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Sauron [17]
3 years ago
6

An uncrewed mission to the nearest star, Proxima Centauri, is launched from the Earth's surface as a projectile with an initial

speed of 44.4 km/s, just enough for the spacecraft to escape the Earth's gravity and leave the solar system. Ignoring air resistance and the Earth's rotation, what is the speed of the spacecraft when it is more than halfway to the star? Assume we are ignoring the effect of the Sun on the spacecraft.
Physics
1 answer:
Anna [14]3 years ago
6 0

Answer:

42.96 km/s

Explanation:

From the conservation of Energy

(PE+KE)_i=(PE+KE)_f\\\Rightarrow -\frac{GmM}{R}+\frac{1}{2}mv_i^2=0+\frac{1}{2}mv_f^2

Mass gets cancelled

-\frac{GM}{R}+\frac{1}{2}v_i^2=0+\frac{1}{2}v_f^2\\\Rightarrow -2\frac{GM}{R}+v_i^2=v_f^2\\\Rightarrow -v_e^2+v_i^2=v_f^2\\\Rightarrow v_f=\sqrt{v_i^2-v_e^2}

v_e=\sqrt{\frac{2Gm}{R}} = Escape velocity of Earth = 11.2 km/s

v_i = Velocity of projectile = 44.4 km/s

v_f=\sqrt{44.4^2-11.2^2}\\\Rightarrow v_f=42.96\ km/s

The velocity of the spacecraft when it is more than halfway to the star is 42.96 km/s

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A 5 kg block is pushed across a table by a horizontal force of 40 N with an acceleration of 5 m/s^2. What is the frictional forc
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When a cube is inscribed in a sphere of radius r, the length Lof a side of the cube is . If a positive point charge Qis placed a
Nana76 [90]

Answer:

  Ф_cube /Ф_sphere = 3 /π

Explanation:

The electrical flow is

      Ф = E A

where E is the electric field and A is the surface area

Let's shut down the electric field with Gauss's law

       Фi = ∫ E .dA = q_{int} / ε₀

the Gaussian surface is a sphere so its area is

        A = 4 π r²

the charge inside is

        q_{int} = Q

we substitute

       E 4π r² = Q /ε₀

       E = 1 / 4πε₀   Q / r²

To calculate the flow on the two surfaces

* Sphere

       Ф = E A

        Ф = 1 / 4πε₀  Q / r² (4π r²)

        Ф_sphere = Q /ε₀

* Cube

Let's find the side value of the cube inscribed inside the sphere.

In this case the radius of the sphere is half the diagonal of the cube

          r = d / 2

We look for the diagonal with the Pythagorean theorem

         d² = L² + L² = 2 L²

         d = √2 L

         

we substitute

          r = √2 / 2 L

          r = L / √2

          L = √2  r

now we can calculate the area of ​​the cube that has 6 faces

          A = 6 L²

          A = 6 (√2  r)²

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the flow is

          Ф = E A

          Ф = 1 / 4πε₀  Q/r²  (12r²)

          Ф_cubo = 3 /πε₀  Q

the relationship of these two flows is

         Ф_cube /Ф_sphere = 3 /π

8 0
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