How to find out the angle of helix of a srew thread?
Q. How to find out the angle of helix of a srew thread?
Asked by UNIVERSAL - Thu Oct 5 13:35:12 2006 - - 3 Answers - 0 Comments
A. I found this wonderful website with math from the beginners level to the top. I hope you find the answers you need thyere.
Answered by Wanda C - Thu Oct 5 13:44:08 2006
Q. How to find out the angle of helix of a srew thread?
Asked by UNIVERSAL - Thu Oct 5 13:35:12 2006 - - 3 Answers - 0 Comments
A. I found this wonderful website with math from the beginners level to the top. I hope you find the answers you need thyere.
Answered by Wanda C - Thu Oct 5 13:44:08 2006
what is the helix angle of 1 5BA thread?
Q. what is the helix angle of 1 5BA thread?
Asked by wishywashy - Tue Jan 15 08:37:40 2008 - - 3 Answers - 0 Comments
A. *47.5 degrees is the helix angle of 1 5BA thread. [The shortening at the crest and root is given by 0.2681683p with a radius of 0.1808346p at both crest and root. The angle of the thread is 47.5 degrees which is different from that used by Whitworth (55 degrees) and ISO Metric (60 degrees) so BA fasteners are never interchangeable with Whitworth/Metric ones even when the pitch and diameter seem similar (except 0BA - 6 mm) . Standard BA sizes are defined between 0BA and 22BA. The very small sizes are not used very often and other standards tend to be used e.g. NHS.] edited: yes, thats true:47.5 deg is the included angle. for ba thread:: -P = Pitch = 1/Number of threads per inch (tpi) h = Triangular height = 1.1363365 x P d = Actual… [cont.]
Answered by alpha b - Tue Jan 15 09:00:26 2008
Q. what is the helix angle of 1 5BA thread?
Asked by wishywashy - Tue Jan 15 08:37:40 2008 - - 3 Answers - 0 Comments
A. *47.5 degrees is the helix angle of 1 5BA thread. [The shortening at the crest and root is given by 0.2681683p with a radius of 0.1808346p at both crest and root. The angle of the thread is 47.5 degrees which is different from that used by Whitworth (55 degrees) and ISO Metric (60 degrees) so BA fasteners are never interchangeable with Whitworth/Metric ones even when the pitch and diameter seem similar (except 0BA - 6 mm) . Standard BA sizes are defined between 0BA and 22BA. The very small sizes are not used very often and other standards tend to be used e.g. NHS.] edited: yes, thats true:47.5 deg is the included angle. for ba thread:: -P = Pitch = 1/Number of threads per inch (tpi) h = Triangular height = 1.1363365 x P d = Actual… [cont.]
Answered by alpha b - Tue Jan 15 09:00:26 2008
A 5.0g sphere on 1.0m long thread has a charge of 100 nC. If it makes an angle of 10 degree to the vertical.?
Q. A 5.0g sphere on 1.0m long thread has a charge of 100 nC. If it makes an angle of 10 degree to the vertical. What is the strength of the electric field?
Asked by I S - Tue Jan 23 22:53:42 2007 - - 1 Answers - 0 Comments
A. Sum the forces acting on the object. The force of the electric field must be related to the horizontal component of the tension in the thread. Once you get that equal, you should be able to solve it using equations from your book.
Answered by polevaulter1000 - Tue Jan 23 22:59:03 2007
Q. A 5.0g sphere on 1.0m long thread has a charge of 100 nC. If it makes an angle of 10 degree to the vertical. What is the strength of the electric field?
Asked by I S - Tue Jan 23 22:53:42 2007 - - 1 Answers - 0 Comments
A. Sum the forces acting on the object. The force of the electric field must be related to the horizontal component of the tension in the thread. Once you get that equal, you should be able to solve it using equations from your book.
Answered by polevaulter1000 - Tue Jan 23 22:59:03 2007
Can I add a wide angle lens to my canon fs100?
Q. I have a wide angle lens from my sony mavica, Sony vcl-mhg07. Is there any way to get it to work on my Canon FS100? There isn't any threading on the Canon to put a step up ring on but maybe there is another way to do it?? Anyone know?
Asked by Janice C - Tue May 26 16:36:43 2009 - - 1 Answers - 0 Comments
A. With my favorite browser and search engine, I did a search using "FS100 fisheye".
Answered by Little Dog - Tue May 26 17:05:49 2009
Q. I have a wide angle lens from my sony mavica, Sony vcl-mhg07. Is there any way to get it to work on my Canon FS100? There isn't any threading on the Canon to put a step up ring on but maybe there is another way to do it?? Anyone know?
