uh..8 seconds?
To find the rate at which Jon must let out the string, you can use the Pythagorean theorem to relate the horizontal and vertical distances. The kite, string, and ground form a right triangle. At this point, the string length is 500 feet, and the vertical height from Jon to the kite remains at 300 feet. We can then compute the rate at which Jon should let out the string.
The A string is the second string and the second thickest string.
The A string is the second string and the second thickest string.
A hockey puck of mass m = 0.25 kg is tied to a string and is rotating horizontally in a circle of radius R = 1.0 m on top of a frictionless table.
F sharp
first string are the better players they start, second string backs them up.
Yes
second thickest if its a 4 string bass or 6 string guitar
on a normal guitar the second string but on a 7 string it has two.. just like the 6 strings its the second but it also has a low B for the top string.
The Second String - 1915 is rated/received certificates of: UK:U
first finger on the A string, or second finger on the G string.
You've specified that the two forces act horizontally, but you've said nothing about how they're arranged relative to the string. -- Are the forces pulling in opposite directions on opposite ends of the string ? -- Are the forces both acting at the same end of the string, with the other end tied to a wall ? -- Are the forces in the same plane ? If so, is the string also in that plane ? I can give you configurations in which the forces are horizontal, and the tension in the string is anything you want between zero and 2,020 N. In fact, if both ends of the string are fastened to walls, and the horizontal forces push on the center of the string and perpendicular to it, then you can create any tension you want in the string, up to the breaking limit of the walls.
43.3