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Nataly_w [17]
3 years ago
14

Which statement describes absolute and apparent brightness?

Physics
2 answers:
maria [59]3 years ago
8 0

♡ hello there ヾ(°∇° ) ♡

   Your answer is...

       ★ ... <u>D</u> !!! ★

×║hope this helps,║×

➶ have a nice day! ➷

       ↬ ɢᴏᴛʜɪx ♡

xeze [42]3 years ago
7 0

The answer is: Absolute brightness is the actual amount of light produced by the star, whereas apparent brightness changes with distance from the observer. Hope this helps, and have a great day

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What is the pressure drop due to the bernoulli effect as water goes into a 3.00-cm-diameter nozzle from a 9.00-cm-diameter fire
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Answer:

\Delta P=1581357.92\ Pa

Explanation:

Given:

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  • volume flow rate, \dot{V}=40\ L.s^{-1}=0.04\ m^3.s^{-1}

<u>Now, flow velocity in hose:</u>

v_h=\frac{\dot V}{\pi.D^2\div 4}

v_h=\frac{0.04\times 4}{\pi\times 0.09^2}

v_h=6.2876\ m.s^{-1}

<u>Now, flow velocity in nozzle:</u>

v_n=\frac{\dot V}{\pi.d^2\div 4}

v_n=\frac{0.04\times 4}{\pi\times 0.03^2}

v_n=56.5884\ m.s^{-1}

We know the Bernoulli's equation:

\frac{P_1}{\rho.g}+\frac{v_1^2}{2g}+Z_1=\frac{P_2}{\rho.g}+\frac{v_2^2}{2g}+Z_2

when the two points are at same height then the eq. becomes

\frac{P_1}{\rho.g}+\frac{v_1^2}{2g}=\frac{P_2}{\rho.g}+\frac{v_2^2}{2g}

\Delta P=\frac{\rho(v_n^2-v_h^2)}{2}

\Delta P=\frac{1000(56.5884^2-6.2876^2)}{2}

\Delta P=1581357.92\ Pa

8 0
3 years ago
An object attached to an ideal spring oscillates with an angular frequency of 2.81 rad/s. the object has a maximum displacement
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x = A cos(ωt + Ф)

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x = A = A cos(ωt + Ф) = A cos(Ф)
Ф = 0

at t = 1.42, with Ф = 0:
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U = 1/2 k x² = 1/2 k [A cos(ωt)]²
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