Log in
Physics
0
answers
0
watching
78
views
23 Nov 2019

http://www.andyslife.org/misc/Classes/Statics/12th%20Edition/95630-HIBBELER_ISM_03.pdf

from teh link above i am having trouble near the bottom , ican't seem to solve equation 1,2, and 3 to get tehre answer



Watch
0
answers
0
watching
78
views

For unlimited access to Homework Help, a Homework+ subscription is required.

Related textbook solutions

Physics: Principles with Applications
7th Edition, 2013
Giancoli
ISBN: 9780321625922

Related questions

A 70 kg crate is dragged across a floor by pulling on aropeattached to the crate and inclined 19° above thehorizontal.(a) If the coefficient of staticfriction is 0.54,what minimum force magnitude is required from therope to start thecrate moving? (b) Ifμk =0.36, what is the magnitude of the initialacceleration (m/s^2) ofthe crate?
The quesion is above. I have solved the equation more thanonceand got the wrong answer. I believe the problem I am having istheequations I am using. I believe if I was using the correctformulasI'd get the problem right.

The wires leading to and from a 0.12 mm diameter lightbulbfilament are 1.5 mm in diameter. The wire to the filament carries acurrent with a current density of 4.5 x 10^5 A/m^2. A) What is thecurrent in the filament? B) What is the current density in thefilament?

I know that J=I/A

I also know that current is conserved, therefore the currentthrough the wire should be the current running through thefilament.

Rearranging the above equation I= J*A solving for the current inthe wires I get:

I=J*A

I=(4.5A/m^2)* π(1.5*10^-3)^2=3.18 but this is wrong....andthat's where I am stuck. I am being told that the answer to a) is0.80 A.....What am I doing wrong?

<DIV>Suppose in the figure that the four identical currents i = 6 A, into or out of the page as shown, form a square of side 118 cm. What is the force per unit length (magnitude and direction) on the wire in the bottom left hand corner? (N/m) Take the positive y direction as up and the positive x direction as to the right.<BR><BR>I do know all of the answers are supposed to be in N/m<BR><BR>http://loncapa2.physics.sc.edu/res/sc/gblanpied/courses/usclib/hrw8/hrwpictures/30-42.jpg<BR><BR>(a)What is the magnitude of the force per unit length?<BR><BR>(b)What is the x component of the force per unit length?<BR><BR>(c)What is the y component of the force per unit length?<BR><BR><BR><BR><BR>******** This method was previously submitted and the answers were found incorrect so im reposting. Thanx guys.<BR><BR>first find magnetic field.<BR>then use F=BiL<BR>B=uI/2piR<BR>u is 4piX10^-7<BR>I is current, 6am<BR>R is distance from between teh wire.<BR>u get 1.02X10^-6 from top left or bot right wire. And 7.19X10^-7 for the top right.<BR>then plug into equation<BR>L is 1 cuz it says unit length.<BR>So u get x-component=1.02X10^-6X6+7.19X10^-7X6XCos45=9.17X10^-6 N<BR>y-component=-1.02X10^-6X6+7.19X10^-7X6Xsin45=-3.07X10^-6 N<BR>Then u use pythragram theorem<BR>square root of Fx^2+Fy^2=ur magnitude.<BR>which is 1.02X10^-5N.<BR>the answer to b, and C is already above.<BR>b is Fx or ur x component.<BR>c is Fy or ur y component. </DIV>

Weekly leaderboard

Home
Homework Help 3,900,000
Physics 110,000
Start filling in the gaps now
Log in