The radius of a solid sphere is twice the radius of a second solid sphere. What is the ratio of?
Q. The radius of a solid sphere is twice the radius of a second solid sphere. What is the ratio of a) their volumes, b) their surface areas?
Asked by Jennifer T - Mon May 5 22:30:10 2008 - - 2 Answers - 0 Comments
A. vb/vx=(rb/rs)^3 vb:vs=9:1 ab/as=(rb/rs)^2 ab:as=4:1
Answered by someone else - Mon May 5 22:33:32 2008
Q. The radius of a solid sphere is twice the radius of a second solid sphere. What is the ratio of a) their volumes, b) their surface areas?
Asked by Jennifer T - Mon May 5 22:30:10 2008 - - 2 Answers - 0 Comments
A. vb/vx=(rb/rs)^3 vb:vs=9:1 ab/as=(rb/rs)^2 ab:as=4:1
Answered by someone else - Mon May 5 22:33:32 2008
How to calculate the volume of a sphere based on percentage of surface area?
Q. How would you calculate the volume of a sphere based off of a percentage of the surface area if the volume of the sphere is known but only in terms of the surface area? Example: A sphere is measured to displace 280,000 cm^3 of water but its surface area is only submerged 72% under water. What is the total volume of the sphere? Please show work and explain and you will receive points. Could you please show work on how you obtained your answer so I can see if it is right? Thank you.
Asked by C_Rock136 - Fri Feb 22 16:49:04 2008 - - 2 Answers - 0 Comments
A. First the formula for the volume of a sphere is: V = (4/3)(pi)(r^3), and the area of a sphere is: A = 4*(pi)(r^2), which makes it possible to rewrite the two formulae as: V = 4/3(pi)(r^3) = 1/3(A*r) and A = 3(V/r) Now, i know that having 72% of the sphere submerged is multiplying the A by .72. 28000 = 1/3(.72A*r) Solving for A, you get: A = 3*28000/(r*.72) and then equate that to A = (3V)/r (3*28000)/(.72r) = (3V)/(r) 28000/.72 = V V = 388,888.888 cm^3
Answered by NBL - Fri Feb 22 17:57:44 2008
Q. How would you calculate the volume of a sphere based off of a percentage of the surface area if the volume of the sphere is known but only in terms of the surface area? Example: A sphere is measured to displace 280,000 cm^3 of water but its surface area is only submerged 72% under water. What is the total volume of the sphere? Please show work and explain and you will receive points. Could you please show work on how you obtained your answer so I can see if it is right? Thank you.
Asked by C_Rock136 - Fri Feb 22 16:49:04 2008 - - 2 Answers - 0 Comments
A. First the formula for the volume of a sphere is: V = (4/3)(pi)(r^3), and the area of a sphere is: A = 4*(pi)(r^2), which makes it possible to rewrite the two formulae as: V = 4/3(pi)(r^3) = 1/3(A*r) and A = 3(V/r) Now, i know that having 72% of the sphere submerged is multiplying the A by .72. 28000 = 1/3(.72A*r) Solving for A, you get: A = 3*28000/(r*.72) and then equate that to A = (3V)/r (3*28000)/(.72r) = (3V)/(r) 28000/.72 = V V = 388,888.888 cm^3
Answered by NBL - Fri Feb 22 17:57:44 2008
How fast is the sphere moving when it reaches the bottom of the incline?
Q. A solid sphere of uniform density rolls without slipping from the top to the bottom of a 30 incline, a distance 2.6m. The sphere has a mass of 2.5kg and a radius of 28cm. It starts from rest at the top of the incline.