Asked by Janice C - Tue May 26 16:36:43 2009 - - 1 Answers - 0 Comments
A. With my favorite browser and search engine, I did a search using "FS100 fisheye".
Answered by Little Dog - Tue May 26 17:05:49 2009
Can you put a a wide angle lens on a Canon FS100?
Q. I was wondering one more thing, what is the thread size for the lens if I were to buy a step down ring?
Asked by THE BEAST - Mon Jun 22 19:10:47 2009 - - 1 Answers - 0 Comments
A. Read the camcorder manual or visit the Canon product site... To answer your questions: Yes 37mm diameter.
Answered by Little Dog - Mon Jun 22 19:17:20 2009
Q. I was wondering one more thing, what is the thread size for the lens if I were to buy a step down ring?
Asked by THE BEAST - Mon Jun 22 19:10:47 2009 - - 1 Answers - 0 Comments
A. Read the camcorder manual or visit the Canon product site... To answer your questions: Yes 37mm diameter.
Answered by Little Dog - Mon Jun 22 19:17:20 2009
Can the Canon G10 also take wide angle/fisheye lenses?
Q. I know there is a teleconverter you can get for this camera while using the conversion lens adaptor LA-DC58K. That adaptor has a 58mm thread size. I have a Raynox 58mm fisheye but I'm wondering if this would work with the G10. I'm not to sure if it will focus right or not. Anyone have experiences with something like this? Just tell me what you think please.
Asked by 3CCD - Mon Oct 13 10:19:40 2008 - - 2 Answers - 0 Comments
A. To get the full fisheye effect you need the lens close to the sensor, putting a wide angle adapter onto the end of an existing lens wont do it without a lot of vignetting, like a circular image on the sensor instead of the usual rectangular one. Notice the Raynox site doesn't claim 180 field of view, it says 180 across the diagonal which is not the same thing at all. It also says that it broadens the view by 37%, the widest setting on the G10 is 28mm doing the maths 37% of 28mm equals = 10.36mm, 28 - 10.36 = 17.64mm. So the adapter will make your lens in effect a 17.64mm hardly fisheye. 10mm is reckoned to be good for a 180 field of view, the real Fisheye effect, but that's with a 35mm sensor, your smaller sensor is in effect cropping… [cont.]
Answered by screwdriver - Mon Oct 13 11:27:59 2008
Q. I know there is a teleconverter you can get for this camera while using the conversion lens adaptor LA-DC58K. That adaptor has a 58mm thread size. I have a Raynox 58mm fisheye but I'm wondering if this would work with the G10. I'm not to sure if it will focus right or not. Anyone have experiences with something like this? Just tell me what you think please.
Asked by 3CCD - Mon Oct 13 10:19:40 2008 - - 2 Answers - 0 Comments
A. To get the full fisheye effect you need the lens close to the sensor, putting a wide angle adapter onto the end of an existing lens wont do it without a lot of vignetting, like a circular image on the sensor instead of the usual rectangular one. Notice the Raynox site doesn't claim 180 field of view, it says 180 across the diagonal which is not the same thing at all. It also says that it broadens the view by 37%, the widest setting on the G10 is 28mm doing the maths 37% of 28mm equals = 10.36mm, 28 - 10.36 = 17.64mm. So the adapter will make your lens in effect a 17.64mm hardly fisheye. 10mm is reckoned to be good for a 180 field of view, the real Fisheye effect, but that's with a 35mm sensor, your smaller sensor is in effect cropping… [cont.]
Answered by screwdriver - Mon Oct 13 11:27:59 2008
A small plastic ball of mass 6.60 10-3 kg and charge +0.11 mC is suspended from an insulating thread and hangs
Q. A small plastic ball of mass 6.60 10-3 kg and charge +0.11 mC is suspended from an insulating thread and hangs between the plates of a capacitor (see the drawing). The ball is in equilibrium, with the thread making an angle of 30.0 with respect to the vertical. The area of each plate is 0.0160 m2. What is the magnitude of the charge on each plate? The string is hanging from the ceiling with a 30 degree angle between the center (between the plates) and the string which causes the positive small plastic ball to be right beside the negative plate. There are also lines going from the positive plate to the negative plate.
