Answer:
Explanation:
Given
mass of boy=36 kg
length of swing=3.5 m
Let T be the tension in the swing
At top point
where v=velocity needed to complete circular path
Th-resold velocity is given by
So apparent weight of boy will be zero at top when it travels with a velocity of
To get the velocity at bottom conserve energy at Top and bottom
At top
Energy at Bottom
Comparing two as energy is conserved
Apparent weight at bottom is given by
Answer:
x = 2.044 m
Explanation:
given data
initial vertical component of velocity = Vy = 2sin18
initial horizontal component of velocity = Vx = 2cos18
distance from the ground yo = 5m
ground distance y = 0
from equation of motion
solving for t
t = 1.075 sec
for horizontal motion
x = 2cos18*1.075
x = 2.044 m
Answer:
13.309 m/s²
Explanation:
Length from shoulder to hand, l = 30 cm = 0.3 m
initial velocity, u = 1 m/s
final velocity, v = 2.5 m/s
time, t = 3 s
Let the tangential acceleration is a.
by using first equation of motion
v = u + at
2.5 = 1 + 3 a
a = 0.5 m/s²
Let the centripetal acceleration is a'.
a' = v'²/l
a' = 2 x 2 / 0.3
a' = 13.3 m/s²
The tangential acceleration and the centripetal acceleration are both perpendicular to each other. So, the net acceleration is given by
A = 13.309 m/s²