Recent content by wadini

is it correct that I use the equation T=NIABsin() ? where N= 101 turns I = 3.1 A A= .97m * .23m B= 0.62T and sin()= sin(70) ??

why would I multiply the number of turns in the coil? Also, I did that but I still did not get the right answer. (101)(3.1 A)(.2231m^2)(.62T)sin(70)= 40.6967 N*m

okay so what I inititally did was T=NIABsin() so I assumed there was only one loop so I plugged in T= (1)(3.1A)(.97*.23)sin(70) and I got T=.4029385043 but that is not correct so what am I doing wrong? Am I just using the totally wrong equation? What am I missing?

oh sorry! that was not part of the problem. There is no radius given

A rectangular loop consists of 101 closely wrapped turns of wire and has dimnesions .97m by .23m. The loop is hinged along the y-axis and its plane makes an angle of 70 degrees with the x axis. A uniform magnetic field of .62 T is directed along the x-axis and the current in the loop is 3.1...

I don't think you square d. I think it is just d. It doesn't make sense if I want to get it in units of N*m

When a certain air filled parallel plate capacitor is connected cross a battery, it acquires a charge (on each plate) of 110 micro Columbs. While the battery connection is maintained, a dielectric slab is inserted into and fills the region between the plates. This results in the accumulation of...

Two point charges of magnitude 5.21 nC and -2.24 nC are separated by 29.6 cm. The columb content is 8.98755*10 ^9 N*m^2/C^2. What is the electric potential at a point midway between the charges? Answer in untis of V I got the answer for this part= 180.36 V The second part of the question...

sorry Vintial = **

Vi- square root 2GMmH/(1.74*10^6)2 7.35 *10^22 kg is the mass of the moon Mm Radius of the moon Re=1.74*10^6 g= 6/67*10^-11 height reached by the projectile h= 260km = 270* 10^3 I plugged all of that into the equation above and got the answer previously stated but it is wrong.

A projectile launched vertically from the surface of the Moon rises to an altitude of 270 km. What was the projectile's initial speed in m/s? okay so I keep getting this answer wrong but I am pretty sure I am doing it correctly... this is what I am doing: 2*6.67*7.35*270/3.02 and...

and how am I supposed to get the time ratio from all this?

so if I drew this correctly the first and second velocities relative to the water is the same...but I don't really know. I don't understand what I am doing. Why would the speed relative to the water be the same...shouldn't they be different because skier number 2 is going at an angle and gets to...

okay so it is something about all the points of the triangle meeting at the same point which works because the skiers do eventually meet at the same point but one just gets there faster than the other.

I will google it.