When cutting Acme threads, what angle do I set my compound at?
Q. I am cutting 1/2"-10 Acme threads for a course project in college. I remember cutting all standard threads at 29 degrees on the compound axis and my teacher quoted I should do about 14 deg. for the Acme but it doesn't sit right with me even though that is about half the 29 degree angle for an Acme thread. What is the best angle to set the compound at? I am thinking about putting it on 14.5 deg. but I don't know if that will cause more trouble than it will prevent.
Asked by Steve - Thu May 31 21:52:33 2007 - - 2 Answers - 0 Comments

A. I believe you are right, it is 29 degrees for a standard Acme cut. 14.5 for each side of the thread will produce the acme shape. The less angle, the stronger the thread but the lower the friction. See the attached link on various threads...
Answered by SilverRAM - Thu May 31 22:48:34 2007

Finding the angle of the thread?
Q. The question gives me a small sphere with m = 2*10^-3 g and carrying a charge of 5.2*10^-8. It hangs from a thread near a charged conducting sheet with a charge density of -2.9*10^-9 C/m^2. Not sure what equation to use considering my teacher doesn't know how to teach so advice would help. I tried using that equation but I keep getting that theta is a negative angle in degrees. This answer isn't right so I'm not sure what I am doing wrong. I keep getting -4.346*10^-4 degrees.
Asked by HaloPrincess - Thu Feb 14 23:01:37 2008 - - 2 Answers - 1 Comments

A. I think this is a two part question. The first part is to figure out how much force the sphere has on it towards the conducting sheet. I think that is E=F/q so F=Eq F=-2.9*10^-9C/m^2*5.2*10^ -8 , you may want to check this formula as I am not sure of this part. The second part is to draw a diagram of all the forces acting on the sphere, there will be the force from part one (towards the sheet), the weight=mg and the force exerted by the string, which will cancel out the other two. You can then find that angle of the string. Hope that helps
Answered by Bill M - Fri Feb 15 01:28:17 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

What is the tension (in N) in each thread?
Q. A sign is hanging from the ceiling on two threads of equal length, which make an angle q of 139.5 with each other. What is the tension (in N) in each thread, assuming that the force of gravity acting on the sign is 44.6 N?
Asked by Master H - Sun Feb 21 22:40:00 2010 - - 1 Answers - 0 Comments

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

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

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

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

What is the magnitude of the charge on each plate?
Q. A small plastic ball of mass 6.20x10^-3kg and charge +0.19 micro coulomb is suspended from an insulating thread and hangs between the plates of a capacitor. 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 0.0128m^2.
Asked by Brandy J - Wed Jun 6 18:42:28 2007 - - 1 Answers - 0 Comments

A. ive got half of this problem done. i drew a diagram, used some trigonometry to calculate the horizontal force acting on the mass, by first working out its downward force due to gravity. but i cant remember any formulas linking force, charge, and plate area. i just know the force exerted by the plates.
Answered by fpa06mr - Wed Jun 6 18:58:35 2007

Thread size 1 1/2 " x 16 TPI?
Q. What type of thread is it? Its from the rear brake hubs of a FX (1948) Holden. We are currently trying to machine up a hub removal tool but can't work out what thread it is (eg.UNF,BSF,etc). The angle of thread seems to be 55degrees. Our "Zues" book at work doesn't have any details. The thread diameter is 1 1/2". The TPI is 16 thread per inch
Asked by myself - Wed Aug 8 00:06:55 2007 - - 2 Answers - 0 Comments

A. I own a shop, and have an extensive background in machine building. 1 1/2 - 16 is either a NF or A.S.M.E. std. machine screw. An NC is 6 T.P.I, and as a rule the NF is double, or more. If you draw it to scale, you can easily come up with the exact pitch. I'm not convinced 55 degrees will work out, but if you draw 16 intervals in an inch, the pitch will fall where it has to by connecting the lines on the drawing. You must have the correct diameter of the O.D. of the top & bottom of the threads to come up with the pitch on the threads. Like I said; I have doubts about the thread pitch you have calculated, but by drawing it to scale, you can use your tools to see the correct pitch, and radius of the bottom of the threads. The only… [cont.]
Answered by musiclicker - Wed Aug 8 08:44:57 2007

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

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

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

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

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

Are Prakticar / Pentacon lenses compatible with any other brand's digital SLR camera?
Q. My Praktica BC1 model came with zoom and wide angle lenses. They have M42 x 1 thread and I would like to use them with a new Digital SLR camera. Do any of the modern Digital SLR cameras (Canon, Minolta, Sony, etc) have the same thread, and so may be compatible with the lenses I have? Thank you.
Asked by Sammy - Mon Feb 18 17:22:09 2008 - - 2 Answers - 0 Comments