Asked by hikari takishima - Sat Sep 20 09:11:44 2008 - - 3 Answers - 0 Comments
A. since the sphere rolls without slipping,, therefore v=w*r also height of the incline=h*sin30=1.3m so conserving energy we have Mgh = 1/2Mv^2 + 1/2 I w^2 also moment of inertia for solid sphere is 2/5Mr^2 this will get simplified to : Mgh = 7/10 Mv^2 v^2 = 10/7 gh v= 18.5 m/s
Answered by arpit g - Sat Sep 20 09:37:57 2008
Q. A solid sphere of uniform density rolls without slipping from the top to the bottom of a 30 incline, a distance 2.6m. The sphere has a mass of 2.5kg and a radius of 28cm. It starts from rest at the top of the incline.
Asked by hikari takishima - Sat Sep 20 09:11:44 2008 - - 3 Answers - 0 Comments
A. since the sphere rolls without slipping,, therefore v=w*r also height of the incline=h*sin30=1.3m so conserving energy we have Mgh = 1/2Mv^2 + 1/2 I w^2 also moment of inertia for solid sphere is 2/5Mr^2 this will get simplified to : Mgh = 7/10 Mv^2 v^2 = 10/7 gh v= 18.5 m/s
Answered by arpit g - Sat Sep 20 09:37:57 2008
How much work is required to completely submerge the sphere?
Q. A sphere has a negligible weight and is flowing in a large freshwater lake. How much work is required to completely submerge the sphere? Proof of calculation needed + explanation
Asked by lime_greenluv - Mon Mar 9 07:42:43 2009 - - 3 Answers - 0 Comments
A. Mass of water displaced x time to displace this water. I'm not doing it for you, as you wouldn't learn anything, would ya?
Answered by Yawn - Mon Mar 9 07:47:51 2009
Q. A sphere has a negligible weight and is flowing in a large freshwater lake. How much work is required to completely submerge the sphere? Proof of calculation needed + explanation
Asked by lime_greenluv - Mon Mar 9 07:42:43 2009 - - 3 Answers - 0 Comments
A. Mass of water displaced x time to displace this water. I'm not doing it for you, as you wouldn't learn anything, would ya?
Answered by Yawn - Mon Mar 9 07:47:51 2009
What is the volume of a sphere and its error?
Q. I have a sphere that I would like to know the volume of. I measure it's diameter with a ruler to be 0.11 m and estimate my value for the diameter to be accurate to 0.005m. Recalling that the volume of a sphere is (4/3) pi r^3 , calculate the volume of the sphere and its error. ___ How do you calculate the error for this?
Asked by YanksFan - Tue Sep 8 19:55:18 2009 - - 1 Answers - 0 Comments
A. The total differential. V=(4/3) R dV = 4 R dR R = D/2 = (0.11)/2 = 0.055m dR = dD/2 = (0.005)/2 = 0.0025m dV = 4 (0.055) (0.0025) = 9.5 x 10^-5 m
Answered by Oprah - Tue Sep 8 20:13:25 2009
Q. I have a sphere that I would like to know the volume of. I measure it's diameter with a ruler to be 0.11 m and estimate my value for the diameter to be accurate to 0.005m. Recalling that the volume of a sphere is (4/3) pi r^3 , calculate the volume of the sphere and its error. ___ How do you calculate the error for this?
Asked by YanksFan - Tue Sep 8 19:55:18 2009 - - 1 Answers - 0 Comments
A. The total differential. V=(4/3) R dV = 4 R dR R = D/2 = (0.11)/2 = 0.055m dR = dD/2 = (0.005)/2 = 0.0025m dV = 4 (0.055) (0.0025) = 9.5 x 10^-5 m
Answered by Oprah - Tue Sep 8 20:13:25 2009
What happens to light inside a perfectly reflective sphere?
Q. If you had a hollow sphere that was perfectly reflective on the inside, and completely inescapable (ie, nothing can escape the interior - this being due to a perfect seal, not because of gravity), and you had a means to turn on and off a light source inside the sphere, and you have a way of observing what happens inside the sphere, what happens when you turn on the light, and then turn it off? Is light forever present in the sphere? And what if we make the assumption that none of the light is lost as heat? Does turning on the light source for longer periods of time add "more" light to the inside - how could this quantity be measured?
