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8_murik_8 [283]
3 years ago
13

Calculate the radius of a circular orbit for which the period is 1 day​

Engineering
1 answer:
astraxan [27]3 years ago
8 0

Answer:

(T²/D³)sys1 = (T²/D³)sys2

sys1 = earth-moon

sys2 = earth-sat

(27.33day)²/(3.8e8m)³ = (1day)²/D³

D = cbrt(7.3e22m³) = 4.2e7 m

Explanation:

exchange in capacity whilst a satellite tv for pc differences altitude How plenty paintings could be executed to flow the satellite tv for pc into yet another around orbit it is bigger above the outdoors of the Earth? satellite tv for pc exchange in capacity with top Assuming the satellite tv for pc is to be boosted to a clean top r? Gravitational capacity capacity (to center of earth) new orbit(2) has a greater robust PE than old one(a million), so exchange is helpful PE = G m?m?/r earth GM = 3.98e14 ?PE = (GM)(m)(a million/r? – a million/r?) KE additionally differences. Get speed at each and every top. New orbit(2) has decrease speed, so exchange is damaging v = ?(GM/R) V? = ?(GM/r?) V? = ?(GM/r?) ?KE = –½m(V?² – V?²) ?KE = –½mGM(a million/r? – a million/r?) including the two ?E = (GM)(m)(a million/r? – a million/r?)– ½mGM(a million/r? – a million/r?) ?E = ½mGM(a million/r? – a million/r?) ?E = ½(3.98e14)(7500) [a million/(0.5e7) –a million/(3.3e7) ] ?E = ½(3.98e14)(7500)(1e-7) [a million/(0.5) –a million/(3.3) ] ?E = ½(3.98e7)(7500) [2 – 0.303 ] ?E = ½(3.98e7)(7500)(a million.70) ?E = 2.04e11 Joules edit, corrected .

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For laminar flow over a flat plate, the local heat transfer coefficient hx is known to vary as x−1/2, where x is the distance fr
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h_x=cx^{-1/2}

We integrate between 0 to x for obtain the value of the coefficient, so\bar{h}_x =\frac{1}{x} \int\limit^x_0 h_x dx\\\bar{h}_x = \frac{c}{x} \int\limit^x_0 \frac{1}{\sqrt{x}}dx\\\bar{h}_x = \frac{c}{c} (2x^{1/2})\\\bar{h}_x = 2cx^{-1/2}

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A hole of diameter D = 0.25 m is drilled through the center of a solid block of square cross section with w = 1 m on a side. The
attashe74 [19]

Answer:

q=4.013\:\:kW\\\\T_1=278.91\:\:^{\circ}C\\\\T_2=275.82\:\:^{\circ}C

Explanation:

R_{conv,1}=(h_1\pi D_1L)^{-1}=(50*0.25*2)^{-1}=0.01273\:\:K/W\\\\R_{conv,2}=(h^2*4wL)^{-1}=(4*4*1)^{-1}=0.0625\:\:K/W\\\\R_{cond(2D)}=(Sk)^{-1}=(8.59*150)^{-1}=0.00078\:\:K/W

So, heat rate can be calculated as follows:

q=\frac{T_{\infty,1}-T_{\infty,2}}{R_{conv,1}+R_{conv,2}+R_{cond(2D)}} =\frac{330-25}{0.076} =4.013\:\:kW

Surface temperatures can be calculated as follows:

T_1=T_{\infty,1}-qR_{conv,1}=330-51.09=278.91\:\:^{\circ}C\\\\T_2=T_{\infty,2}+qR_{conv,2}=25+250.82=275.82\:\:^{\circ}C

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4 years ago
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