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2 years ago

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

Engineering

1 answer:

astraxan [27]2 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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D: Neither Technician A nor Technician B are correct.

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2 years ago

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Over [174]

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3 years ago

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2 years ago

A ceramic matrix composite contains internal flaws as large as 0.001 cm in length. The plane strain fracture toughness of the co

murzikaleks [220]

Since the applied stress required for failure due to crack propagation is still higher than 550 MPa, the ceramic is expected to fail due to overload and not because of the flaws

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$k_{IC}$ = $f$ б $\sqrt{\pi a}$

F=geometry factor of the flaw

б=Stress applied

$k_{IC}$ =Fracture toughness

a=Flaw size

<u>Given that </u>

Internal Flaw,a=0.001cm

Fracture Toughness $k_{IC}$ =45MPa $\sqrt{m}$

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An internal Flaw i s 0.001 cm

2a=0.001cm

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3 years ago

I have a molten Ni-Cu alloy with 60 wt%Ni. (ii) I cool it down to a temperature where I have both solid and liquid phases. At th

SIZIF [17.4K]

Answer:

The percentage of the remaining alloy would become solid is 20%

Explanation:

Melting point of Cu = 1085°C

Melting point of Ni = 1455°C

At 1200°C, there is a 30% liquid and 70% solid, the weight percentage of Ni in alloy is the same that percentage of solid, then, that weight percentage is 70%.

The Ni-Cu alloy with 60% Ni and 40% Cu, and if we have the temperature of alloy > temperature of Ni > temperature of Cu, we have the follow:

60% Ni (liquid) and 40% Cu (liquid) at temperature of alloy

At solid phase with a temperature of alloy and 50% solid Cu and 50% liquid Ni, we have the follow:

40% Cu + 10% Ni in liquid phase and 50% of Ni is in solid phase.

The percentage of remaining alloy in solid is equal to

Solid = (10/50) * 100 = 20%

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3 years ago

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

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