Asked by comusbassington - Tue Jul 3 16:41:48 2007 - - 16 Answers - 2 Comments
A. You get light when it is turned on and darkness when the light is turned off. The light is a stream of photons. So as long as it the light source is on the light stream will illuminate the inside and reflect all around. When you turn the light off the stream of photons will stop and so the light will go out. Photons don't flow in a stream; photons act like waves, but are actually particles. Photons are unique particles that don't obey all the laws. According to the laws of physics you can't create or destroy anything, you only change its form. When you turn on the light you are turning electrical energy into light, or using some similar operation. The photons are there and they exist, but they move at random. They don't have any… [cont.]
Answered by Dan S - Tue Jul 3 16:49:51 2007
Q. If you had a hollow sphere that was perfectly reflective on the inside, and completely inescapable (ie, nothing can escape the interior - this being due to a perfect seal, not because of gravity), and you had a means to turn on and off a light source inside the sphere, and you have a way of observing what happens inside the sphere, what happens when you turn on the light, and then turn it off? Is light forever present in the sphere? And what if we make the assumption that none of the light is lost as heat? Does turning on the light source for longer periods of time add "more" light to the inside - how could this quantity be measured?
Asked by comusbassington - Tue Jul 3 16:41:48 2007 - - 16 Answers - 2 Comments
A. You get light when it is turned on and darkness when the light is turned off. The light is a stream of photons. So as long as it the light source is on the light stream will illuminate the inside and reflect all around. When you turn the light off the stream of photons will stop and so the light will go out. Photons don't flow in a stream; photons act like waves, but are actually particles. Photons are unique particles that don't obey all the laws. According to the laws of physics you can't create or destroy anything, you only change its form. When you turn on the light you are turning electrical energy into light, or using some similar operation. The photons are there and they exist, but they move at random. They don't have any… [cont.]
Answered by Dan S - Tue Jul 3 16:49:51 2007
How fast will the electron be moving when it reaches the surface of the sphere?
Q. An electron starts from rest 2.82 cm from the center of a uniformly charged sphere of radius 2.17 cm. If the sphere carries a total charge of 1.11 10-9 C, how fast will the electron be moving when it reaches the surface of the sphere?
Asked by eltel2910 - Wed Oct 4 12:11:23 2006 - - 1 Answers - 0 Comments
A. assuming the charges on the sphere don't move and the excess charge is +ve then the electron will be doing about 1.4e6 m/second by then
Answered by MikeY - Wed Oct 4 13:27:44 2006
Q. An electron starts from rest 2.82 cm from the center of a uniformly charged sphere of radius 2.17 cm. If the sphere carries a total charge of 1.11 10-9 C, how fast will the electron be moving when it reaches the surface of the sphere?
Asked by eltel2910 - Wed Oct 4 12:11:23 2006 - - 1 Answers - 0 Comments
A. assuming the charges on the sphere don't move and the excess charge is +ve then the electron will be doing about 1.4e6 m/second by then
Answered by MikeY - Wed Oct 4 13:27:44 2006
What is the difference between a colony, a sphere of influence, a protectorate and a territory?
Q. What is the difference between a colony, a sphere of influence, a protectorate and a territory? Like in history and colonial powers and stuff...anyone?
Asked by E Biz - Sat Apr 12 23:26:51 2008 - - 4 Answers - 0 Comments
A. A Territory is an integral possession of the couhtry, with aspects of self-government but with the executive appointed by the central government. In US history, Territories were expected to become States whenevr they were deemed to have sufficient population. A Protectorate is a sovereign country under the protection of another country. Usuallly this entails not only guaranteeing the protectorate's territory against encroachment by other powers, but some degree of supervision of at least the Protectorate's foreign affairs so that it doesn't provoke a crisis. Some coutries have been "Protected" right out of existance, such as Hawaii. Others have undergone a period of "getting on their feet" and gone onto become fully independent countries,… [cont.]
