This i believe is true , brainlest ?
Answer:
A. As the ropes are horizontal the child has travelled 2m of vertical displacement from his lowest position.
Gpe @ A=mgh=40*9.81*2=784.8J
B. At 30degree vertical angle the vertical displacement from lowest position is given by
2-2cos(30)=2-1.73=0.27m
Gpe @B= 40*9.81*0.27=106 J
C: at the bottom of circular arc it's Gpe is zero relative to lowest position as bottom of arc itself is lowest position.
Initial angular velocity = 0 (starting from rest)
Final angular velocity = 30 rad/s
Distance traveled = (20 rev)*(2π rad/rev) = 40π rad
Let the angular acceleration be α rad/s².
Then
(30 rad/s)² = (0 rad/s)² + 2*(α rad/s²)*(40π rad)
α = 3.58 rad/s²
Answer: 3.58 rad/s² (nearest hundredth)
In this case, the movement is uniformly delayed (the final
rapidity is less than the initial rapidity), therefore, the value of the
acceleration will be negative.
1. The following equation is used:
a = (Vf-Vo)/ t
a: acceleration (m/s2)
Vf: final rapidity (m/s)
Vo: initial rapidity (m/s)
t: time (s)
2. Substituting the values in the equation:
a = (5 m/s- 27 m/s)/6.87 s
3. The car's acceleration is:
a= -3.20 m/ s<span>^2</span>