Explanation:
answer is attached, kindly see answer
<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:
21.3 V, 1.2 A
Explanation:
1.
These resistors are in series, so the net resistance is:
R = R₁ + R₂ + R₃
R = 20 + 30 + 45
R = 95
So the current is:
V = IR
45 = I (95)
I = 9/19
So the voltage drop across R₃ is:
V = IR
V = (9/19) (45)
V ≈ 21.3 V
2.
First, we need to find the equivalent resistance of R₂ and R₃, which are in parallel:
1/R₂₃ = 1/R₂ + 1/R₃
1/R₂₃ = 1/10 + 1/10
R₂₃ = 5
Now we find the overall resistance by adding the resistors in series:
R = R₁ + R₂₃ + R₄
R = 10 + 5 + 10
R = 25
So the current through R₁ is:
V = IR
30 = I (25)
I = 1.2 A
answer:
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