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insens350 [35]
3 years ago
6

If the center of mass passes outside the area of support of an object, what will happen to it?

Physics
1 answer:
defon3 years ago
6 0

Answer:

If a vertical line extending down from an object's CG extends outside its area of support, the object will topple

Explanation:

We can understand better this situation using a diagram with the forces acting on it.

In the attached image we can see that when the gravity center is bouncing outside from the area of the pedestal, the object will be out of balance and will fall.

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Two dogs are pulling on a chew toy. One dog pulls the chew toy with 64 N [E] and
PIT_PIT [208]

Answer:

Eastward, at 11 m/s^2

Explanation:

64N-31N=unbalanced force of 33N

F=ma

33N=(3kg)a

a=11m/s^2 to the East

3 0
2 years ago
When do dementia come about
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3 years ago
2) A skier stands at rest and begins to ski downhill with an acceleration of 3.0 m/s² {downhill). What is
olga2289 [7]

Answer:

her displacement <em>s=337.5m</em>

Explanation:

check out the above attachment ☝️

7 0
2 years ago
Read 2 more answers
The table below shows the level of carbon dioxide in the atmosphere for a period of 50 years.
yan [13]

Answer: The level of CO2 has risen.

Explanation:

From the table shown, we can see that the quantity of CO₂ in the atmosphere has steadily risen since the year 1960 going from 317 CO₂PPM in that year to 390 CO₂PPM in 2010.

This is a cause for alarm because with so much carbon dioxide in the atmosphere, there will be an even greater greenhouse effect that will contribute to global warming.

6 0
2 years ago
Two astronauts, each with a mass of 50 kg, are connected by a 7 m massless rope. Initially they are rotating around their center
kiruha [24]

Answer:

The angular  velocity is w_f =  1.531 \ rad/ s

Explanation:

From the question we are told that

     The mass of each astronauts is  m =  50 \ kg

      The initial  distance between the two  astronauts  d_i  =  7 \  m

Generally the radius is mathematically represented as r_i  =  \frac{d_i}{2} = \frac{7}{2}  =  3.5 \  m

      The initial  angular velocity is  w_1 = 0.5 \  rad /s

       The  distance between the two astronauts after the rope is pulled is d_f =  4 \  m

Generally the radius is mathematically represented as r_f  =  \frac{d_f}{2} = \frac{4}{2}  =  2\  m

Generally from the law of angular momentum conservation we have that

           I_{k_1} w_{k_1}+ I_{p_1} w_{p_1} = I_{k_2} w_{k_2}+ I_{p_2} w_{p_2}

Here I_{k_1 } is the initial moment of inertia of the first astronauts which is equal to I_{p_1} the initial moment of inertia of the second astronauts  So

      I_{k_1} = I_{p_1 } =  m *  r_i^2

Also   w_{k_1 } is the initial angular velocity of the first astronauts which is equal to w_{p_1} the initial angular velocity of the second astronauts  So

      w_{k_1} =w_{p_1 } = w_1

Here I_{k_2 } is the final moment of inertia of the first astronauts which is equal to I_{p_2} the final moment of inertia of the second astronauts  So

      I_{k_2} = I_{p_2} =  m *  r_f^2

Also   w_{k_2 } is the final angular velocity of the first astronauts which is equal to w_{p_2} the  final angular velocity of the second astronauts  So

      w_{k_2} =w_{p_2 } = w_2

So

      mr_i^2 w_1 + mr_i^2 w_1 = mr_f^2 w_2 + mr_f^2 w_2

=>   2 mr_i^2 w_1 = 2 mr_f^2 w_2

=>   w_f =  \frac{2 * m * r_i^2 w_1}{2 * m *  r_f^2 }

=>    w_f =  \frac{3.5^2 *  0.5}{  2^2 }

=>   w_f =  1.531 \ rad/ s

       

3 0
3 years ago
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