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marta [7]
3 years ago
10

A wheel of diameter 50.0 cm starts from rest and rotates with a constant angular acceleration of 2.00 rad/s^2. At the instant th

e wheel has computed its second revolution, compute the radial acceleration of a point on the rim in two ways.
(1)Using the relationship .
arad=w2r
(2)from the relationship
arad=v2/r
Mathematics
1 answer:
Kruka [31]3 years ago
8 0
R = 25 cm = 0.25 m
α = 2.00 rad /s²
T(heta) = 4 π
ω = T(heta)/ t
ω = α t
α t = 4 π / t
t ² = 2 π
t = √ ( 2π )  = 2.5 s
ω = α t = 2 · 2.5 = 5 rad/s
v = ω r = 5 · 0.25 = 1.25 m/s
(1)    a (rad) = ω² r = 25 · 0.25 = 6.25 m/s²
(2)    a (rad) = v² / r = ( 1.25 )² / 0.25 = 0.15625 / 0.25 = 6.25 m/s²
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2 years ago
Plz help me, thank you
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Answer:

P=40(1.03526)^{t}

Step-by-step explanation:

<u>Exponential Growth </u>

The natural growth of some magnitudes can be modeled by the equation:

P=P_o(1+r)^{t}

Where P is the actual amount of the magnitude, Po is its initial amount, r is the growth rate and t is the time.

The initial number of bacteria is Po=40 and it doubles (P=2Po) at t=20 min. With that point we can find the value of r:

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Simplifying:

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Solving for 1+r:

1+r=\sqrt[20]{2}

1+r=1.03526

The exponential function that models the situation is:

\mathbf{P=40(1.03526)^{t}}

4 0
2 years ago
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