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time should you wait between pushes is 2.83 sec.
the question is incomplete, full statement is-
A 24 kg child sits on a 2.0-m-long rope swing. You are going to give the child a small, brief push at regular intervals. If you want to increase the amplitude of her motion as quickly as possible, how much time should you wait between pushes?
<h3>What is Amplitude?</h3>In physics, amplitude refers to the greatest displacement or distance that a point on a vibrating body or wave may move relative to its equilibrium location. It is equivalent to the vibration path's half-length.
regular interval - at similarly spaced intervals: having the same interval of time between occurrences From 4 a.m. to midnight, the buses operate at regular intervals. The boards are positioned at regular intervals, with an equal amount of space between each.
The length of swing, l = 2.1 m
The time between the pushes is nothing but the Time period
and is given by the formula,
= 2 * 3.14 ( 2.0/ 9.8 ) ^ (1/2)
= 2.83 sec
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The current in the circuit is 5 A
Explanation:
The intensity of current is given by the equation:
where
I is the current
q is the amount of charge passing through a given point of the circuit in a time interval of t
For the cell in this problem, we have
q = 150 C is the charge
t = 30 s is the time interval
Substituting into the equation, we f ind
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Answer:
The attached diagram explains the system,
Sum of Fy = 0
N=9.81
N - mgCos60 = 0
F= ukN= (0.53)(9.81) =
F= 5.12 N
So
F.d= 1/2(mv.v) - mgdsin60
-5.12*0.5 = 0.5*v^2 - 2*(9.81)*(0.5*sin60)
(a) v = 2.436 m/s
For deflection
-F.x = 1/2(mv.v) - mgxsin60 + 1/2 (k*x*x)
by solving for with values of v, m, g, F, k
800x^2 - 11.87 x - 5.938 = 0
by solving the quadratic equation
x = 0.093, -0.079
(b) x = 0.093 m
correct Answer is 0.093m
Explanation:
