Recent content by dewdrop714

thanks for your help! =]

i tried to solve the problem...can someone please check my work? i think i may have made a mistake somewhere along the way I = 300 is given for the question B for Point A = [μ0I / 2pi]*[(1/x) - (1/r+x)] = (4pi *10^-7)(300) / (2)(pi) * (1/1m) - (1/1.5m) = 1.97*10^-4 T B for Point B =...

Don't know where to begin...mag field question 1. Homework Statement http://s685.photobucket.com/albums/vv212/dewdr0p714/?action=view&current=physicsexam2q2.jpg Each support string connects to a corner of the rectangular ceiling. The current carrying wires are .5m apart. The current...

thank you so much! =]

is the angle 90? so then sin of 90 = 1?

stuck on light intensity problem... 1. Homework Statement http://s685.photobucket.com/albums/vv212/dewdr0p714/?action=view&current=physicsexam2q5withoutphotocell.jpg A conducting rod slides down 2 frictionless copper tracks at a constant velocity. There is a .50 T magnetic field...

so m = 4? or m = 0? and is the largest angle you can have 180 degrees? I am still very confused on this question...

1. Homework Statement a) Exactly 4 bright fringes are seen on either side of the central maximum when a diffraction grating is illuminated with yellow-green light of a wavelength of 570 nm. What is the maximum number of lines/meter for the grating? 2. Homework Equations sin(theta) =...

Well 4186 is spec heat of water and 2093 is spef heat of ice. I also converted the temps to kelvins. Then I input them into the equations...I did Q1= (800)(4186)(305.93)=102*10^7 and Q2=(800)(2093)(287.93)=482*10^6. Then i got Q= Q1-Q2 = 542*10^6.

thank you for your help i figured it out! =]

Hmm I don't quite understand what you mean. I get the part about q being proportional to v so .75q~.75v. But being that the equation is q=q0(1-e^-t/RC), what would q0 be? And the whole RC time constant thing confuses me completely...like how can t=RC?

Oh thanks i didn't know that. I posted it on photobucket and here's the link: http://s685.photobucket.com/albums/vv212/dewdr0p714/?action=view&current=5circuittakehometest1.jpg

Well 4186 is spec heat of water and 2093 is spef heat of ice. I also converted the temps to kelvins. Then I input them into the equations...I did Q1= (800)(4186)(305.93)=102*10^7 and Q2=(800)(2093)(287.93)=482*10^6. Then i got Q= Q1-Q2 = 542*10^6.

1. The problem statement http://s685.photobucket.com/albums/vv212/dewdr0p714/?action=view&current=5circuittakehometest1.jpg 1) Determine the equivalent capacitance of the circuit 2) How many RC time constants are needed for the charge on the capacitors to reach 75% of their final...

The problem statement: To save on heating costs, the owner of a greenhouse keeps 800kg of water around in barrels. During a winter day, the water is heated by the sun to 50 degrees Fahrenheit. During the night the water freezes into ice at 32 degrees Fahrenheit in 10 hours. An electrical...