fima

This page lists questions and answers that were posted by visitors named fima.

Questions

The following questions were asked by visitors named fima.

Answers

The following answers were posted by visitors named fima.

well, A is not the answer, I was thinking about C, but I'm not sure

from ELLI A its right, but just found; b) p=m*v, so v(t)=(integral Ff(t) from (the point where it starts to move) to t) /m the point where it starts to move: beta*t^2-m*g*mu=0 t=sqrt(m*g*mu/beta) v(5)=[beta/3*5^3-m*g*mu*5-beta/3*sqrt(m*g*mu/beta)^3+m*g*mu...

are you sure about d??? I thought the wind will increase the frequency of the source Andy, for C) f'=f(1- vobs/vsound)

so the radius is not important??? and we'll have to consider v_1=0 and v_2'=0

so the radius is not important??? and we'll have to consider v_1=0 and v_2'=0 v_1'=mhammer*vhammer/Mmerry???!!!!

ok, so considering b=1 and m=2 vf t=5= 5.33 P= F(4)*v(4)=85.28 WEll, thanks @BW, by the way, have you donde the ruler problem?

I got those new equations from jiskha . com / display.cgi?id=1386491909 But I'm not sure about those, I'm still waiting for someone to confirm.... any

and the question for the merry? I'm missing the ruler and the merry.

that was really easy, thanks!!! any hint for the ruler, so I can try it out??

apply conservation of energy: U=m*g*hcm EK= 1/2*I*w^2, I=1/3*m*L^2 Eini= mg(L/2) + 0 Efin= mg(L/2)cos30 + 1/2*I*w^2 solve for Eini=Efin -> w= any one knows how to aswer the other questions?

apply conservation of energy: U=m*g*hcm EK= 1/2*I*w^2, I=1/3*m*L^2 Eini= mg(L/2) + 0 Efin= mg(L/2)cos30 + 1/2*I*w^2 solve for Eini=Efin -> w= any one knows how to aswer the other questions?

apply conservation of energy: U=m*g*hcm EK= 1/2*I*w^2, I=1/3*m*L^2 Eini= mg(L/2) + 0 Efin= mg(L/2)cos30 + 1/2*I*w^2 solve for Eini=Efin -> w= any one knows how to aswer the other questions?

apply conservation of energy: U=m*g*hcm EK= 1/2*I*w^2, I=1/3*m*L^2 Eini= mg(L/2) + 0 Efin= mg(L/2)cos30 + 1/2*I*w^2 solve for Eini=Efin -> w= any one knows how to answer the other questions?

Thanks, That is v(4) for the next questions, but the procedure is correct

apply conservation of energy: U=m*g*hcm EK= 1/2*I*w^2, I=1/3*m*L^2 Eini= mg(L/2) + 0 Efin= mg(L/2)cos30 + 1/2*I*w^2 solve for Eini=Efin -> w= any one knows how to answer the other questions?

apply conservation of energy: U=m*g*hcm EK= 1/2*I*w^2, I=1/3*m*L^2 Eini= mg(L/2) + 0 Efin= mg(L/2)cos30 + 1/2*I*w^2 solve for Eini=Efin -> w= any one knows how to answer the other questions?

apply conservation of energy: U=m*g*hcm EK= 1/2*I*w^2, I=1/3*m*L^2 Eini= mg(L/2) + 0 Efin= mg(L/2)cos30 + 1/2*I*w^2 solve for Eini=Efin -> w= any one knows how to answer the other questions?

apply conservation of energy: U=m*g*hcm EK= 1/2*I*w^2, I=1/3*m*L^2 Eini= mg(L/2) + 0 Efin= mg(L/2)cos30 + 1/2*I*w^2 solve for Eini=Efin -> w= any one knows how to answer the other questions?

apply conservation of energy: U=m*g*hcm EK= 1/2*I*w^2, I=1/3*m*L^2 Eini= mg(L/2) + 0 Efin= mg(L/2)cos30 + 1/2*I*w^2 solve for Eini=Efin -> w= any one knows how to answer the other questions?

apply conservation of energy: U=m*g*hcm EK= 1/2*I*w^2, I=1/3*m*L^2 Eini= mg(L/2) + 0 Efin= mg(L/2)cos30 + 1/2*I*w^2 solve for Eini=Efin -> w= any one knows how to answer the other questions?

apply conservation of energy: U=m*g*hcm EK= 1/2*I*w^2, I=1/3*m*L^2 Eini= mg(L/2) + 0 Efin= mg(L/2)cos30 + 1/2*I*w^2 solve for Eini=Efin -> w= any one knows how to answer the other questions?