Answered by Sydney Michelle - Sat Apr 12 23:54:35 2008
Q. What is the difference between a colony, a sphere of influence, a protectorate and a territory? Like in history and colonial powers and stuff...anyone?
Asked by E Biz - Sat Apr 12 23:26:51 2008 - - 4 Answers - 0 Comments
A. A Territory is an integral possession of the couhtry, with aspects of self-government but with the executive appointed by the central government. In US history, Territories were expected to become States whenevr they were deemed to have sufficient population. A Protectorate is a sovereign country under the protection of another country. Usuallly this entails not only guaranteeing the protectorate's territory against encroachment by other powers, but some degree of supervision of at least the Protectorate's foreign affairs so that it doesn't provoke a crisis. Some coutries have been "Protected" right out of existance, such as Hawaii. Others have undergone a period of "getting on their feet" and gone onto become fully independent countries,… [cont.]
Answered by Sydney Michelle - Sat Apr 12 23:54:35 2008
What is the sphere's angular velocity at the bottom of the incline?
Q. An 8.40-cm-diameter, 300 sphere is released from rest at the top of a 2.00-m-long, 17.0 incline. It rolls, without slipping, to the bottom. What fraction of its kinetic energy is rotational?
Asked by juan - Wed Apr 1 20:22:20 2009 - - 1 Answers - 0 Comments
A. Here to calculate the angular velocity, and the rotational kinetic energy, you must use: radius = 8.4 cm, not half of that The reason being that the center of rotation for a rolling sphere is the contact point with the incline, NOT the center of the sphere.
Answered by PhysicsDude - Sun Apr 5 18:00:07 2009
Q. An 8.40-cm-diameter, 300 sphere is released from rest at the top of a 2.00-m-long, 17.0 incline. It rolls, without slipping, to the bottom. What fraction of its kinetic energy is rotational?
Asked by juan - Wed Apr 1 20:22:20 2009 - - 1 Answers - 0 Comments
A. Here to calculate the angular velocity, and the rotational kinetic energy, you must use: radius = 8.4 cm, not half of that The reason being that the center of rotation for a rolling sphere is the contact point with the incline, NOT the center of the sphere.
Answered by PhysicsDude - Sun Apr 5 18:00:07 2009
Which of the following statements about the celestial sphere is false?
Q. Which of the following statements about the celestial sphere is false? a. The Sun moves along a path on the celestial sphere called the ecliptic. b. Where the ecliptic intersects the celestial equator are points called equinoxes. c. The coordinates on the celestial sphere are right-ascension and declination d. Right-ascension lines are similar to longitude lines on the Earth's surface. e. The Earth's axis is tilted 23.5 degrees with respect the axis line connecting the celestial poles.
Asked by resident59 - Sat Jun 6 05:28:03 2009 - - 4 Answers - 0 Comments
A. a. The Sun moves along a path on the celestial sphere called the ecliptic - true b. Where the ecliptic intersects the celestial equator are points called equinoxes. - true c. The coordinates on the celestial sphere are right-ascension and declination -true d. Right-ascension lines are similar to longitude lines on the Earth's surface. - true e. The Earth's axis is tilted 23.5 degrees with respect the axis line connecting the celestial poles. - false - with respect to the plane of the earth's orbit, or the ecliptic.
Answered by Elizabeth H - Sat Jun 6 06:36:12 2009
Q. Which of the following statements about the celestial sphere is false? a. The Sun moves along a path on the celestial sphere called the ecliptic. b. Where the ecliptic intersects the celestial equator are points called equinoxes. c. The coordinates on the celestial sphere are right-ascension and declination d. Right-ascension lines are similar to longitude lines on the Earth's surface. e. The Earth's axis is tilted 23.5 degrees with respect the axis line connecting the celestial poles.
