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Masteriza [31]
3 years ago
6

What is the type of relationship between the height (H) and the atmospheric density, and which segment can give the most accurat

e approximations for the interpolated values?
(picture attached below)

Physics
1 answer:
vovikov84 [41]3 years ago
3 0

Answer:

B.  inverse plot, 0.51 kilograms/meter3

Explanation: First of all, we note that the relationship between the altitude and the atmospheric density is an inverse relationship. In fact, an inverse relationship is a relationship between the x-variable and the y-variable of the form

Therefore, as the x increases, the y decreases, and as the x decreases, they increases. This is exactly what occurs with the altitude and the atmospheric density in this plot: as the altitude increases, the density decreases, and vice-versa.

Moreover, we can infer the value of the atmospheric density at an altitude of 1,291 km. This point is located between point A (2550 km) and point B(1000 km), so the density must have a value between 0.30 kg/m^3 and 0.54 kg/m^3, so the correct choice is

B.  inverse plot, 0.51 kilograms/meter3

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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
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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

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(ω, ω0)∫dω/ω² = (t, 0)∫dt × tan θ

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ω = 2.37/(1 - (2.37 × 0.22 × tan 57))

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