A 0.75 kg rock is projected from the edge of the top of a building with an initial velocity of 11.9 m/s at an angle 59◦ above th
1 answer:
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let the length of the beam be "L"
from the diagram
AD = length of beam = L
AC = CD = AD/2 = L/2
BC = AC - AB = (L/2) - 1.10
BD = AD - AB = L - 1.10
m = mass of beam = 20 kg
m₁ = mass of child on left end = 30 kg
m₂ = mass of child on right end = 40 kg
using equilibrium of torque about B
(m₁ g) (AB) = (mg) (BC) + (m₂ g) (BD)
30 (1.10) = (20) ((L/2) - 1.10) + (40) (L - 1.10)
L = 1.98 m
Answer : The final energy of the system if the initial energy was 2000 J is, 3500 J
Solution :
(1) The equation used is,
where,
= final internal energy
= initial internal energy
q = heat energy
w = work done
(2) The known variables are, q, w and
initial internal energy = = 2000 J
heat energy = q = 1000 J
work done = w = 500 J
(3) Now plug the numbers into the equation, we get
(4) By solving the terms, we get
(5) Therefore, the final energy of the system if the initial energy was 2000 J is, 3500 J