apply conservation of energy: U=m*g*hcm EK= 1/2*I*w^2, I=1/3*m*L^2 Eini= mg(L/2) + 0 Efin= mg(L/2)cos30 + 1/2*I*w^2 solve for Eini=Efin -> w= any one knows how to answer the other questions?

the equation is Ok, and the w value got green checked. So YES, I'm sure about A. But anyone knows how to do B and C. And please, try to post the procedure, because we all have different data.

following those equations I got a green mark for A. But any idea about how b and c can be done?

the equation is Ok, and the w value got green checked. So YES, I'm sure about A. And I did for conservation of energy. But anyone knows how to do B and C. And please, try to post the procedure, because we all have different data.

to solve the geometrical equations, I need alpha, can I calculate alpha like this: torque at bottom: mgl/2*sin(theta)=Ic*alpha, Ic=1/3*3*l^2, though should I add a "-"???

just checked and green mark, except for Fy. considering m=0.2, L=0.15, g=10 andDamon eq. ax = -L/2*sin T w^2 + L/2 cos T alpha ay = -L/2*cos T w^2 + L/2 cos T alpha torque=mgL/2 sin T Icm= 1/3 L^2 alpha=torque.bottom/Icm with my data alpha=50,w=5.17, ax=2...

ERROR I used: ay= -L/2*cos T w^2 - L/2 cos T alpha, though I'm thinking there might be an error with the signs

I wrote for mass=200gr L=15cm, and I'm sure about those values, because I got Fx, but I'm having problems with Fy

@KEITANAKO, following your suggestions, I replaced in the torq. eq and got Fy=0.76, though if snyone can confirm that value for MASS=200gr and L=15cm, because I don't waste my last chance

by the way, have you done the 4th questions?

for me it was 3.2, I think data are variables. By the way, have you done question 4? because I did B=uo N/L I, I from the 3rd quest. but for "a big t" and N/L= loop density, but that wans't the aswer, any hint?

for me it was 3.2, I think data are variables. By the way, have you done question 4? because I did B=uo N/L I, I from the 3rd quest. but for "a big t" and N/L= loop density, but that wans't the aswer, any hint?

and the first questoin?

3.2 By the way, have you done question 4? because I did B=uo N/L I, I from the 3rd quest. but for "a big t" and N/L= loop density, but that wans't the aswer, any hint? And the first question???

for me it was 3.2, I think data are variables. By the way, have you done question 4? because I did B=uo N/L I, I from the 3rd qst but for "a big t" and N/L= loop density, but that wans't the answer, any hint? and the fisrt question?

draw the forces, there's repulsion between the two equal wire against the third one We have: F = [uo.I1.I2/(2.pi.d)]*Ltotal 2*F*sin(a)= m*g = lamba*Ltotal*g sin(a)=h/0.5d Replace the values and you'll I. By the way, have you done the one with the thick me...

draw the forces, there's repulsion between the two equal wire against the third one We have: F = [uo.I1.I2/(2.pi.d)]*Ltotal 2*F*sin(a)= m*g = lamba*Ltotal*g sin(a)=h/0.5d Replace the values and you'll I. By the way, have you done the one with the thick me...

Hello, can someone please explain the one with the thick metal slab???

Hi Lora, I1 and I2 are the currents, so one is 200A, and the second is the one you are looking for. So, you have to solve the equation. And forget about the sign, the repulsion force created by the 2 wire are in equilibrium with he weight. Well, and the q...

that's what I thought, but then, the procedure indicated is not convincing. Has anyone check the one with the 1mm diameter wire and current, which is the max B? I'm not sure how to calculate that.

L total is just reference, since you don't have the lenght of the wire, but you need it for the total mass, and the Force is by lenght unit, it will go, just write it down. F/L=uo*I1*I2/(2*pi*d), that L will be Ltotal. and consdier that the three wire hav...

Hello, so any idea for the original question???? Has anyone checked the answer of the problem?

Yep, that's the value. @Lora, did you get the answer? Yesterday I could get into edx, I=63.7

PLEASE!! anyone has a hint on the question posted by @klaus??? By the way, any idea on: A square loop of wire of side L with total resistance R moves at constant speed v into a region of uniform magnetic field B pointing perpendicular to the plane of the...

I'm on that question too, but i'm not sure how to start that problem.... and the question with the moving rod???!!!

so b it is???? absolutely sure?

Hi Elena, I got that too, hope it's the right answer