Asked by ladydragonaap - Thu Jun 7 12:47:12 2007 - - 2 Answers - 0 Comments
A. Draw a FBD of ball. You should find the magnitude of the (electric) force acting on the ball as F = w tan 30 , where w is the weight of the ball, namely, 6.6 E-3 kg 9.8 m/s = 64.68 E-3 N. Thus, F = 37.34 E-3 N. Electric force on the ball is just F = qE. Find E. You should get 340 N/C, approx. Apply now Gauss' law to the positive capacitor plate. Neglect fringe (edge) effects. The field points everywhere to the right, in the region between plates. Zero anywhere else. Then, Q = EA = 8.854 E-12 340 1.6 E-2 = 48 E-6 = 48 ueC.
Answered by Jicotillo - Thu Jun 7 14:55:57 2007
Q. A small plastic ball of mass 6.60 10-3 kg and charge +0.11 mC is suspended from an insulating thread and hangs between the plates of a capacitor (see the drawing). The ball is in equilibrium, with the thread making an angle of 30.0 with respect to the vertical. The area of each plate is 0.0160 m2. What is the magnitude of the charge on each plate? The string is hanging from the ceiling with a 30 degree angle between the center (between the plates) and the string which causes the positive small plastic ball to be right beside the negative plate. There are also lines going from the positive plate to the negative plate.
Asked by ladydragonaap - Thu Jun 7 12:47:12 2007 - - 2 Answers - 0 Comments
A. Draw a FBD of ball. You should find the magnitude of the (electric) force acting on the ball as F = w tan 30 , where w is the weight of the ball, namely, 6.6 E-3 kg 9.8 m/s = 64.68 E-3 N. Thus, F = 37.34 E-3 N. Electric force on the ball is just F = qE. Find E. You should get 340 N/C, approx. Apply now Gauss' law to the positive capacitor plate. Neglect fringe (edge) effects. The field points everywhere to the right, in the region between plates. Zero anywhere else. Then, Q = EA = 8.854 E-12 340 1.6 E-2 = 48 E-6 = 48 ueC.
Answered by Jicotillo - Thu Jun 7 14:55:57 2007
What is the speed of the ball when it is in circular motion?
Q. A small metal ball is suspended from the ceiling by a thread of negligible mass. The ball is then set in motion in a horizontal circle so that the thread describes a cone. The acceleration of gravity is 9.8 m/s^2. Thread - 2.7m Angle - 25 degrees Mass of Ball - 6.3kg thank you.
Asked by pleasehlep>>>ME - Thu Jan 14 17:54:15 2010 - - 1 Answers - 0 Comments
A. 2.7cos(25) = 2.447m 2.7-2.447 =0.253m PE = mgh = 6.3 x 9.8 x 0.253 = 15.62 J PE = KE KE = 1/2mv^2 15.62/(1/2m) = v^2, sq-rt = v = 2.23 m/s
Answered by Physicsquest - Thu Jan 14 18:09:48 2010
Q. A small metal ball is suspended from the ceiling by a thread of negligible mass. The ball is then set in motion in a horizontal circle so that the thread describes a cone. The acceleration of gravity is 9.8 m/s^2. Thread - 2.7m Angle - 25 degrees Mass of Ball - 6.3kg thank you.
Asked by pleasehlep>>>ME - Thu Jan 14 17:54:15 2010 - - 1 Answers - 0 Comments
A. 2.7cos(25) = 2.447m 2.7-2.447 =0.253m PE = mgh = 6.3 x 9.8 x 0.253 = 15.62 J PE = KE KE = 1/2mv^2 15.62/(1/2m) = v^2, sq-rt = v = 2.23 m/s
Answered by Physicsquest - Thu Jan 14 18:09:48 2010
How can I solve this physics question involving Coulomb's law and statics?
Q. I keep getting this question wrong, perhaps someone could show me how to do it? Two small spheres of 15g each are suspended from a common point by threads of length 35cm. Each thread makes an angle with the vertical of 20 degrees. Each sphere carries the same charge. Find the magnitude of this charge. Please help! Thanks.
Asked by elemental_funk - Fri Nov 13 01:33:54 2009 - - 1 Answers - 0 Comments
A. From Conservation of energy and given "small spheres" Ug = Ue U = mgL(1-cos(20)) = kq^2/r r = 2 L Sin(20) q = Sqrt(m g L (1 - Cos(20)) r / k) q = 2.87297287074558E-07 C q = 3E-07 C If you need more help, let me know.
