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Vitek1552 [10]
2 years ago
12

A gravitational force of 2.54 x 10^-10 is exerted by two objects that are 0.25 m apart. If the first object has a mass of 0.35 k

g, what is the mass of the second object?
Physics
1 answer:
Murrr4er [49]2 years ago
3 0

Answer :

what is the answer options?

Explanation:

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A migrating robin flies due north with a speed of 12 m/s relative to the air. The air moves due east with a speed of 6.3 m/s rel
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consider east-west direction along X-axis  and north-south direction along Y-axis

V_{ra} = velocity of migrating robin relative to air = 12 j m/s

(where "j" is unit vector in Y-direction)

V_{ag} = velocity of air relative to ground = 6.3 i m/s

(where "i" is unit vector in X-direction)

V_{rg} = velocity of migrating robin relative to ground = ?

using the equation

V_{rg} = V_{ra} + V_{ag}

V_{rg} = 12 j + 6.3 i

V_{rg} = 6.3 i + 12 j

magnitude : sqrt((6.3)² + (12)²) = 13.6 m/s

direction : tan⁻¹(12/6.3) = 62.3 deg north of east

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A thin, uniform rod is bent into a square of side length a. If the total mass is M, find the moment of inertia about an axis thr
Papessa [141]

Answer:

The  moment of inertia about an axis through the center and perpendicular to the plane of the square is

    I_s =  \frac{Ma^2}{3}

Explanation:

From the question we are told that

   The length of one side of the square is  a

   The total mass of the square is  M

Generally the mass of one size of the square is mathematically evaluated as

    m_1 = \frac{M}{4}

Generally the moment of inertia of one side of the square is mathematically represented as

        I_g =  \frac{1}{12}  *  m_1 * a^2

Generally given that m_1 = m_2 = m_3 = m_4 = m it means that this moment inertia evaluated above apply to every side of the square  

Now substituting for  m_1

  So

       I _g=  \frac{1}{12}  *  \frac{M}{4} * a^2

Now according to  parallel-axis theorem the moment of inertia of one side of the square about an axis through the center and perpendicular to the plane of the square is mathematically represented as

      I_a =  I_g + m [\frac{q}{2} ]^2

=>    I_a =  I_g + {\frac{M}{4} }* [\frac{q}{2} ]^2

substituting for I_g

=>    I_a =  \frac{1}{12}  *  \frac{M}{4} * a^2 + {\frac{M}{4} }* [\frac{q}{2} ]^2

=>    I_a = \frac{Ma^2}{48} + \frac{Ma^2}{16}

=>    I_a = \frac{Ma^2}{12}

Generally the moment of inertia of the square about an axis through the center and perpendicular to the plane of the square is mathematically represented as

      I_s = 4 * I_a

=>   I_s = 4 * \frac{Ma^2}{12}

=>   I_s =  \frac{Ma^2}{3}

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Answer:

A force of μk⋅60N is needed to keep a 60-newton rubber block moving across level, dry asphalt in a straight line at a constant speed.

Explanation:

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