The equilibrium position for a pendulum is straight down. If it moves through that position every second then its period is actually 2 seconds. This is because the period is how long it takes to go from one extreme and back again. It will pass through the equilibrium point twice when doing this. Once on the way down and again on the way back.
Answer:
I know 1, that is in the case of a burning of a candle.
Explanation:
The first shell has 2 in and the second shell has 8 in therefore ypuvwont need to add anymore :-) so the answer is 0
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When you sketch the problem, it would look like that shown in the picture. The velocity the airplane should achieve must be 120 km/h in order for it to lift off the ground and take-off. Before this, it has to build up speed in order to reach the final velocity from rest. In rectilinear motion, one of the useful equations used is
2ax = vf² - vi²
For consistency, let's convert km/h to m/s.
120 km/h * (1000 m/1 km) * (1 h/3600 s) = 33.33 m/s
Substituting the values,
2a(280 m) = (33.33 m/s)² - 0²
a = 1.984 m/s
The minimum acceleration is 1.984 m/s.