Answered by Al P - Fri Nov 13 02:08:30 2009
Q. I keep getting this question wrong, perhaps someone could show me how to do it? Two small spheres of 15g each are suspended from a common point by threads of length 35cm. Each thread makes an angle with the vertical of 20 degrees. Each sphere carries the same charge. Find the magnitude of this charge. Please help! Thanks.
Asked by elemental_funk - Fri Nov 13 01:33:54 2009 - - 1 Answers - 0 Comments
A. From Conservation of energy and given "small spheres" Ug = Ue U = mgL(1-cos(20)) = kq^2/r r = 2 L Sin(20) q = Sqrt(m g L (1 - Cos(20)) r / k) q = 2.87297287074558E-07 C q = 3E-07 C If you need more help, let me know.
Answered by Al P - Fri Nov 13 02:08:30 2009
how to find radius of the circular motion?
Q. A stone of mass 0.48 kg is attached to an extremely light thread, whose mass can be considered negligible, and swung around a fixed vertical axis Y so that the stone undergoes circular motion in a horizontal plane. The length of thread from the stone to the axis of rotation Y is L = 2.4 m and the thread makes an angle of = 71.0 to the vertical.
Asked by blue01 - Wed Mar 25 03:59:59 2009 - - 1 Answers - 0 Comments
A. R/L = sin(71) R = L sin(71)
Answered by Retsum - Wed Mar 25 05:25:39 2009
Q. A stone of mass 0.48 kg is attached to an extremely light thread, whose mass can be considered negligible, and swung around a fixed vertical axis Y so that the stone undergoes circular motion in a horizontal plane. The length of thread from the stone to the axis of rotation Y is L = 2.4 m and the thread makes an angle of = 71.0 to the vertical.
Asked by blue01 - Wed Mar 25 03:59:59 2009 - - 1 Answers - 0 Comments
A. R/L = sin(71) R = L sin(71)
Answered by Retsum - Wed Mar 25 05:25:39 2009
what is the angle?
Q. 2 balls have the same mass of 3.4 g. They are hung by string of length L = .80 m from the same point. When the balls are given equal charges so that q equals q2 = q, each string hangs at a 10 degreee angle from the vertical. a. find q's magnitude i did this, and got q = 1.12 x 10^-8 C. I just need help with part b. b. if the strings are shortened, L = 0.3500 m, and the charge remains the same, what will be the new angle each thread would make with the vertical? ten points for best answer, I am really stuck.
Asked by Ginger - Wed Apr 16 00:56:02 2008 - - 1 Answers - 0 Comments
A. I do not get the same answer as yours for the first part. Did you remember that the distance between the charges is d = 2*L*sinoe? tanoe = Fe / m*g Fe = K*q^2 / d^2 = K*q^2 / 4*L^2*sin^2oe q= [m*g*4*L^2*sin^2oe*tanoe / K q = 2*L*sinoe [m*g*tanoe / k] = 2*0.8*.174* [.0034*9.8*0.176 / 9*10^9] This comes out 2.24 * 10^-7 C Take the formula from above for the new length, L' and the new angle oe' tanoe' = [K*q^2 / d = K*q^2 / m*g*4*L'^2*sin^2oe' put the value for q^2 from above and cancel terms to get sin^2oe' * tanoe' = (L/L')^2 * sinoe * tanoe sin^2oe' * tanoe' = (0.8/0.35)^2 * 0.174^2 * 0.176 sin^2oe' * tanoe' = 5.22 * 0.302 * 0.176 = 0.277 You can solve this by approximation, since oe' is small, sinoe' ~ tanoe' ~ oe' (radians) [cont.]
Answered by gp4rts - Wed Apr 16 02:02:14 2008
Q. 2 balls have the same mass of 3.4 g. They are hung by string of length L = .80 m from the same point. When the balls are given equal charges so that q equals q2 = q, each string hangs at a 10 degreee angle from the vertical. a. find q's magnitude i did this, and got q = 1.12 x 10^-8 C. I just need help with part b. b. if the strings are shortened, L = 0.3500 m, and the charge remains the same, what will be the new angle each thread would make with the vertical? ten points for best answer, I am really stuck.
