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Block Plane
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Blocking A Volleyball Terms To Know, Part 4
In part four of the volleyball blocking study guide I talk about the terms the double block and using the block.
Double block - A double block is formed when two players manage to travel along the net to get in front of and form a block against a hitter in a rally. With two players blocking against a hitter, the hitter has less options for open spaces on the court to hit to and is forced to hit to the area of the court the double block is not covering.
The hitter may choose to hit down the line if the double block is covering their cross court, or they will hit hard cross court if the double block is taking away the line. Other options include the dink or the "tip", where the hitter at the height of their jump will tip the ball softly over the block so the ball lands in an open area of the court, out of the block's reach and away from the defense.
The hitter may also choose to aim for the "seam" of the block which is the space that's created between the arms of the two opposing blockers, if for some reason the blockers don't "close the block."
The hitter can aim for a hole in the block or "use the block" which means a hitter can hit towards the hole that a blocker has created by not keeping their hands close enough to each other or close enough to their blocking partner's hands.
Using the block - When a hitter decides to "use the block' or "use the blockers hands" or "wipe the block" - this means they will wipe the outside hands of the blocker with the ball in order to make a point. This is a practiced skill that is very effective for smaller outside hitters who face big blockers and need a hitting option they can rely on when they can't hit over or hit the ball past the block. "Wiping the block" technique is based on a spiker learning how to spike in a way that they aim the ball for the outside hand of the blocker who's closest to the antenna, so when the ball is deflected off of the outside hand of the blocker it bounces outside the court.
This skill is very effective for an outside hitter to learn because it's very hard to defend a ball that's already on it's way outside of the court's boundary lines.
To avoid "being used" an outside blocker should train to go up to block in a way that their outside hand, the one closest to the antenna, is turned into the court, this way while they block and their hands penetrate the plane of the net, that outside hand is positioned to push the ball back into the opposing team's court.
About the Author
And now I invite you to get Free video instruction from Olympians and pro players on blocking a volleyball from April Chapple, creator of the volleyball news hub Volleyball Voices.com.
A block slides down an inclined plane, starting from rest and being pushed with a constant acceleration?
A block slides down an inclined plane, starting from rest and being pushed with a constant acceleration of 4.55 m/s2 over a distance of 19.4 cm. It then decelerates at a constant rate of 1.06 m/s2 (because of friction), until it again comes to rest. Find the total time the block is in motion.
_______ s
Please let me know how you solved this, Thanks!
They threw in a few extra items just to confuse you. Does it really matter that the block slides down an inclined plane? No. Let's take it one step at a time. Begin with the block being accelerated at the rate of 4.55m/s^2 over a distance of 19.4cm (or 0.194m). How long did it take for that to take place? They were nice enough to give us the acceleration, "a", and the distance that the block travelled. Just plug into the equation to find the time.
s = (1/2)(a)(t^2)
0.194m = 0.5(4.55m/s^2)(t^2)
0.292s = t
That's part of the total time that we're looking for, and it will allow us to figure out how fast the block is moving at this point from the equation
v = at
We know that a=4.55m/s^2 and that t=.292sec, so
v = 4.55m/s^2(0.292s) = 1.33m/s (rounded value)
Now we're almost there. If it's moving at 1.33m/s and decelerating at the rate of 1.06m/s^2, how long does it take to come to a stop? We can use the same equation that we just used above
v = at
1.33m/s = (1.06m/s)t
1.25s = t (rounded again)
The total time the block is in motion is the sum of the acceleration time, 0.292s, and the deceleration time, 1.25s. Your final answer will depend upon how you round the intermediate results. We could have taken a number of shortcuts, but I hope that taking the long way will make the method more clear to you.