Asked by resident59 - Sat Jun 6 05:28:03 2009 - - 4 Answers - 0 Comments
A. a. The Sun moves along a path on the celestial sphere called the ecliptic - true b. Where the ecliptic intersects the celestial equator are points called equinoxes. - true c. The coordinates on the celestial sphere are right-ascension and declination -true d. Right-ascension lines are similar to longitude lines on the Earth's surface. - true e. The Earth's axis is tilted 23.5 degrees with respect the axis line connecting the celestial poles. - false - with respect to the plane of the earth's orbit, or the ecliptic.
Answered by Elizabeth H - Sat Jun 6 06:36:12 2009
What are some tips for how to cover a sphere with fondant?
Q. I have never covered a sphere in fondant...can you help me learn how? I can cover a round cake... Thanks!
Asked by SAH - Fri Dec 14 15:06:38 2007 - - 2 Answers - 0 Comments
A. Firstly roll out the fondant dough so it is large enough that you can cover it. Have the ball on some sort of spike or base to keep it from rolling. I'd attach a couple of heavy dowels with screws to a board that would keep the sphere slightly above the work surface. You might need to make "washers" out of some heavy plastic such as a shortening can lid. This would distribute the weight if your working with cake. The smoothing of the fondant without the weight pulling the dough too thin or tearing is the trickiest part. Just take your time and imagine it being finished. If you can do a regular cake you can do shapes.
Answered by lemonlimesherbet - Fri Dec 14 18:08:46 2007
Q. I have never covered a sphere in fondant...can you help me learn how? I can cover a round cake... Thanks!
Asked by SAH - Fri Dec 14 15:06:38 2007 - - 2 Answers - 0 Comments
A. Firstly roll out the fondant dough so it is large enough that you can cover it. Have the ball on some sort of spike or base to keep it from rolling. I'd attach a couple of heavy dowels with screws to a board that would keep the sphere slightly above the work surface. You might need to make "washers" out of some heavy plastic such as a shortening can lid. This would distribute the weight if your working with cake. The smoothing of the fondant without the weight pulling the dough too thin or tearing is the trickiest part. Just take your time and imagine it being finished. If you can do a regular cake you can do shapes.
Answered by lemonlimesherbet - Fri Dec 14 18:08:46 2007
How many degrees of rotation are there in a sphere?
Q. I know that a 2-Dimensional circle has 360 degrees of rotation, but how many are there the circle's 3-Dimensional counterpart - the sphere?
Asked by Brngths - Sat Aug 11 13:52:56 2007 - - 4 Answers - 0 Comments
A. The clarification of your question indicates that you are after the concept of "solid angle". Compare this to the usual concept of "circular" or "planar" angle: A planar angle (measured in radians "rad") between two straight lines originating at the center of a circle of unit radius is the length of the circular arc between those two lines (more precisely, from one line, chosen as "reference", to the other, as an angle is assigned a positive sign if you rotate counterclockwise and a negative sign otherwise). Similarly, a solid angle (measured in steradians "sr") is assigned to the cone generated by half a straight line originating at the center of a sphere of unit radius with one point of that line moving in a closed loop at the surface… [cont.]
Answered by DrGerard - Sun Aug 12 03:46:17 2007
Q. I know that a 2-Dimensional circle has 360 degrees of rotation, but how many are there the circle's 3-Dimensional counterpart - the sphere?
Asked by Brngths - Sat Aug 11 13:52:56 2007 - - 4 Answers - 0 Comments
A. The clarification of your question indicates that you are after the concept of "solid angle". Compare this to the usual concept of "circular" or "planar" angle: A planar angle (measured in radians "rad") between two straight lines originating at the center of a circle of unit radius is the length of the circular arc between those two lines (more precisely, from one line, chosen as "reference", to the other, as an angle is assigned a positive sign if you rotate counterclockwise and a negative sign otherwise). Similarly, a solid angle (measured in steradians "sr") is assigned to the cone generated by half a straight line originating at the center of a sphere of unit radius with one point of that line moving in a closed loop at the surface… [cont.]
