Answer
given,
Weight of the child = 110 N
length of the swing,L = 2 m
now, calculating the potential energy when the string is horizontal
Potential energy = m g h
now, h = L (1 - cos θ) where θ is the angle made by the string with the vertical.
PE = m g L (1 - cos θ)
when rope is horizontal θ = 90°
PE = 110 x 2 (1 - cos 90°)
PE = 220 J
now, calculating potential energy when string made 25° with horizontal
PE = m g L (1 - cos θ)
when rope is horizontal θ = 25°
PE = 110 x 2 (1 - cos 25°)
PE = 20.61 J
Answer:
the tension in each side of the cable is 3677.57 N
Explanation:
given data
traffic light = 20 kg
cable between two poles = 30 m
sag in the cable = 0.40 m
solution
by the free body diagram
tan θ = 
.............1
θ = 1.527 °
and
tension = mg
The net force is along x - axis is express as
T2 cosθ = T1 cosθ .................2
so T2 - T1 ..............3
and
when we take it along y axis that is express as
( T1 + T2) sinθ = mg ...................4
so by equation 3 we put here
2 × T1 sin(1.527) = 20 × 9.8
T1 = 3677.57 N
Magnitude of acceleration = (change of speed) / (time for the change) =
(12 m/s - 0) / (3 sec) =
12/3 = <em>4 m/s²</em>
What's a challenge question ? Have we all passed the event horizon
and been spaghettified without knowing it ?
Acceleration=9.81m/s^2
initial velocity=0m/s
time=.28s
We have to find final velocity.
The equation we use is
Final velocity=initial velocity+acceleration x time
Vf=0m/s+(9.81m/s^2)(.28s)
Vf=2.7468m/s
We would round this to:
Vf (final velocity)=2.7m/s
The answer is A sorry i would of expalined why but i have to go :c