Asked by Ginger - Wed Apr 16 00:56:02 2008 - - 1 Answers - 0 Comments
A. I do not get the same answer as yours for the first part. Did you remember that the distance between the charges is d = 2*L*sinoe? tanoe = Fe / m*g Fe = K*q^2 / d^2 = K*q^2 / 4*L^2*sin^2oe q= [m*g*4*L^2*sin^2oe*tanoe / K q = 2*L*sinoe [m*g*tanoe / k] = 2*0.8*.174* [.0034*9.8*0.176 / 9*10^9] This comes out 2.24 * 10^-7 C Take the formula from above for the new length, L' and the new angle oe' tanoe' = [K*q^2 / d = K*q^2 / m*g*4*L'^2*sin^2oe' put the value for q^2 from above and cancel terms to get sin^2oe' * tanoe' = (L/L')^2 * sinoe * tanoe sin^2oe' * tanoe' = (0.8/0.35)^2 * 0.174^2 * 0.176 sin^2oe' * tanoe' = 5.22 * 0.302 * 0.176 = 0.277 You can solve this by approximation, since oe' is small, sinoe' ~ tanoe' ~ oe' (radians) [cont.]
Answered by gp4rts - Wed Apr 16 02:02:14 2008
A small metallic bob is suspended from the ceiling by a thread of negligible mass.?
Q. The ball is then set in motion in a horizontal circle so that the thread describew a cone. The length of the string, angle between the axis of rotation and the string, and the mass of the bob are given. Find the magnitude of the angular momentum???
Asked by Vladimir G - Sun Nov 5 12:07:38 2006 - - 3 Answers - 0 Comments
A. The answer depends on what point you use as the center. anyway,L=r(cross)p where r is the radius vector for the path, and p is the linear momentum. But, to use this formula, we need to find the linear momentum of the bob. Since we know its mass, we need to find its velocity. There are three forces of the string. tension, gravity, and centripetial force. since the object is not accelerating, the sum of these is 0. So, Ft+Fc+Fg=0 Or, Ft=-Fc-Fg If y is the vertical axis and x is the horisontal axis, then the y compontent of the force of tension is Ft*cos(angle)=-mg and the x component is Ft*sin(angle)=-mv^2/r dividing the x component by the y component gives tan(angle)=v^2/rg so, v=sqrt(r*g*tan(angle)) so, the linear momentum is p= [cont.]
Answered by Shadow Fish - Sun Nov 5 12:57:28 2006
Q. The ball is then set in motion in a horizontal circle so that the thread describew a cone. The length of the string, angle between the axis of rotation and the string, and the mass of the bob are given. Find the magnitude of the angular momentum???
Asked by Vladimir G - Sun Nov 5 12:07:38 2006 - - 3 Answers - 0 Comments
A. The answer depends on what point you use as the center. anyway,L=r(cross)p where r is the radius vector for the path, and p is the linear momentum. But, to use this formula, we need to find the linear momentum of the bob. Since we know its mass, we need to find its velocity. There are three forces of the string. tension, gravity, and centripetial force. since the object is not accelerating, the sum of these is 0. So, Ft+Fc+Fg=0 Or, Ft=-Fc-Fg If y is the vertical axis and x is the horisontal axis, then the y compontent of the force of tension is Ft*cos(angle)=-mg and the x component is Ft*sin(angle)=-mv^2/r dividing the x component by the y component gives tan(angle)=v^2/rg so, v=sqrt(r*g*tan(angle)) so, the linear momentum is p= [cont.]
Answered by Shadow Fish - Sun Nov 5 12:57:28 2006
how do I thread an iv catheter?
Q. once you insert the angiocath and see the flash of blood what do you do next. after this point i always mess things up. once i see the flash and try to push the catheter off it doesn't seem like it wants to go. after you see the flash of blood do you insert the needle anymore or lower the needle angle?
Asked by dreamgirleternity1 - Wed Jul 16 23:03:01 2008 - - 2 Answers - 0 Comments
A. you hold the angio cath in place and slowly advance it as you slowly remove the needle.If you leave the needle in place you'll run the risk of blowing the vein. Be ready once that needle is out, the blood will be a flowin'. Your timing should be such that the cath is all the way in just as the needle comes out..don't forget to pop off the tourniquet..then attach the iv line or lock..i have seen people float a cath in..they get their flash back insert the cath a centimeter or so, remove the needle attach the fluid, and float the cath in while running the fluid. Good Luck!
Answered by B-Rabbit - Wed Jul 16 23:18:31 2008
Q. once you insert the angiocath and see the flash of blood what do you do next. after this point i always mess things up. once i see the flash and try to push the catheter off it doesn't seem like it wants to go. after you see the flash of blood do you insert the needle anymore or lower the needle angle?