Answered by DrGerard - Sun Aug 12 03:46:17 2007
What exactly would happen if you had a hollow sphere with mirror on the interior side?
Q. If you shined a high-energy laser or a light somehow at the interior side of a closed sphere, how would the light reflect?
Asked by mooseyhootington - Wed Jul 30 03:57:12 2008 - - 1 Answers - 0 Comments
A. The light will be reflected back and forth within the sphere. However the energy of the light will be gradually absorbed by the atoms of the mirror and converted into heat which would then be dissipated through the wall of the sphere and eventually the inside of the sphere becomes dark.
Answered by draco4843 | - Wed Jul 30 08:38:12 2008
Q. If you shined a high-energy laser or a light somehow at the interior side of a closed sphere, how would the light reflect?
Asked by mooseyhootington - Wed Jul 30 03:57:12 2008 - - 1 Answers - 0 Comments
A. The light will be reflected back and forth within the sphere. However the energy of the light will be gradually absorbed by the atoms of the mirror and converted into heat which would then be dissipated through the wall of the sphere and eventually the inside of the sphere becomes dark.
Answered by draco4843 | - Wed Jul 30 08:38:12 2008
What is the radius of the metal sphere in meter?
Q. A metal sphere centered at the origin carries a surface charge of charge density 16.4 nC/m2. At r = 4 m, the potential is 650 V and the magnitude of the electric field is 250 V/m. The permittivity of free space is 8.85 10 12 V. Determine the radius of the metal sphere. Answer in units of m.
Asked by Nina - Fri Feb 20 13:53:49 2009 - - 2 Answers - 3 Comments
A. AS electric field is not zero th epoint of observation is beyond teh radius of the sphere. Let it be R. So total charge, Q is given by Q = 4x3.14x16.4x10^-9R^2 = [2.06x10^-7]R^2 Coulomb Potential,V at r distance away from centre is given by V= 9x10^9x(2.06x10^-7xR^2)/4 =650 (given). So (R^2 = 2600/[1.854x10^3]= 1.402 or R = 1.18 m
Answered by Let'slearntothink - Fri Feb 20 14:15:32 2009
Q. A metal sphere centered at the origin carries a surface charge of charge density 16.4 nC/m2. At r = 4 m, the potential is 650 V and the magnitude of the electric field is 250 V/m. The permittivity of free space is 8.85 10 12 V. Determine the radius of the metal sphere. Answer in units of m.
Asked by Nina - Fri Feb 20 13:53:49 2009 - - 2 Answers - 3 Comments
A. AS electric field is not zero th epoint of observation is beyond teh radius of the sphere. Let it be R. So total charge, Q is given by Q = 4x3.14x16.4x10^-9R^2 = [2.06x10^-7]R^2 Coulomb Potential,V at r distance away from centre is given by V= 9x10^9x(2.06x10^-7xR^2)/4 =650 (given). So (R^2 = 2600/[1.854x10^3]= 1.402 or R = 1.18 m
Answered by Let'slearntothink - Fri Feb 20 14:15:32 2009
What if a deep space probe slammed into the celestial sphere?
Q. What if Aristotle and Ptolemy were right? What if the Voyagers 1 and 2 crash into the celestial sphere? How would that change our space program and our view of our place in the universe?
Asked by Link - Sat Aug 25 14:53:12 2007 - - 5 Answers - 0 Comments
A. Aristotle and Ptolemy didn't say there was a single celestial sphere. They said there were many spheres (one for each visible solar system object), at varying distances from the Earth, and all concentric. All the way out to the one holding the "stationary" stars. If they were right, then Voyager 2 (the first of the two to launch) would have proved it 30 years ago, when it would have shattered the one containing the Moon. Interestingly enough, that didn't happen. Nor did it happen later when they passed through the "spheres" of Mercury, Venus, the Sun, Mars, Jupiter, etc., etc. Nor when numerous other probes passed through various "spheres" over the twenty years before them. To hold that the solar system is composed of mostly empty… [cont.]
