -found the work of friction=-799.68 J after that I am unsure of where to go since I don't have a force applied