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viktelen [127]
3 years ago
13

A 39.5 kg child stands at the center of a 125 kg playground merry-go-round which rotates at 3.10 rad/s. If the child moves to th

e edge of the merry-go-round, what is the new angular velocity of the system? Model the merry-go-round as a solid disk
Physics
1 answer:
Neporo4naja [7]3 years ago
3 0
Using the law of conservation of angular momentum, we have 

<span>I1 w1 = I2 w2 </span>

<span>ie., m1r^2/2 x w1 = ( m1r^2/2 + m2r^2 ) w2 </span>

<span>ie., new angular velocity w2 = m1 w1 / ( m1+ 2m2) = 125 x 3.1 / ( 125 + 2 x39.5 ) </span>

<span>= 1.8995 = 1.9 rad /sec ( nearly )</span>
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Five different forces act on an object. Is it possible for the net force on the object to be zero?
bazaltina [42]
No because there must be an even # if their is an even amount one of the forces isn’t being cancelled
4 0
3 years ago
jet is flying at 500 mph east relative to the ground. A Cessna is flying at 150 mph 60° north of west relative to the ground. Wh
Greeley [361]

Answer:

C. 590 mph

\vert v_{cj}\vert=589.49\ mph

Explanation:

Given:

  • velocity of jet, v_j=500\ mph
  • direction of velocity of jet, east relative to the ground
  • velocity of Cessna, v_c=150\ mph
  • direction of velocity of Cessna, 60° north of west

Taking the x-axis alignment towards east and hence we have the velocity vector of the jet as reference.

Refer the attached schematic.

So,

\vec v_j=500\ \hat i\ mph

&

\vec v_c=150\times (\cos120\ \hat i+\sin120\ \hat j)

\vec v_c=-75\ \hat i+75\sqrt{3}\ \hat j\ mph

Now the vector of relative velocity of Cessna with respect to jet:

\vec v_{cj}=\vec v_j-\vec v_c

\vec v_{cj}=500\ \hat i-(-75\ \hat i+75\sqrt{3}\ \hat j )

\vec v_{cj}=575\ \hat i-75\sqrt{3}\ \hat j\ mph

Now the magnitude of this velocity:

\vert v_{cj}\vert=\sqrt{(575)^2+(75\sqrt{3} )^2}

\vert v_{cj}\vert=589.49\ mph is the relative velocity of Cessna with respect to the jet.

8 0
3 years ago
if a bowling ball hits a wall a force of 6 N, the wall exerts a force of how much back. on the bowling ball
grigory [225]
It would exert the same back right?
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3 years ago
Karen runs sets in basketball practice. She starts from a line runs 2.0 m, returns to the line, runs 4.0 m, to the line, runs 6.
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D. distance = 23 m, displacement = + 1 m

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Let's remind the difference between distance and displacement:

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In this case, the distance covered by Karen is given by the sum of all its movements:

distance = 2.0 m + 2.0 m+4.0 m+4.0 m +6.0 + 5.0 m=23.0 m

The displacement instead is given by the difference between the final point (1.0 m in front of the starting line) and the starting point (the starting line, 0 m):

displacement = +1.0 m-0 m=+1 m

8 0
3 years ago
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<span>Chemical Energy is converted into Mechanical Energy.

</span>When gasoline is burned in a car engine, chemical energy is converted into mechanical energy.
8 0
3 years ago
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