Answered by skeptik - Mon Aug 27 21:18:37 2007
Q. What if Aristotle and Ptolemy were right? What if the Voyagers 1 and 2 crash into the celestial sphere? How would that change our space program and our view of our place in the universe?
Asked by Link - Sat Aug 25 14:53:12 2007 - - 5 Answers - 0 Comments
A. Aristotle and Ptolemy didn't say there was a single celestial sphere. They said there were many spheres (one for each visible solar system object), at varying distances from the Earth, and all concentric. All the way out to the one holding the "stationary" stars. If they were right, then Voyager 2 (the first of the two to launch) would have proved it 30 years ago, when it would have shattered the one containing the Moon. Interestingly enough, that didn't happen. Nor did it happen later when they passed through the "spheres" of Mercury, Venus, the Sun, Mars, Jupiter, etc., etc. Nor when numerous other probes passed through various "spheres" over the twenty years before them. To hold that the solar system is composed of mostly empty… [cont.]
Answered by skeptik - Mon Aug 27 21:18:37 2007
What is a good public sphere topic for a speech?
Q. I have to do a group speech in my Comm 100 class and it has to be a topic on the public sphere, and I have no clue what a topic on that would be. Any suggestions?
Asked by kldavis89 - Wed Sep 12 00:43:37 2007 - - 1 Answers - 0 Comments
A. Why not do one on "The Benefits of safety belt" or "Seat belt safety".I got an "A" with that topic. Everything you need is on the web .Goodluck!
Answered by izuma4 - Wed Sep 12 00:52:05 2007
Q. I have to do a group speech in my Comm 100 class and it has to be a topic on the public sphere, and I have no clue what a topic on that would be. Any suggestions?
Asked by kldavis89 - Wed Sep 12 00:43:37 2007 - - 1 Answers - 0 Comments
A. Why not do one on "The Benefits of safety belt" or "Seat belt safety".I got an "A" with that topic. Everything you need is on the web .Goodluck!
Answered by izuma4 - Wed Sep 12 00:52:05 2007
How to determine the outside diameter of a sphere?
Q. Supposed the system is in equilibrium. Determine the diameter of the outside sphere if the submerged sphere has a diameter of 35cm (RD=1.5) the answer should be 24.27cm , but i kept on having 40cm. help? whats your solution?
Asked by Let M - Sun Jul 6 05:22:17 2008 - - 1 Answers - 0 Comments
A. i get 21? ar you sure its 24?
Answered by unknown - Sun Jul 6 06:29:01 2008
Q. Supposed the system is in equilibrium. Determine the diameter of the outside sphere if the submerged sphere has a diameter of 35cm (RD=1.5) the answer should be 24.27cm , but i kept on having 40cm. help? whats your solution?
Asked by Let M - Sun Jul 6 05:22:17 2008 - - 1 Answers - 0 Comments
A. i get 21? ar you sure its 24?
Answered by unknown - Sun Jul 6 06:29:01 2008
Can I still use light gates when measuring a sphere falling through honey to find its viscosity?
Q. For my A level physics project, I'm measuring the viscosity of honey with a 'falling sphere' method. I'm wondering whether the light gates will register the sphere falling through the honey which will have a colour of its own - has anyone done something similar to this themselves? And could there be any accurate alternatives to using light gates? Thanks.
Asked by Sam - Sat Oct 31 17:21:51 2009 - - 1 Answers - 0 Comments
A. Sam, sounds like a great project. Light gates should work through honey, as long as the light source is strong enough. One problem you will have is alignment - making sure the spheres actually block the path of the light as they move past. I suspect that you can get equally good results simply timing by hand. When timing by hand you will want to make the distance the balls fall through as large as possible (this makes the inaccuracies when timing by hand less important) - see if you can borrow a large graduated cylinder. If you use a small sphere, then it will take a long time to fall, so that also helps. When you make the measurement make sure that you give the sphere time to reach its "terminal velocity" in the fluid, as this is… [cont.]
