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steposvetlana [31]
1 year ago
6

you are the science officer on a visit to a distant solar system. prior to landing on a planet you measure its diameter to be 1.

8 × 107 m and its rotation period to be 22.3 hours. you have previously determined that the planet orbits 8.6 × 1011 m from its star with a period of 402 earth days. once on the surface you find that the acceleration due to gravity is 63.6 m/s2. what are the mass of (a) the planet and (b) the star? (g
Physics
1 answer:
Eduardwww [97]1 year ago
3 0

The mass of the star of the distant solar system is 82 * 10²² kg and the mass of the planet of the distant solar system is 8.58 * 10²⁵ kg

The planet stays in its orbit and moves in a centripetal motion due to the gravitational force of attraction between the planet and the star.

Fc = m v² / r

Fg = G M m / r²

Fc = Fg

m v² / r = G M m / r²

v² = G M / r

M = r v² / G

g = Acceleration due to gravity

G = Gravitational constant

M = Mass of star

r = Distance between both bodies

G = 6.67 * 10⁻¹¹ N m² / kg²

r = 8.6 * 10¹¹ m

T = 402 days = 3.47 * 10⁷ s

v = 2 π r / T

v = 2 * 3.14 * 8.6 * 10¹¹ / 3.47 * 10⁷

v = 15.56 * 10⁴ m / s

Ms = 8.6 * 10¹¹ * 15.56 * 10⁴ / ( 6.67 * 10⁻¹¹ )

Ms = 20.1 * 10²⁶ kg

F = G M m / r²

F = m g

Equating,

m g = G M m / r²

Mp = g r² / G

Mp = 63.6 * ( 1.8 * 10⁷ / 2 )² / 6.67 * 10⁻¹¹

Mp = 8.58 * 10²⁵ kg

To know more about Orbital velocity

brainly.com/question/15886625

#SPJ4

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The period is simply the inverse of the frequency, therefore:

T = 1 / f

T = 1 / 775 Hz

T = 0.001290 s    → possible answer

T = 1.29 × 10⁻³ s   → possible answer

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4 0
2 years ago
A hockey puck is struck so that it slides at a constant speed and strikes the far side of the rink, 58.2 m away. The shooter hea
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Answer:

v = 33.66 m/s

Explanation:

Let hockey puck is moving at constant speed v

so here we have

d = vt

so time taken by the puck to strike the wall is given as

t = \frac{58.2}{v}

now time taken by sound to come back at the position of shooter is given as

t_2 = \frac{58.2}{340}

t_2 = 0.17s

so we know that total time is 1.9 s

1.9 = t + t_2

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7 0
3 years ago
A rocket is fired straight up. It contains two stages (Stage 1 and Stage 2) of solid rocket fuel that are designed to burn for 1
Archy [21]

Answer:

a)  y= 3.5 10³ m, b)   t = 64 s

Explanation:

a) For this exercise we use the vertical launch kinematics equation

Stage 1

          y₁ = y₀ + v₀ t + ½ a t²

          y₁ = 0 + 0 + ½ a₁ t²

Let's calculate

         y₁ = ½ 16 10²

         y₁ = 800 m

At the end of this stage it has a speed

        v₁ = vo + a₁ t₁

        v₁ = 0 + 16 10

        v₁ = 160 m / s

Stage 2

        y₂ = y₁ + v₁ (t-t₀) + ½ a₂ (t-t₀)²

        y₂ = 800 + 150 5 + ½ 11 5²

        y₂ = 1092.5 m

Speed ​​is

        v₂ = v₁ + a₂ t

        v₂ = 160 + 11 5

        v₂ = 215 m / s

The rocket continues to follow until the speed reaches zero (v₃ = 0)

         v₃² = v₂² - 2 g y₃

         0 = v₂² - 2g y₃

         y₃ = v₂² / 2g

         y₃ = 215²/2 9.8

         y₃ = 2358.4 m

The total height is

          y = y₃ + y₂

          y = 2358.4 + 1092.5

          y = 3450.9 m

           y= 3.5 10³ m

b) Flight time is the time to go up plus the time to go down

Let's look for the time of stage 3

          v₃ = v₂ - g t₃

          v₃ = 0

          t₃ = v₂ / g

          t₃ = 215 / 9.8

          t₃ = 21.94 s

The time to climb is

          t_{s} = t₁ + t₂ + t₃

          t_{s} = 10+ 5+ 21.94

          t_{s} = 36.94 s

The time to descend from the maximum height is

          y = v₀ t - ½ g t²

When it starts to slow down it's zero

         y = - ½ g t_{b}²

         t_{b}  = √-2y / g

         

        t_{b} = √(- 2 (-3450.9) /9.8)

        t_{b} = 26.54 s

Flight time is the rise time plus the descent date

        t = t_{s} + t_{b}

        t = 36.94 + 26.54

        t =63.84 s

        t = 64 s

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