The book's potential energy will be converted entirely to kinetic energy at that point in the fall, since we can neglect losses due to air resistance. The potential energy is given by PE=mgh, where m=1.2kg, h=0.9m, and g=9.81m/s^2. This is the acceleration due to gravity.
This gives:
PE=1.2*0.9*9.81=10.59J
This is also the kinetic (mechanical) energy right before the landing.
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
use the formula: v^2=(3kT)/m
Where:
<em>v is the velocity of a molecule</em>
<em>k is the Boltzmann constant (1.38064852e-23 J/K)</em>
<em>T is the temperature of the molecule in the air</em>
<em>m is the mass of the molecule</em>
For an H2 molecule at 20.0°C (293 K):
v^2 = 3 × 1.38e-23 J/K × 293 K / (2.00 u × 1.66e-27 kg/u)
v^2 = 3.65e+6 m^2/s^2
v = 1.91e+3 m/s
For an O2 molecule at same temp.:
v^2 = 3 × 1.38e-23 J/K × 293 K / (32.00 u × 1.66e-27 kg/u)
v^2 = 2.28e+5 m^2/s^2
v = 478 m/s
Therefore, the ratio of H2:O2 velocities is:
1.91e+3 / 478 = 4.00