Asked by dreamgirleternity1 - Wed Jul 16 23:03:01 2008 - - 2 Answers - 0 Comments
A. you hold the angio cath in place and slowly advance it as you slowly remove the needle.If you leave the needle in place you'll run the risk of blowing the vein. Be ready once that needle is out, the blood will be a flowin'. Your timing should be such that the cath is all the way in just as the needle comes out..don't forget to pop off the tourniquet..then attach the iv line or lock..i have seen people float a cath in..they get their flash back insert the cath a centimeter or so, remove the needle attach the fluid, and float the cath in while running the fluid. Good Luck!
Answered by B-Rabbit - Wed Jul 16 23:18:31 2008
is there a wide angle lens converter (auxiliary lens) that i could use at 18mm and not vignette?
Q. i have a 18-70mm lens and i would like to know if there is an auxiliary lens i could use with it and not vignette at 18mm? or will none not vignette at 18mm ? i have an Opteka 0.35x and i have to use it at around 29 mm for it to not vignette(although, it doesnt fit my lens without an adapter, i accedently bought the 52mm instead of the 55mm thread )
Asked by Zenge - Thu Jul 29 16:57:15 2010 - - 1 Answers - 0 Comments
A. Most add on, filter type wide angle adapters were made during the film camera era, so they are not set up to be used with the smaller sensors on most digital SLRs. These filters were never meant to be a "professional" solution, and more of a gimmick overall. The reason you are finding you can only use it at 29mm without vignetting is that that was the widest lens most average people could afford with the old film SLRs, so the filters were designed work with focal lengths no wider than 28mm. I have never heard of an adapter that was designed to be used with any lens wider than 28mm
Answered by David L - Fri Jul 30 01:47:01 2010
Q. i have a 18-70mm lens and i would like to know if there is an auxiliary lens i could use with it and not vignette at 18mm? or will none not vignette at 18mm ? i have an Opteka 0.35x and i have to use it at around 29 mm for it to not vignette(although, it doesnt fit my lens without an adapter, i accedently bought the 52mm instead of the 55mm thread )
Asked by Zenge - Thu Jul 29 16:57:15 2010 - - 1 Answers - 0 Comments
A. Most add on, filter type wide angle adapters were made during the film camera era, so they are not set up to be used with the smaller sensors on most digital SLRs. These filters were never meant to be a "professional" solution, and more of a gimmick overall. The reason you are finding you can only use it at 29mm without vignetting is that that was the widest lens most average people could afford with the old film SLRs, so the filters were designed work with focal lengths no wider than 28mm. I have never heard of an adapter that was designed to be used with any lens wider than 28mm
Answered by David L - Fri Jul 30 01:47:01 2010
How would u find the centre of gravity of a chain of given mass rotating at constant velocity.?
Q. A closed chain of mass "m" is attached to a vertical shaft by means of a thread, rotating with a constant angular velocity "w" and thread forms an angle "a" with the vertical.Find the distance between chain's centre of gravity and rotation axis.Please answer me!!!
Asked by MCM - Fri Dec 16 10:12:48 2005 - - 2 Answers - 0 Comments
A. if i knew the answer to that i wouldnt waste my time on this stupid board and i would have a real job
Answered by midnightprophet - Fri Dec 16 10:19:05 2005
Q. A closed chain of mass "m" is attached to a vertical shaft by means of a thread, rotating with a constant angular velocity "w" and thread forms an angle "a" with the vertical.Find the distance between chain's centre of gravity and rotation axis.Please answer me!!!
Asked by MCM - Fri Dec 16 10:12:48 2005 - - 2 Answers - 0 Comments
A. if i knew the answer to that i wouldnt waste my time on this stupid board and i would have a real job
Answered by midnightprophet - Fri Dec 16 10:19:05 2005
two identical small insulating balls are suspended by separate 0.25m threads that are attached?
Q. two identical small insulating balls are suspended by separate 0.25m threads that are attached to a common point on the ceiling. Each ball has a mass of 8.0 x 10^-4 kg. Initially the balls are uncharged and hang straight down. They are then given identical positive charges and, as a result, spread apart with an angle of 36 degrees between the threads. Dtermine (a) the charge on each ball and (b) the tension in the thread.
Asked by Marie - Wed Feb 17 20:54:36 2010 - - 1 Answers - 0 Comments
A. Electrostatic force Fe = the horizontal component of the tension on the thread. Fe = mg*tan(theta) r = 2*0.25*sin(theta) A. To find the charge q, use the electrostatic force formula: Fe = kq^2/r^2 ==> q = sqrt(Fe*r^2/k) B. Tension = sqrt((mg)^2+Fe^2)
Answered by kirchwey - Thu Feb 18 09:28:02 2010
Q. two identical small insulating balls are suspended by separate 0.25m threads that are attached to a common point on the ceiling. Each ball has a mass of 8.0 x 10^-4 kg. Initially the balls are uncharged and hang straight down. They are then given identical positive charges and, as a result, spread apart with an angle of 36 degrees between the threads. Dtermine (a) the charge on each ball and (b) the tension in the thread.
