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Darya [45]
3 years ago
12

Your teacher (175kg) and the lab (15kg) are 2m apart. What is the gravitational force between them? Draw a FBD, and label

Physics
1 answer:
Pachacha [2.7K]3 years ago
7 0
The gravitational force between two objects is given by
F=G \frac{m_1 m_2 }{r^2}
where
G is the gravitational constant
m1 and m2 are the masses of the two objects
r is the separation between the two objects

In this problem, m_1 = 175 kg, m_2 =15 kg and r=2 m, therefore the gravitational force between the two objects is
F=(6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2}) \frac{(175 kg)(15 kg)}{(2m)^2}=4.38 \cdot10^{-8} N
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If the gymnast mass were doubled, her height (h) from the top of the board would be as follows,

с  Stay the same

Explanation:

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  • The greater the force of gravity, it would give a direct impact on the object's acceleration; thus considering only a force, the heavier the object is, it would accelerate faster. But an acceleration depends upon the two factors which are  force and mass.
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A car of mass 998 kilograms moving in the positive y–axis at a speed of
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The total momentum should come out to be  <span>2.0 x 10^4 kilogram meters/second </span>
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3 years ago
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Explanation:

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3 years ago
Your mother is sure that you were driving too fast because she knows
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3 years ago
If the briefcase hits the water 6.0 s later, what was the speed at which the helicopter was ascending?
vovikov84 [41]

Complete Question

In an action movie, the villain is rescued from the ocean by grabbing onto the ladder hanging from a helicopter. He is so intent on gripping the ladder that he lets go of his briefcase of counterfeit money when he is 130 m above the water. If the briefcase hits the water 6.0 s later, what was the speed at which the helicopter was ascending?

Answer:

The speed of the helicopter is u  =  7.73 \  m/s

Explanation:

From the question we are told that

   The height at which he let go of the brief case is  h =  130 m  

    The  time taken before the the brief case hits the water is  t =  6 s

Generally the initial speed of the  briefcase (Which also the speed of the helicopter )before the man let go of it is  mathematically evaluated using kinematic equation as

      s = h+  u t +  0.5 gt^2

Here s  is the distance covered by the bag at sea level which is zero

      0 = 130+  u * (6) +  0.5  *  (-9.8) * (6)^2

=>    0 = 130+  u * (6) +  0.5  *  (-9.8) * (6)^2

=>   u  =  \frac{-130 +  (0.5 * 9.8 *  6^2) }{6}

=>   u  =  7.73 \  m/s

     

7 0
3 years ago
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