The 'period' of a pendulum . . . the time it takes to go back and forth once, and return to where it started . . . is
T = 2π √(length/gravity)
For this pendulum,
T = 2π √(0.24m / 9.8 m/s²)
T = 2π √0.1565 s²
T = 0.983 second
If you pull it to the side and let it go, it hits its highest speed at the BOTTOM of the swing, where all the potential energy you gave it has turned to kinetic energy. That's 1/4 of the way through a full back-and-forth cycle.
For this pendulum, that'll be (0.983s / 4) =
<em>(A). T = 0.246 second</em> <em><===</em>
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Notice that the formula T = 2π √(length/gravity) doesn't say anything about how far the pendulum is swinging. For small angles, it doesn't make any difference how far you pull it before you let it go . . . the period will be the same for tiny swings, little swings, and small swings. It doesn't change if you don't pull it away too far. So . . .
<em>(B).</em> The period is the same whether you pulled it 3.5 or 1.75 . <em>T = 0.246 s.</em>
Answer:
14.715 m
Explanation:
Assume that the acceleration due to gravity is 9.81 m/s^2 downwards, take downwards as positive
First second:
v = u + at
v = 9.81 m/s
Second second:
s = ut + (1/2)at^2
s = 9.81(1) + (1/2)(9.81)(1)^2
s = 14.715 m
Answer:
Explanation:
As we know that
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so the height above window is given as
Answer:
1 and 2
Explanation:
It is given that, force exerted by air is negligible in any way.
Also, it is given that channel is in the shape of a segment of a circle with its center at O.
When it is within the frictionless channel at position 'Q',
A gravity exerts force on the ball in the downward direction. On the other hand, the channel pointing from Q to O also exerts a force on the ball.
However, there is no any force in the direction of motion. On the other hand, he channel pointing from O to Q does not exert a force on the ball.
