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kupik [55]
3 years ago
8

The space shuttle usually orbited Earth at altitudes of around 300.0 km. 1) Determine the time for one orbit of the shuttle abou

t Earth. (Express your answer to three significant figures.) h You currently have 0 submissions for this question. Only 10 submission are allowed. You can make 10 more submissions for this question. 2) How many sunrises per day did the astronauts witness
Physics
1 answer:
777dan777 [17]3 years ago
3 0

Answer:

Explanation:

distance of shuttle from centre of the earth = radius of the orbit

= 6300 + 300 = 6600 km

= 6600 x 10³

Formula of time period of the satellite

T = 2π R /v₀ , v₀ is orbital velocity

v₀ = √gR , ( if height  is small with respect to radius )

T = 2π R /√gR

= 2π√ R /√g

= 2 x 3.14 x √ 6600 x 10³ / √9.8

= 2 x 3.14 x 256.9 x 10 / 3.13

= 5154.41 s

= 5154.41 / 60 minutes

= 85.91 m

85.9 minutes.

2 ) No of sunrise per day = no of rotation per day

= 24 x 60 / 85.9

= 16.76

or 17 sunrises.

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The emf induced in a coil that is rotating in a magnetic field will be at a maximum at which moment?
adelina 88 [10]
TLDR: It will reach a maximum when the angle between the area vector and the magnetic field vector are perpendicular to one another.

This is an example that requires you to investigate the properties that occur in electric generators; for example, hydroelectric dams produce electricity by forcing a coil to rotate in the presence of a magnetic field, generating a current.

To solve this, we need to understand the principles of electromotive forces and Lenz’ Law; changing the magnetic field conditions around anything with this potential causes an induced current in the wire that resists this change. This principle is known as Lenz’ Law, and can be described using equations that are specific to certain situations. For this, we need the two that are useful here:

e = -N•dI/dt; dI = ABcos(theta)

where “e” describes the electromotive force, “N” describes the number of loops in the coil, “dI” describes the change in magnetic flux, “dt” describes the change in time, “A” describes the area vector of the coil (this points perpendicular to the loops, intersecting it in open space), “B” describes the magnetic field vector, and theta describes the angle between the area and mag vectors.

Because the number of loops remains constant and the speed of the coils rotation isn’t up for us to decide, the only thing that can increase or decrease the emf is the change in magnetic flux, represented by ABcos(theta). The magnetic field and the size of the loop are also constant, so all we can control is the angle between the two. To generate the largest emf, we need cos(theta) to be as large as possible. To do this, we can search a graph of cos(theta) for the highest point. This occurs when theta equals 90 degrees, or a right angle. Therefore, the electromotive potential will reach a maximum when the angle between the area vector and the magnetic field vector are perpendicular to one another.

Hope this helps!
6 0
3 years ago
IMPORTANTT!!! SOLVEE!!! and EXPLAIN!!!
Pepsi [2]
My answer -

I believe that the answer is (A).

P.S

Happy to help you have an AWESOME!! day
5 0
3 years ago
Which characteristic should a good scientific question have?
Brilliant_brown [7]
The best answer to this question is D. a god scientific question should be based on observations instead of just imagination.

Hope this helps.
7 0
3 years ago
Read 2 more answers
A +12 μC charge and -8 μC charge are 4 cm apart. Find the magnitude and direction of the E-field at the point midway between t
Natasha_Volkova [10]

Answer:

Explanation:

Given

Charge of first Particle q_1=+12\ \mu C

Charge of second Particle q_2=-8\ \mu C

distance between them d=4\ cm

k=9\times 10^{9}

magnetic field due to first charge at mid-way between two charged particles is

E_1=\frac{kq_1}{r^2}

r=\frac{d}{2}=\frac{4}{2}=2\ cm

E_1=\frac{9\times 10^9\times 12\times 10^{-6}}{(2\times 10^{-2})^2}

E_1=27\times 10^7\ N/C (away from it)

Electric field due to q_2=-8\ \mu C

E_2=\frac{kq_2}{r^2}

E_2=-\frac{9\times 10^9\times 8\times 10^{-6}}{(2\times 10^{-2})^2}

E_2=-18\times 10^7\ N/C(towards it)

E_{net}=E_1+E_2

E_{net}=9\times 10^7\ N/C(away from first charge)        

8 0
3 years ago
A scuba diver is sitting on a boat while waiting to go on a dive and sees light reflected from the water's surface. At what angl
LUCKY_DIMON [66]

Answer:

θ_p = 53.0º

Explanation:

For reflection polarization occurs when a beam is reflected at the interface between two means, the polarization in total when the angle between the reflected and the transmitted beam is 90º

Let's write the transmission equation

     n1 sin θ₁ = ne sin θ₂

The angle to normal (vertcal) is

    180 = θ2 + 90 + θ_p

    θ₂ = 90 - θ_p

Where θ₂ is the angle of the transmitted ray θ_p is the angle of the reflected polarized ray

We replace

     n1 sin θ_p = n2 sin (90 - θ_p)

Let's use the trigonometry relationship

    Sin (90- θ_p) = sin 90 cos θ_p - cos 90 sin θ_p = cos θ_p

In the law of reflection  incident angle equals reflected angle,  

    ni sin θ_p = ns cos θ_p

    n₂ / n₁ = sin θ_p / cos θ_p

    n₂ / n₁ = tan θ_p

    θ_p = tan⁻¹ (n₂ / n₁)

Now we can calculate it

The refractive index of air is 1 (n1 = 1) the refractive index of seawater varies between 1.33 and 1.40 depending on the amount of salts dissolved in the water

n₂ = 1.33

      θ_p = tan⁻¹ (1.33 / 1)

      θ_p = 53.0º

n₂ = 1.40

      θ_p = tan⁻¹ (1.40 / 1)

      Tep = 54.5º

4 0
3 years ago
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