Asked by Marie - Wed Feb 17 20:54:36 2010 - - 1 Answers - 0 Comments
A. Electrostatic force Fe = the horizontal component of the tension on the thread. Fe = mg*tan(theta) r = 2*0.25*sin(theta) A. To find the charge q, use the electrostatic force formula: Fe = kq^2/r^2 ==> q = sqrt(Fe*r^2/k) B. Tension = sqrt((mg)^2+Fe^2)
Answered by kirchwey - Thu Feb 18 09:28:02 2010
which thread braks first? (Mechanics)?
Q. when given a certain mass hanged by lets say two threads...and lets say that your given that each thread can sustain a maximum tension of 25 N (one of the threads is always horizontal while the other is at an angle to the vertical axis)...how can u predict which one will break first if the other is given the maximum 25 N???
Asked by IEE6000 - Thu Nov 29 16:07:23 2007 - - 1 Answers - 0 Comments
A. The one at an angle to vertical Tv, will always break first since Tv*sin( )=Th, where Th is the horizontal. By the way Tv*cos( )=m*g j
Answered by odu83 - Mon Dec 3 14:05:54 2007
Q. when given a certain mass hanged by lets say two threads...and lets say that your given that each thread can sustain a maximum tension of 25 N (one of the threads is always horizontal while the other is at an angle to the vertical axis)...how can u predict which one will break first if the other is given the maximum 25 N???
Asked by IEE6000 - Thu Nov 29 16:07:23 2007 - - 1 Answers - 0 Comments
A. The one at an angle to vertical Tv, will always break first since Tv*sin( )=Th, where Th is the horizontal. By the way Tv*cos( )=m*g j
Answered by odu83 - Mon Dec 3 14:05:54 2007
Help calculating tension and magnitude of the centripetal force!!!?
Q. A stone of mass 0.20 kg is attached to an extremely light thread, whose mass can be considered negligible, and swung around a fixed vertical axis Y so that the stone undergoes circular motion in a horizontal plane. The length of thread from the stone to the axis of rotation Y is L = 2.6 m and the thread makes an angle of = 73.0 to the vertical. What is the tension in the thread? What is the magnitude of the centripetal force on the stone?
Asked by blue01 - Thu Mar 26 04:20:09 2009 - - 2 Answers - 0 Comments
A. draw a diagram of a stone with a downward force for gravity, a horizontal force (to the right) for the centripetal force, and a diagonal force (up to the left) for the tension force of the string you know the magnitude of the downward force (W = mg) then you can calculate the tension in the string because you know the vertical component of the tension must be equivalent to the stone's weight and you know the direction of that vector then you can calculate the magnitude of the centripetal force by finding the horizontal component of the tension that's probably not what your teacher wants you to do, but I'd probably argue that it's a valid solution
Answered by jayjayem13 - Thu Mar 26 05:00:43 2009
Q. A stone of mass 0.20 kg is attached to an extremely light thread, whose mass can be considered negligible, and swung around a fixed vertical axis Y so that the stone undergoes circular motion in a horizontal plane. The length of thread from the stone to the axis of rotation Y is L = 2.6 m and the thread makes an angle of = 73.0 to the vertical. What is the tension in the thread? What is the magnitude of the centripetal force on the stone?
Asked by blue01 - Thu Mar 26 04:20:09 2009 - - 2 Answers - 0 Comments
A. draw a diagram of a stone with a downward force for gravity, a horizontal force (to the right) for the centripetal force, and a diagonal force (up to the left) for the tension force of the string you know the magnitude of the downward force (W = mg) then you can calculate the tension in the string because you know the vertical component of the tension must be equivalent to the stone's weight and you know the direction of that vector then you can calculate the magnitude of the centripetal force by finding the horizontal component of the tension that's probably not what your teacher wants you to do, but I'd probably argue that it's a valid solution
Answered by jayjayem13 - Thu Mar 26 05:00:43 2009
how to thread an ancient empisal sewing machine?
