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algol [13]
4 years ago
14

A square steel bar has a length of 9.7 ft and a 2.9 in by 2.9 in cross section and is subjected to axial tension. The final leng

th is 9.70710 ft . The final side length is 2.89933 in . What is Poisson's ratio for the material? Express your answer to three significant figures.
Physics
1 answer:
alisha [4.7K]4 years ago
5 0

The Poisson's ratio definition is given as the change in lateral deformation over longitudinal deformation. Mathematically it could be expressed like this,

\upsilon = \frac{\text{Lateral Strain}}{\text{Longitudinal Strain}}

\upsilon = \frac{(\delta a/a)}{(\delta l/l)}

Replacing with our values we would have to,

\upsilon = \frac{(2.89933-2.9/2.9)}{(9.70710-9.7/97)}

\upsilon = 0.1977

Therefore Poisson's ratio is 0.1977.

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A 9.5 V battery supplies a 3.5 mA current to a circuit for 7.0 h. Part A How much charge has been transferred from the negative
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Answer:

88.2 C

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The amount of charge passed is from the negative to the positive terminal

shall be given by:

Q = I * t = 3.5mA * 7h * 3600s/h = 88.2 C

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The flywheel is rotating with an angular velocity ω0 = 2.37 rad/s at time t = 0 when a torque is applied to increase its angular
nika2105 [10]

Answer:

ω = 12.023 rad/s

α = 222.61 rad/s²

Explanation:

We are given;

ω0 = 2.37 rad/s, t = 0 sec

ω =?, t = 0.22 sec

α =?

θ = 57°

From formulas,

Tangential acceleration; a_t = rα

Normal acceleration; a_n = rω²

tan θ = a_t/a_n

Thus; tan θ = rα/rω² = α/ω²

tan θ = α/ω²

α = ω²tan θ

Now, α = dω/dt

So; dω/dt = ω²tan θ

Rearranging, we have;

dω/ω² = dt × tan θ

Integrating both sides, we have;

(ω, ω0)∫dω/ω² = (t, 0)∫dt × tan θ

This gives;

-1[(1/ω_o) - (1/ω)] = t(tan θ)

Thus;

ω = ω_o/(1 - (ω_o × t × tan θ))

While;

α = dω/dt = ((ω_o)²×tan θ)/(1 - (ω_o × t × tan θ))²

Thus, plugging in the relevant values;

ω = 2.37/(1 - (2.37 × 0.22 × tan 57))

ω = 12.023 rad/s

Also;

α = (2.37² × tan 57)/(1 - (2.37 × 0.22 × tan 57))²

α = 8.64926751525/0.03885408979 = 222.61 rad/s²

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