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Tabletop Hockey


Stiga

Hockey


Stiga NHL Stanley Cup Hockey Table Game (Detroit Red Wings / Toronto Maple Leafs)
(Sports) Stiga
Release date: 2010-11-19

1.65-inch legs; measures 37 x 3 x 19.75 inches (W x H x D)
High-quality table hockey game with official NHL team colors and logos
Expertly crafted 3D players outfitted in Maple Leaf and Red Wing uniforms


Price: $112.95 $84.89

Answers

Where can I find the Excalibur 612 Tabletop Air Hockey table?

I tried ebay, Amazon, and like all other big name online stores. Can anyone help or find another tabletop air hockey for the same price?


http://toys.findandcompare.com/4_2438/ai r-hockey-tables.html

Table top hockey


Stiga table top hockey. The puck is missing but a pool ball makes a good substitute. Actually I was just experimenting with iMovie. Shot by a ...

Do you own a tabletop hockey game? What brand and from what year?


David E - could you score with your goalie, too? LOL


I had a table top game that was Flyers vs Islanders, so it must have been about 1981 or 1982. The stupid support for the overhead dome got warped though, so when you would have a face off the puck would practically land against the boards. I spent so much time just taking target practice that no one would play me because I could literally score with any player from anywhere on the ice.

In the early 90s my cat managed to lose all three of my pucks and I haven't played it since.

*edit*
Actually you could. He moved side to side but also in and out of the crease by about an inch. If you got the puck in the middle of the goalie and gave it a sharp push, you could send it the length of the ice. Obviously, it wasn't as an accurate a shot as the players with the sticks.

SureShot Hockey
Poof Slinky

Price: $27.99 $17.00

Ages 5 and Up
Enjoy the thrill of hockey in your own home!
Fast Action Fun!

A flat puck (mass M) is rotated in a cicle on a frictionless air-hockey tabletop, and is held in this orbit?

by a light cord connected to a dangling block (mass m) through a central hole. Show that the speed of the pick is given by:

v= sq. rt. of (mgR/M)

PRETTY TOUGH
PLEASE SHOW WORK. THANK YOU


tension in the string = mg
v^2=mgR/M so mg=Mv^2 / R

if Mv^2/R corresponds to your formula for centripetal force required to constrain the puck in this orbit then , providing that R is the length of string sticking out of the hole onto the table top you have solved it.

Stiga High Speed Table Top Hockey Game
Stiga USA

Price: $108.04 $109.95

Dimensions: 37in L x 19.75in W x 3.25in H
Detailed International Hockey Men - Teams: Sweden and Finland
Playing Surface Features Ice Sheet Technology for Faster Game Play!

Physics question involving moving around raw formulas?

A flat puck (mass M) is rotated in a circle on a frictionless air-hockey tabletop, and is held in this orbit by a light cord connected to a dangling block (mass m) through a central hole. Show that the speed of the puck is given by by v=sqrt(mgR/M).
Can someone please tell me how to get his? It's using the formulas for Uniform Circular Motion.


a = v²/R
fc = M*a = Mv²/R

fm = m*g

For equilibrium, fc = fm or

Mv²/R = m*g → v² = mgR/M

I need some help with this physics question?

A flat puck (mass M) is rotated in a circle on a frictionless air-hockey tabletop, and is held in this orbit by a light cord connected to a dangling block (mass m) throught a central hole. Show that the speed of the puck is given by:
v (velocity) = squareroot (mgR/M)

I'm not quite sure what the question is asking for. Thank you for your help.


The dangling mass will not move and the puck will keep rotating in its orbit, . This implies that the centripetal force that keeps the puck in a circular orbit is just the weight of the dangling mass, this is

Fc = mg

Now, centripetal force on the puck is equal to the mass of the puck times the centripetal acceleration, or

Fc = M ac

centripetal acceleration is related to the tangential velocity of the puck as

ac = V^2 / R

where R is the radius of the orbit. Then we have

Fc = M ac =M omega^2 R = M V^2 / R

and, thus

M V ^2 / R = mg

and, solving for V, we get

V = SQRT ( m g R / M )

QED.


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