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goldenfox [79]
3 years ago
10

What is the altitude of the Sun at noon on December 22, as seen from a place on the Tropic of Cancer?

Physics
1 answer:
scZoUnD [109]3 years ago
4 0

Answer:

108.217 °

Explanation:

Day of year = 356 = d (Considering year of 365 days)

Latitude of Tropic of Cancer = 23.5 °N

Declination angle

δ = 23.45×sin[(360/365)(d+284)]

⇒δ = 23.45×sin[(360/365)(356+284)]

⇒δ = 5.2832 °

Altitude angle at solar noon

90+Latitude-Declination angle

= 90+23.5-5.2832

= 108.217 °

∴ Altitude angle of the Sun as seen from the tropic of cancer on December 22 is 108.217 °

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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

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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