Recent content by cestlavie

Okay, this is what I was missing! So the second force in my and @erobz's example is Newton's third law in application. But because elastic force is applied on both ends, it would be incorrect to add ##50~N## twice as it's the same force. Thank you, everyone, for responding!

$$F=kx$$ $$k=\frac F x= \frac {50+50~N} {5+5~ cm}= \frac {100~N} {10~cm}= 10~N/{cm}$$ However, the answer is ##5~N/cm##, because the force on the spring is ##50~N##. I am having trouble understanding why the force isn't ##50~N## + ##50~N##. The diagram looks as though the spring is experiencing...

The answer is 1.1 J, but I don't know how to get there. The only equation I can think of that might be related to this is Intensity, which I've added above. I could find area, using .0004m as the diameter, and energy using 2.0 E 46 J, but I get stuck on energy.

@TSny Got it. So they used ##B=\frac {k^a} {2}## for the average value, resulting in the answer with 0.5.

@TSny My professor formatted y as a subscript -_-. That explains the cubed value, but I still don't see where 0.5 comes from. If ##B=ky## and ##A=a^2##: ##Flux = BA = (ka)(a^2) = ka^3##

@TSny I used ##a^2## for A and ##k## for B. B is technically ##k_y## but the answer options do not include the subscript.

I know the answer is ##ka^3/2##. I got ##ka^2## and I don't know how to get the right answer. I saw an explanation using integrals, but my class is algebra-based. My attempt: ##Flux=ABcos\theta##. I figure ##cos\theta## is 1 becuase the angle between the magnetic field and the normal to the...

I feel all kinds of stupid. I tried doing a different set of numbers and realized that my calculator was not in degrees. Thank you @Keith_McClary for your help.

Oops! I forgot to add the square root! But my answer is for s, not s squared. ##s^2= \frac {\tau} {NIBsin\theta}## ##s=\sqrt {\frac {\tau} {NIBsin\theta}}=0.0632m = 6.32 cm##

If ##\tau= 0.0727, N=60, i=1.3, B=1.0,## and ##\theta=15##, I tried the following calculation: ##\tau=NIABsin\theta## ##\tau=NIs^2Bsin\theta## ##s^2=\frac {\tau} {NIBsin\theta}=\frac {.0727} {60*1.3*1*sin(15)}=0.0632 m=6.32 cm## The answer is probably right in front of me, but I don't know what...

I appreciate that you guys work people through a problem instead of just giving them the answer. Thanks for having me!