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Whitepunk [10]
2 years ago
14

The mass of Mars is 6. 42 × 1023 kg. Its moon Phobos is 9378 kilometers away from Mars and has a mass of 1. 06 × 1016 kg. What i

s the gravitational force between Mars and Phobos? 5. 16 × 1015 N 5. 16 × 1021 N 4. 84 × 1022 N 4. 84 × 1025 N.
Physics
1 answer:
statuscvo [17]2 years ago
6 0

The gravitational force between two massive objects is directly proportional to the product of masses. The value of the gravitational force between Mars and Phobos is5. 16 \times  10^2^1 \rm \ N.

<h2></h2><h2>Newton's law of universal gravitation: </h2>

F=G{\dfrac{m_1m_2}{r^2}}

Where,

F = Force

G = Gravitational constant = \bold {6.67 x 10^{-11 }}N.m²/kg²

m_1 = Mass of Mars =  \bold {6. 42 \times  10^{23 }\rm \ kg}

m_2    = Mass of Phobos = \bold {1.06 \times  10^{16 }\rm \ kg}

r = Distance between centers of the masses = 9378 km

Put the values in the formula,

F=\bold {6.67 x 10^{-11 }}{\dfrac{\bold {6. 42 \times  10^{23 }\rm \ kg}\times \bold {1.06 \times  10^{16 }\rm \ kg}}{9378 ^2}}\\\\F= 5. 16 \times  10^2^1 \rm \ N

Therefore, the value of the gravitational force between Mars and Phobos is5. 16 \times  10^2^1 \rm \ N.

Learn more about Newton's law of universal gravitation: brainly.com/question/858421

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Andru [333]

Answer:

The ball will have an upward velocity of 6 m/s at a height of 5.51 m.

Explanation:

Hi there!

The equations of height and velocity of the ball are the following:

y = y0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

y = height at time t.

y0 = initial height.

v0 = initial velocity.

t = time.

g = acceleration due to gravity (-9.81 m/s² considering the upward direction as positive).

v = velocity of the ball at time t.

Placing the origin at the throwing point, y0 = 0.

Let´s use the equation of velocity to obtain the time at which the velocity is 12.0 m/s / 2 = 6.00 m/s.

v = v0 + g · t

6.00 m/s = 12.0 m/s -9.81 m/s² · t

(6.00 - 12.0)m/s / -9.81 m/s² = t

t = 0.612 s

Now, let´s calculate the height of the baseball at that time:

y = y0 + v0 · t + 1/2 · g · t²     (y0 = 0)

y = 12.0 m/s · 0.612 s - 1/2 · 9.81 m/s² · (0.612 s)²

y = 5.51 m

The ball will have an upward velocity of 6 m/s at a height of 5.51 m.

Have a nice day!

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The letter B is the letter that represents the location of the resister in the diagram.

Explanation:

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3 years ago
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xeze [42]

Answer:

Explanation:

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2 years ago
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horizontal clothesline is tied between 2 poles, 14 meters apart. When a mass of 3 kilograms is tied to the middle of the clothes
lbvjy [14]

Answer:

The tension is  T =  103.96N

Explanation:

The free body diagram of the question is shown on the first uploaded image From the question we are told that

           The distance between the two poles is D =14 m

          The mass tied between the two cloth line is  m = 3Kg

         The distance it sags is d_s = 1m

The objective of this solution is to obtain the magnitude of the tension on the ends of the  clothesline

Now the sum of the forces on the y-axis is zero assuming  that the whole system is at equilibrium

       And this can be mathematically represented as

                             \sum F_y = 0

 To obtain \theta we apply SOHCAHTOH Rule

 So    Tan \theta = \frac{opp}{adj}

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            = tan^{-1} [\frac{1}{7}]

          =8.130^o

=>  \  \ \ \ \ \ \ \ 2T sin\theta -mg =0

=>  \  \ \ \ \ \ \ \ T =\frac{mg}{2 sin\theta}

=>  \  \ \ \ \ \ \ \ T = \frac{3 * 9.8 }{2 sin \theta }

=>  \  \ \ \ \ \ \ \  T =\frac{29.4}{2sin(8.130)}

=>  \  \ \ \ \ \ \ \  T = 103.96N

             

                 

5 0
2 years ago
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