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11111nata11111 [884]
2 years ago
12

The international Space Station (ISS) orbits the Earth once every 90 mins at an altitude of 409 km. How high would it have to be

in order to be to be in geosynchronous orbit
Physics
1 answer:
Oksi-84 [34.3K]2 years ago
6 0

It would have to be 36,719 Km high in order to be to be in geosynchronous orbit.

To find the answer, we need to know about the third law of Kepler.

<h3>What's the Kepler's third law?</h3>
  • It states that the square of the time period of orbiting planet or satellite is directly proportional to the cube of the radius of the orbit.
  • Mathematically, T²∝a³
<h3>What's the radius of geosynchronous orbit, if the time period and altitude of ISS are 90 minutes and 409 km respectively?</h3>
  • The time period of geosynchronous orbit is 24 hours or 1440 minutes.
  • As the Earth's radius is 6371 Km, so radius of the ISS orbit= 6371km + 409 km = 6780km.
  • If T1 and T2 are time period of geosynchronous orbit and ISS orbit respectively, a1 and a2 are radius of geosynchronous orbit and ISS orbit, as per third law of Kepler, (T1/T2)² = (a1/a2)³
  • a1= (T1/T2)⅔×a2

           = (1440/90)⅔×6780

           = 43,090 km

  • Altitude of geosynchronous orbit = 43,090 - 6371= 36,719 km

Thus, we can conclude that the altitude of geosynchronous orbit is 36,719km.

Learn more about the Kepler's third law here:

brainly.com/question/16705471

#SPJ4

   

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Two electrons exert a force of repulsion of 1.2 N on each other. How far apart are they? The elementary charge is 1.602 × 10−19
aleksandr82 [10.1K]

Answer:

The answer to your question is distance between these electrons

                                                   = 1.386 x 10⁻¹⁴ m

Explanation:

Data

Force = F = 1.2 N

distance = d = ?

charge = q₁ = q₂ = 1.602 x 10⁻¹⁹ C

K = 8.987 x 10⁹ Nm²/C²

Formula

-To solve this problem use the Coulomb's equation

  F = kq₁q₂ / r²

-Solve for r²

  r² = kq₁q₂ / F

-Substitution

  r² = (8.987 x 10⁹)(1.602 x 10⁻¹⁹)(1.602 x 10⁻¹⁹) / 1.2

- Simplification

  r² = 2.306 x 10⁻²⁸ / 1.2

  r² = 1.922 x 10⁻²⁸

-Result

  r = 1.386 x 10⁻¹⁴ m

7 0
4 years ago
Release an electron initially at rest in the presence of an electric field. The electron tends to go to the region of 1. same el
olga nikolaevna [1]

Answer:

The electron tends to go to the region of 4. higher electric potential.

Explanation:

When a charged particle is immersed in an electric field, it experiences a force given by

F=qE

where

q is the charge of the particle

E is the electric field

The direction of the force depends on the sign of the charge. In particular:

- The force and the electric field have the same direction if the charge is positive

- The force and the electric field have opposite directions if the charge is negative

Therefore, an electron (negative charge) moves in the direction opposite to the electric field lines.

However, electric field lines go from points at higher potential to points at lower potential: so, electrons move from regions at lower potential to regions of higher potential.

Therefore, the correct answer is

The electron tends to go to the region of 4. higher electric potential.

4 0
4 years ago
What does it mean when work is positive?
Degger [83]

Answer:

To have a positive job, the two vectors must have the same direction

Explanation:

Work is a scalar defined as the scalar product of two vectors, the froce and the displacement. To have a positive job, the two vectors must have  

collinear and that their arrows point in the same direction.

You can also appreciate this from the work equation

       W = F. r

bold indicate vector

      W = F  r cos θ

 

With θ the angle between force and displacement

7 0
3 years ago
The maximum displacement in an oscillatory motion is A = 0.49 m. Determine the position x at which the kinetic energy of the par
amid [387]

Answer:

x = 0.40 m

Explanation:

  • When the displacement is maximum, the particle is momentarily at rest, which means that at this point (assuming no friction present) all the mechanical energy is elastic potential, which can be written as follows:

      E_{tot} = U_{o} = \frac{1}{2} *k*A^{2}  (1)

  • Since in absence of friction, total mechanical energy must keep constant, this means that at any time, the sum of the kinetic and potential energy, must be equal to (1), as follows:

       E_{tot} = U_{o} = \frac{1}{2} *k*A^{2}  = (KE)_{f} + U_{f}  (2)

  • If KEf = U/2f, replacing in (2), we get:

      E_{tot} = U_{o} = \frac{1}{2} *k*A^{2}  = (U/2)_{f} + U_{f} =  \frac{3}{2} *U_{f}  (3)

  • At any point, the elastic potential energy is given by the following expression:

       U_{f} = \frac{1}{2} *k*x^{2}   (4)

      where k= spring constant (N/m) and x is the displacement from the

      equilibrium position.

  • Replacing (4) in (3), simplifying and rearranging, we get:

       E_{tot} = U_{o} = \frac{1}{2} *A^{2}  =  \frac{3}{4} *x^{2}   (5)

  • Finally, solving for x, we get:

        x = \sqrt{\frac{2}{3} } * A =  \sqrt{\frac{2}{3} } * 0.49m = 0.40 m  (6)

8 0
3 years ago
Henrietta is going off to her physics class, jogging down the sidewalk at a speed of 4.15 m/sm/s. Her husband Bruce suddenly rea
Darya [45]

Answer:

Explanation:

Distance travelled by Henrietta in 5.5 s = 4.15 x 5.5 = 22.825 m .

Time taken by lunch of bagels to fall vertically by 55.2 m . Let it be t .

s = ut + 1/2 g t²

55.2 = 0 + .5 x 9.8 x t²

t² = 11.26

t = 3.356 s

By the time the lunch of bagels touches the hand of Henrietta , she would have travelled further by distance

s = 3.356 x 4.15 = 13.9 m

She is now at distance of 22.825 + 13.9 = 36.725 m from window .

So lunch of bagels must travel a horizontal distance of 36.725 m in 3.356 s which the time of fall of bagel .

Speed of bagel = distance / time

= 36.725 / 3.356

= 10.94 m /s

b )

Henrietta is 36.725 m from window at the time when she catches the bangel.

6 0
3 years ago
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