Answered by macman - Sat Oct 31 22:18:31 2009
Q. For my A level physics project, I'm measuring the viscosity of honey with a 'falling sphere' method. I'm wondering whether the light gates will register the sphere falling through the honey which will have a colour of its own - has anyone done something similar to this themselves? And could there be any accurate alternatives to using light gates? Thanks.
Asked by Sam - Sat Oct 31 17:21:51 2009 - - 1 Answers - 0 Comments
A. Sam, sounds like a great project. Light gates should work through honey, as long as the light source is strong enough. One problem you will have is alignment - making sure the spheres actually block the path of the light as they move past. I suspect that you can get equally good results simply timing by hand. When timing by hand you will want to make the distance the balls fall through as large as possible (this makes the inaccuracies when timing by hand less important) - see if you can borrow a large graduated cylinder. If you use a small sphere, then it will take a long time to fall, so that also helps. When you make the measurement make sure that you give the sphere time to reach its "terminal velocity" in the fluid, as this is… [cont.]
Answered by macman - Sat Oct 31 22:18:31 2009
What happens when a positively charged rod is held near a metal sphere?
Q. Metal sphere A is initially neutral. A positively charged rod is brought near, but not touching. Is A now positive, negative, or neutral? I would really appreciate anyone who can give me an answer and explain why it works that way!
Asked by headdeskusername - Thu Jan 22 09:50:18 2009 - - 3 Answers - 0 Comments
A. If A is not connected to anything else, A remains neutral overall because of the conservation of charge. But there is separation of charges; negative charges are attracted to the side nearer the rod, and an equal positive charge is repelled to the other side. The ref. describes this phenomenon of charging by induction using a two-sphere system rather than a single sphere. The principle the same for a single sphere.
Answered by kirchwey - Thu Jan 22 10:32:20 2009
Q. Metal sphere A is initially neutral. A positively charged rod is brought near, but not touching. Is A now positive, negative, or neutral? I would really appreciate anyone who can give me an answer and explain why it works that way!
Asked by headdeskusername - Thu Jan 22 09:50:18 2009 - - 3 Answers - 0 Comments
A. If A is not connected to anything else, A remains neutral overall because of the conservation of charge. But there is separation of charges; negative charges are attracted to the side nearer the rod, and an equal positive charge is repelled to the other side. The ref. describes this phenomenon of charging by induction using a two-sphere system rather than a single sphere. The principle the same for a single sphere.
Answered by kirchwey - Thu Jan 22 10:32:20 2009
How do the coordinates of the points inside a sphere change, by increasing the volume of the sphere?
Q. I have a tetrahedral mesh of a sphere, if I increase the volume of the sphere, like by k, what would be the new coordinates of the vertices of tetradera in the sphere? (centroid of the sphere is fixed)
Asked by Jery A - Mon Apr 27 00:30:47 2009 - - 1 Answers - 0 Comments
A. volume of cube = (4/3) pi r ^ 3 Multiplying the volume by factor M would multiply the coordinates on the sphere by M^(1/3)
Answered by Hk - Mon Apr 27 00:39:49 2009
Q. I have a tetrahedral mesh of a sphere, if I increase the volume of the sphere, like by k, what would be the new coordinates of the vertices of tetradera in the sphere? (centroid of the sphere is fixed)
Asked by Jery A - Mon Apr 27 00:30:47 2009 - - 1 Answers - 0 Comments
A. volume of cube = (4/3) pi r ^ 3 Multiplying the volume by factor M would multiply the coordinates on the sphere by M^(1/3)
Answered by Hk - Mon Apr 27 00:39:49 2009
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But the enterprising mathematicians finally found a way to construct the elusive shape by mathematically tinkering with a sphere . ...
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XSportSeeker
Fri, 30 Oct 2009 14:36:39 GM
By design house Seeper, a spherical multitouch display: Very interesting concept! I can't imagine an application for that right now other than in a museum.
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