Recent content by candyq27

Ok so I rearranged the formula to get B=emf/-NAwsinwt = 28/(-160)(.03)(14.96)(sin(14.96*.315)) =28/71.808 =0.38993 Is that correct? Thanks!

Homework Statement I have a graph of an emf of a generator vs time, and i know the cross-sectional area per turn is 0.03m^2 and there are 160 turns. From the wave graph I figured out the frequency of the generator is 2.4Hz and the angular speed is 14.96rad/s. I need to figure out the magnetic...

Well for the second I did 0.105-0.03 and it's incorrect...?

I get that the third sound comes 0.41s because you have to do 0.437-0.03, but I can't figure out how to get the second sound...?

So first it went through the steel handrail so that is 150/5040= 0.02976s, then through the water so 150/1432= 0.1047s, then the air so 150/343=0.4373s. So the first sound happens after 0.03s, then the second sound comes after 0.1047s, then the third sound comes after 0.4373s. Right?

Homework Statement An explosion occurs at the end of a pier. The sound reaches the other end of the pier by traveling through three media: air, fresh water, and a slender metal handrail. The speeds of sound in air, water, and the handrail are 343, 1482, and 5040 m/s, respectively. The sound...

Thanks so is the primary voltage the 122 or the 4270? The equation is either 122/4270=23/x x= 805 OR 4270/122=23/x x= 0.657 The 805 turns makes more sense so that must be right. Thanks!

Homework Statement In some places, insect "zappers," with their blue lights, are a familiar sight on a summer's night. These devices use a high voltage to electrocute insects. One such device uses an ac voltage of 4270 V, which is obtained from a standard 122.0 V outlet by means of a...

Ok so the back emf would be 88.6/3 = 29.5 so then I can use the equation I= (V-E)/R = (130-29.5)/9.42 = 10.7 A Thank you!

Homework Statement A motor is designed to operate on 130 V and draws a current of 13.8 A when it first starts up. At its normal operating speed, the motor draws a current of 4.40 A. (a) What is the resistance of the armature coil? (b) What is the back emf developed at normal speed? (c)...

I grab the wire and point my thumb out and the direction of my fingers is the direction of the field, so it is either to the right or to the left depending on which way I turn my thumb.

Is the induced emf the same thing as the induced current? We learned that the induced current is proportional to the neg slope of the graph. We also learned that the motional emf is -vLB. I don't know what ohmic is??

I'm not sure what the first part means. If i rotate the picture then it can be anywhere. If the current points into the page the field is going to the right and if it points out of the page the field is going to the left. I don't understand how to tell which direction is the correct direction...

That's what I did. My thumb points into the page so my fingers curl to the right

Ok now I'm really confused. I thought that the induced current was proportional to the negative slope of the magnetic flux graph..? Sorry if it's dumb, we just learned this tonight and have this homework due Wednesday so I'm trying to sort through it. Thanks for your help.