A. No DSLR is going to have such a lens mount; indeed, the last new film cameras with this lens mount were made about 35 years ago. You can get lens adapters. I use an EOS-to-M42 adapter on my Canon DSLRs so I can use my 1960s Pentax screwmount lenses on them. Accessories, too, like closeup extension tubes. All the usual things apply: manual focus, stop-down metering, a focal length factor, if applicable. A good lens will take good pictures on a DSLR. A DSLR will also show you exactly how bad a bad lens is.
Answered by laurahal42 - Mon Feb 18 18:02:48 2008

How much charge was placed on each sphere?
Q. Two .20g metal spheres are hung from a common point by nonconducting threads 30cm long. both are given identical charges, and the eectrostatic repulsion forces them apart until the angle between the threads is 20 degrees. how much charge was placed on each sphere? ok how would i start this problem?? can u even do this without knowing the charges? id really appreaciate some help on this stuff, i dont get it. thanks
Asked by Kel - Wed Jul 11 16:13:45 2007 - - 3 Answers - 0 Comments

A. First figure out the force required to hold the spheres at a 20 degree angle, actually 10 degrees for each sphere, based on g(9.8nm/s^2) the mass of the spheres and the length of the string. Then you can use coulombs law to calculate the charge on each sphere, in this case Q1=Q2 so you can just use Q1 squared.
Answered by LG - Wed Jul 11 16:22:26 2007

What is the potential difference in volts? Help !!!?
Q. A small object with a mass of 380 mg carries a charge of 40.0 nC and is suspended by a thread between the vertical plates of a parallel-plate capacitor. The plates are separated by 5.00 cm. If the thread makes an angle of 16.0 with the vertical, what is the potential difference between the plates?
Asked by activegirl - Fri Feb 2 13:15:54 2007 - - 1 Answers - 0 Comments

A. 380mg = 0.380g = 0.000 380 kg = 3.8e-4 kg The horizontal force on the object is 3.8e-4kg*9.8m/s^2*sin(16) = 1.026e-3N The electrical field is given by Coulomb's law in Newton per Coulomb (or equivalent Volts per meter) F = qE E = F/q = 1.026e-3N/4e-8C = 25650 N/C = 25650 V/m Now by multiplying V/m by distance we will get Volts. So the distance between the plate being 5e-2m, the voltage is 25650 V/m * 5e-2m = 1282.5 Volts
Answered by catarthur - Fri Feb 2 22:39:13 2007

What is the charge on each ball?
Q. I got the answer to the tension part of the question. Tension = 4.78e-3 N. Now, what is the charge on each ball? Two identical, small insulating balls are suspended by separate 0.37-m threads that are attached to a common point on the ceiling. Each ball has a mass of 4.4 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 51o between the threads. Determine (a) the charge on each ball and (b) the tension in the threads.
Asked by krystal_michele - Fri Nov 9 21:46:00 2007 - - 1 Answers - 0 Comments

A. By the way, this is a physics question. First let us focus our attention on a single ball. Its weight is W = 4.4x10^-4kg * 9.8m/s^2 = 0.0043N. The 0.37-m thread attached to the ball makes 51 /2 = 25.5 with vertical. The vertical component of tension T in the thread must balance the weight of the ball: T cos(25.5 ) = W, and its horizontal component must be balanced by the charge-charge interaction: T sin(25.5 ) = W tan(25.5 ) = 0.00206N. (a) Since each thread is 0.37m long, the ball-ball distance is: r = 2*0.37m*sin(25.5 ) = 0.32m Let the charge on each ball be q C. We have: q^2/ 4 r^2 = 0.00206N Thus: q = sqrt(4 r^2 * 0.00206N) = 0.32*sqrt(0.00206/8.99e9) C = 1.5x10^-7 C. (b) T = W/cos(25.5 ) = 4.78x10^-3 N.
Answered by Hahaha - Sun Nov 11 16:32:16 2007

What is the strength of the magnetic field?
Q. A long, straight wire with linear mass density of 42 g/m is suspended by threads. There is a uniform magnetic field pointing vertically downward. A 12 A current in the wire experiences a horizontal magnetic force that deflects it to an equilibrium angle of 10 degrees. What is the strength of the magnetic field? I have a pic if that helps but I'm not sure how to set up a link so i could mail it. I appreciate any help.
Asked by Dude - Tue Jul 28 11:18:00 2009 - - 1 Answers - 0 Comments

A. The horizontal force per unit length: Fh = F/L = I x B causes the current carrying wire to be deflected by an angle of 10 deg. due to the interaction with the magnetic field. Therefore, we have a force triangle consisting of the horizontal magnetic force and the tension on the threads supporting the wire. The tension on the wire is resultant of the weight of the wire and the magnetic force acting on the wire. These two component forces create the force triangle with the 10 deg. central angle. Hence, tan@ = Fh/Fv, => where @ = 10 deg. and Fv = g : linear mass density = 42g/m Fh = Fv tan@ = (42)(9.8)tan(10) = 72.57 N Also, the horizontal force per unit length is the Lorentz force due to a current… [cont.]
Answered by Jacy - Tue Jul 28 12:14:48 2009