Q. Have borrowed my mother-in-laws old empisal sewing machine to do an urgent job, as my own machine is broken. This machine has a pulley system, and an unusual threading process. There is no spool holder at the top right hand side, or bobbin winder, instead there are two metal prongs with holes at an angle of 15 degrees on top of the machine above the normal threading section, and there is also a bobbin like area for the thread to pass through before going down to the needle. I need to know the process of where the thread needs to start through this maze down to the needle. I would appreciate a quick response from anyone who can guide me through this process,
Asked by Belinda - Mon Apr 26 20:39:13 2010 - - 2 Answers - 0 Comments
A. Not a whole lot of help, but has a downloadable manual for a machine that seems to match your description. 5GBP via paypal, though. They have manuals for other machines that are free to look at online, but pay to download. The view online appears that the copies are readable, fwiw.
Answered by kay - Tue Apr 27 22:47:07 2010
Q. Have borrowed my mother-in-laws old empisal sewing machine to do an urgent job, as my own machine is broken. This machine has a pulley system, and an unusual threading process. There is no spool holder at the top right hand side, or bobbin winder, instead there are two metal prongs with holes at an angle of 15 degrees on top of the machine above the normal threading section, and there is also a bobbin like area for the thread to pass through before going down to the needle. I need to know the process of where the thread needs to start through this maze down to the needle. I would appreciate a quick response from anyone who can guide me through this process,
Asked by Belinda - Mon Apr 26 20:39:13 2010 - - 2 Answers - 0 Comments
A. Not a whole lot of help, but has a downloadable manual for a machine that seems to match your description. 5GBP via paypal, though. They have manuals for other machines that are free to look at online, but pay to download. The view online appears that the copies are readable, fwiw.
Answered by kay - Tue Apr 27 22:47:07 2010
What is the best off-loom bead-weaving thread?
Q. I have a basic familiarity with stringing beads and would like to move on to off-loom weaving techniques (for example, peyote, herringbone, right-angle weave, brick stitch). My interest lies in seed beads. So my goal is to figure out what I need to get started before I run out and buy things! What is the best off-loom bead-weaving thread for seed beads? I have read a little about silk threads, nylon threads like Nymo, Nylux, and C-lon, and other threads such as Fireline and Power Pro. Can anyone tell me a good thread to start with? Also, I would appreciate hearing where you purchase the thread that you recommend. Thanks so much. I ll choose the most helpful answer for 10 points!
Asked by mtnlady - Sun Apr 25 08:26:07 2010 - - 2 Answers - 0 Comments
A. I have been beading for about 2 years. I have tried Nymo, Silamide, C-lon, and Fireline. Fireline has become my personal favorite. I started with Nymo, but found that I could use only short lengths because it tended to fray after multiple passes through beads. Crystals have a tendency to cut it. The limitations of Fireline is that it is available in only smoke and crystal whereas Nymo is available in a myriad of colors. Fireline's biggest advantages are that crystals will not cut it and it creates a firmer final result if tension is an issue for you. None of these threads are really expensive, so it would not be a strain on your wallet to try them all. I purchase all my thread at ArtBeads.com. Good luck! Della
Answered by Della - Mon Apr 26 09:21:20 2010
Q. I have a basic familiarity with stringing beads and would like to move on to off-loom weaving techniques (for example, peyote, herringbone, right-angle weave, brick stitch). My interest lies in seed beads. So my goal is to figure out what I need to get started before I run out and buy things! What is the best off-loom bead-weaving thread for seed beads? I have read a little about silk threads, nylon threads like Nymo, Nylux, and C-lon, and other threads such as Fireline and Power Pro. Can anyone tell me a good thread to start with? Also, I would appreciate hearing where you purchase the thread that you recommend. Thanks so much. I ll choose the most helpful answer for 10 points!
Asked by mtnlady - Sun Apr 25 08:26:07 2010 - - 2 Answers - 0 Comments
A. I have been beading for about 2 years. I have tried Nymo, Silamide, C-lon, and Fireline. Fireline has become my personal favorite. I started with Nymo, but found that I could use only short lengths because it tended to fray after multiple passes through beads. Crystals have a tendency to cut it. The limitations of Fireline is that it is available in only smoke and crystal whereas Nymo is available in a myriad of colors. Fireline's biggest advantages are that crystals will not cut it and it creates a firmer final result if tension is an issue for you. None of these threads are really expensive, so it would not be a strain on your wallet to try them all. I purchase all my thread at ArtBeads.com. Good luck! Della
Answered by Della - Mon Apr 26 09:21:20 2010
From Yahoo Answer Search: 'thread angle'
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