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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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A 132mm diameter solid circular section

Ganezh [65]

Answer:

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5 0

3 years ago

The dam cross section is an equilateral triangle, with a side length, L, of 50 m. Its width into the paper, b, is 100 m. The dam

lisabon 2012 [21]

Answer:

Explanation:

In an equilateral trinagle the center of mass is at 1/3 of the height and horizontally centered.

We can consider that the weigth applies a torque of T = W*b/2 on the right corner, being W the weight and b the base of the triangle.

The weigth depends on the size and specific gravity.

W = 1/2 * b * h * L * SG

Then

Teq = 1/2 * b * h * L * SG * b / 2

Teq = 1/4 * b^2 * h * L * SG

The water would apply a torque of elements of pressure integrated over the area and multiplied by the height at which they are apllied:

$T1 = \int\limits^h_0 {p(y) * sin(30) * L * (h-y)} \, dy$

The term sin(30) is because of the slope of the wall

The pressure of water is:

p(y) = SGw * (h - y)

Then:

$T1 = \int\limits^h_0 {SGw * (h-y) * sin(30) * L * (h-y)} \, dy$

$T1 = \int\limits^h_0 {SGw * sin(30) * L * (h-y)^2} \, dy$

$T1 = SGw * sin(30) * L * \int\limits^h_0 {(h-y)^2} \, dy$

$T1 = SGw * sin(30) * L * \int\limits^h_0 {h^2 - 2*h*y + y^2} \, dy$

T1 = SGw * sin(30) * L * (h^2*y - h*y^2 + 1/3*y^3)(evaluated between 0 and h)

T1 = SGw * sin(30) * L * (h^2*h - h*h^2 + 1/3*h^3)

T1 = SGw * sin(30) * L * (h^3 - h^3 + 1/3*h^3)

T1 = 1/3 * SGw * sin(30) * L * h^3

To remain stable the equilibrant torque (Teq) must be of larger magnitude than the water pressure torque (T1)

1/4 * b^2 * h * L * SG > 1/3 * SGw * sin(30) * L * h^3

In an equilateral triangle h = b * cos(30)

1/4 * b^3 * cos(30) * L * SG > 1/3 * SGw * sin(30) * L * b^3 * (cos(30))^3

SG > SGw * 4/3* sin(30) * (cos(30))^2

SG > 1/2 * SGw

For the dam to hold, it should have a specific gravity of at leas half the specific gravity of water.

This is avergae specific gravity, including holes.

6 0

2 years ago

The correct area in sq. Inches and sq. Feet is: Select one: a. 966.76 sq. Inches and 8.056 sq. Feet b. 96.676 sq. Inches and 8.0

kogti [31]

Answer:

c. 96.676 sq. Inches and 0.671 sq. Feet

Explanation:

From the list of the given option, we are told to chose the correct area in sq. inches that correspond to sq. Feet.

If we recall from the knowledge of our conversion table that,

1 sq feet = 144 sq inches

Then, let's confirm if the option were true.

a. 966.76 sq. Inches and 8.056 sq. Feet

Assuming

if 1 sq feet = 8.056

in sq inches, we have ( 8.056 × 144 ) sq inches

= 1160.064 sq. inches

So, 1160.064 sq. inches is equal to 8.056 sq. Feet. Then option 1 is wrong

b. 96.676 sq. Inches and 8.056 sq. Feet

if 1 sq feet = 8.056

in sq inches, we have ( 8.056 × 144 ) sq inches

= 1160.064 sq. inches

So, 1160.064 sq. inches is equal to 8.056 sq. Feet. Then option 2 is wrong/

c. 96.676 sq. Inches and 0.671 sq. Feet

if 1 sq feet = 0.671

in sq inches, we have ( 0.671 × 144 ) sq inches

= 96.624 sq. Inches which is closely equal to 96.676 sq. Inches

Therefore, this is the correct answer as it proves that 96.676 sq. Inches = 0.671 sq. Feet

8 0

3 years ago

1.Which thematic group uses technology to direct the behavior of dynamical systems, ensuring that they behave in a predictable m

dsp73

The thematic group governing the behavior of dynamic system is control system, and the one packaging the miniature components to conduct electricity is electronic system. Thus, option E is correct.

<h3>What are thematic groups?</h3>

The thematic groups comprises of group of people that work over the same idea or the concept. The thematic groups working on different projects and ideas comprised to work with different systems.

The control of dynamic system in order to produce the desirable outcome has been the role of the control system.

The packaging of the electronic miniatures in system that performs the role of direction of electricity is the role of electronic system thematic group. Thus, option E is correct.

Learn more about thematic group, here:

brainly.com/question/32731

#SPJ1

8 0

1 year ago

What scale model proves the initial concept?

Tju [1.3M]

Answer: A prototype

Explanation:

5 0

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