Determine the angular velocity of the merry-go-round if a jumps off horizontally in the −n direction with a speed of 2 m/s , mea
1 answer:
by angular momentum conservation we will have
angular momentum of child + angular momentum of merry go round = 0
angular momentum of child = mvR
m = mass of child
R = radius of child
v = speed = 2 m/s
now let's say moment of inertia of merry go round is I
so we will have
so merry go round will turn in opposite direction with above speed
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Answer: see the graph attached (straight line, passing through the origin and positive slope).
Justification:
1) Kinetic energy and temperature are in direct proportion. That means:
i) Being kinetic energy y and temperature x: y α x
ii) That implies: y = kx,where k is the constant of proportionality.
iii) The graph is a line that passes through the origin and has positive slope k (k = y / x).
2) The proportional relationship between kinetic energy (KE) and temperature (T) is shown by the Boltzman law, which states:
Average KE = [3 / 2] KT, where K is Boltzman's constant, whose graph is of the form shown in the figure attached.
Justification:
1) Kinetic energy and temperature are in direct proportion. That means:
i) Being kinetic energy y and temperature x: y α x
ii) That implies: y = kx,where k is the constant of proportionality.
iii) The graph is a line that passes through the origin and has positive slope k (k = y / x).
2) The proportional relationship between kinetic energy (KE) and temperature (T) is shown by the Boltzman law, which states:
Average KE = [3 / 2] KT, where K is Boltzman's constant, whose graph is of the form shown in the figure attached.