<span>A gymnast with mass m1 = 43 kg is on a balance beam that sits on (but is not attached to) two supports. The beam has a mass m2 = 115 kg and length L = 5 m. Each support is 1/3 of the way from each end. Initially the gymnast stands at the left end of the beam.
1)What is the force the left support exerts on the beam?
2)What is the force the right support exerts on the beam?
3)How much extra mass could the gymnast hold before the beam begins to tip?
Now the gymnast (not holding any additional mass) walks directly above the right support.
4)What is the force the left support exerts on the beam?
5)What is the force the right support exerts on the beam?</span>
Answer:
Option C) 2,090 J/(mol K)
Explanation:
Data:
Volume in the beaker = 429 ml
temperature = 20° C
Density = 789 kg/m³
Equilibrium reading = 429
volume change = 29 ml
= 0.029 L
Energy change = mcΔT
U + PΔV
Answer:
4.32
Explanation:
The centripetal acceleration of any object is given as
A(cr) = v²/r, where
A(c) = the centripetal acceleration
v = the linear acceleration
r = the given radius, 1.9 m
Since we are not given directly the centripetal acceleration, we'd be using the value of acceleration due to gravity, 9.8. This means that
9.8 = v²/1.9
v² = 1.9 * 9.8
v² = 18.62
v = √18.62
v = 4.32 m/s
This means that, the minimum speed the cup must have so as not to get wet or any spill is 4.32 m/s
Answer: He has only move 0.2 yards
Explanation: When you subtract 18.3 from 18.5 you get 0.2 and that is how much he's moved