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Vitek1552 [10]
3 years ago
5

What is one advantage a primary source has over a secondary source? A. A primary source always includes several perspectives. B.

A primary source combines several different accounts of an event. C. A primary source offers a firsthand account of an event. D. A primary source generally benefits from having been written after an event.
Physics
1 answer:
Ostrovityanka [42]3 years ago
5 0

Answer: a primary source I typically free of outside Interpretation

Explanation:

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A fly wheel with a diameter of 1.20m is rotating at an angular speed of 200 Rev/min
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you will For this problem, you will need to look up physical parameters for objects in space want to keep about 4 significant fi
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Answer:

a)  r₁ = 3.836 10⁷ m,  b)   F = - 3,390 10⁸ N , c)  R = 120.3 m

Explanation:

a) This is a problem of equilibrium where the force is gravitational, we call F1 the force of the Moon and F2 the force from the earth.

       F₁ -F₂ = 0

       F₁ = F₂

       G m M_{m} / r₁² = G m M_{e} / r₂²

Let's look for the measured distance from a Coordinate System located on the Moon,

         r₂ = D - r₁

Where D is the average distance from Terra to the Moon 3.84 10⁸ m

        M_{m} / r₁² = M_{e} / (D - r₁)²

       (D² - 2 D r₁ + r₁²) = M_{e} /M_{m} r₁²

       (1 - M_{e} / M_{m})r₁² - 2D r₁ + D² = 0

Let's replace and solve the second degree equation

       (1 - 5.98 10²⁴ / 7.36 10²²) r₁² - 2 3.84 10⁸ r₁ + (3.84 10⁸)² = 0

       -80.25 r₁² - 7.68 10⁸ r₁ + 14.75 10¹⁶ = 0

        r₁² + 9.57 10⁶ r₁ - 1.838 10¹⁵ = 0

        r₁ = [-9.57 10⁶ ±√ (91.58 10¹² + 7.352 10¹⁵)] / 2

        r₁ = [-9.57 10⁶ + - 86.28 10⁶] / 2

The results are:

       r₁’= 38.355 10 6 m

       r₁ ’’ = -47.915 10 6 m

We take the positive result that a distance between the moon and the Earth, the equalization point is    r₁ = 3.836 10⁷ m

b) The force at point R = 2 r₁

        R = 2 3.8355 10⁷ = 7.671 10⁷ m

        F = F₁ - F₂

        F = G m  M_{m} / R² - G m  M_{e} / (D- R)²

        F = G m ( M_{m} / R² -  M_{e} / (D-R)²)

   F = m 6.67 10⁻¹¹ (7.36 10²² / (7.671 10⁷)² - 5.98 10²⁴ / (3.84 10⁸ - 0.7671 10⁸)²

        F = m 6.67 10⁻¹¹ (0.125076 10⁸ - 0.63329 10⁸)

        F = m (-3.3897 10⁸) N

The mass m of the rocket must be known, suppose it is worth 1 kg (m = 1 kg)

        F = - 3,390 10⁸ N

c) let's use gravitational force from the moon

        F = G m  M_{m} / R²

        R =√ G m  M_{m} / F

        R = √ (6.67 10⁻¹¹ 1 7.36 10²² / 3.3897 10⁸)

        R = √ (1.4482 10⁴)

        R = 1.2034 10² m

        R = 120.3 m

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