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2 years ago

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"The International Space Station (ISS) orbits at a distance of 350 km above the surface of the Earth. (a) Determine the gravitat

ional field strength that astronauts onboard the ISS would experience. (b) Calculate the orbital velocity of the ISS. (c) Determine the period of the ISS orbit around Earth"

1 answer:

vagabundo [1.1K]2 years ago

3 0

Answer:

(a) g = 8.82158145 $m/s^2$ .

(b) 7699.990192m/s.

(c)5484.3301s = 1.5234 hours.(extremely fast).

Explanation:

(a) Strength of gravitational field 'g' by definition is

$g = \frac{M_{(earth)} }{r^2} G$ , here G is Gravitational Constant, and r is distance from center of earth, all the values will remain same except r which will be radius of earth + altitude at which ISS is in orbit.

r = 6721,000 meters, putting this value in above equation gives g = 8.82158145 $m/s^2$ .

(b) We have to essentially calculate centripetal acceleration that equals new 'g'.

$a_{centripetal}=\frac{V^2}{r} =g$ here g is known, r is known and v is unknown.

plugging in r and g in above and solving for unknown gives V = 7699.990192m/s.

(c) S = vT, here T is time period or time required to complete one full revolution.

S = earth's circumfrence , V is calculated in (B) T is unknown.

solving for unknown gives T = 5484.3301s = 1.5234hours.

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To solve this problem it is necessary to apply the rules and concepts related to logarithmic operations.

From the definition of logarithm we know that,

$Log_{10}(10) = 1$

In this way for the given example we have that a logarithm with base 10 expressed in the problem can be represented as,

$log_{10}(1,000,000)$

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2 years ago

A bowling ball has a mass of 5 kg. A student rolls the bowling ball one time at 1 m/s and a second time and 2 m/s. Compare the m

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2 years ago

A ball is bouncing to a height that is 1/3 of the previous height after every bounce. What kind of sequence is this? How could y

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Answer and Explanation:

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For example let the initial height is 243 fit

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After second bounce 81/3=27 feet

After third bounce 27/3 =9 feet

After fourth bounce 9/3 =3 feet

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2 years ago

brainly an andean condor with a wingspan and a mass soars along a horizontal path. model its wings as a rectangle with a width.

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The difference between the pressure at the top surfaces of the condor's wings and the pressure at the bottom surfaces is 101,204 Pa.

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The difference in pressure between the top and bottom of the wingspan is calculated as follows;

ΔP = P(top) - P(bottom)

<h3>Area of the wingspan</h3>

A = bh

A = 2.7 m x 0.27 m

A = 0.729 m²

<h3>Weight of the Andean condor</h3>

W = mg

W = 9 x 9.8

W = 88.2 N

<h3>Pressure at the top surface of condor's wings</h3>

The pressure at the top surface of condor's wings is due to atmospheric pressure

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<h3>Pressure at the bottom surface of condor's wings</h3>

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ΔP = 101,325 Pa - 120.99 Pa

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The complete question is below;

An Andean condor with a wingspan of 270 cm and a mass of 9.00 kg soars along a horizontal path. Model its wings as a rectangle with a width of 27.0 cm.

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1 year ago

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Now, expression for the transfer of heat per unit cell is as follows.

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Putting the given values into the above formula as follows.

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Here, T is 1 second so, power conducted is equal to heat transferred.

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Thus, we can conclude that 2.92 watt power will be conducted through the rod when it reaches steady state.

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2 years ago