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const2013 [10]
4 years ago
8

You place a 3.0-m-long board symmetrically across a 0.5-m-wide chair to seat three physics students at a party at your house. If

67-kg Dan sits on the left end of the board and 50-kg Tahreen on the right end of the board, where should 54-kg Komila sit to keep the board stable? What assumptions did you make?
Physics
1 answer:
wlad13 [49]4 years ago
3 0

Answer:

  • between locations that are 14 cm outboard of the chair edges
  • the weightless board is centered and end sitters are 25 cm from the ends

Explanation:

We can assume the .5 m-wide chair means that it is comfortable for each student to sit 0.25 m from the end of the board. If the board is centered on the chair, then each student is 1 m from the edge of the chair.

When Dan and Tahreen are seated on the board, their center of mass is ...

  (50 kg×2.5 m)/(50 kg +67 kt) = 1.068 m

to the right of the position where Dan is seated. Since this location is over the chair, the board is stable.

Komila can sit as much as x distance from the chair toward Dan, where ...

  67(1) +54(x) = 50(1.5)

  x = 8/54 ≈ 0.148 . . . . meters

Or, Komila can sit as much as x distance from the chair toward Tahreen, where ...

  67(1.5) = 54(x) +50(1)

  x = 50.5/54 ≈ 0.935 . . . . meters

<u>Scenario 1</u>

Assuming the (weightless) board is centered on the chair, Komila can sit anywhere between 14.8 cm left of the chair and 93.5 cm right of the chair and the board will remain stable. Sitting on the board centered on the chair is a suitable location. The two students sitting on the ends must become (and stay) seated at the same time. They both must be seated 0.25 m from the end of the board for the other dimensions to remain valid.

<u>Scenario 2</u>

Assuming the (weightless) board is located so its left end is 1.068 m from the chair, and Dan and Tahreen are seated 0.25 m from the ends of the board, Komila can sit anywhere within (117/54×.25 m) = 0.54 m of the chair and the board will remain stable. Again, sitting centered on the chair is a suitable location.

__

There does not appear to be any location where Komila can sit and have the board remain stable with only Dan or Tahreen seated on one end (assuming a width of 0.5 m for each sitter).

_____

<em>Comment on the question</em>

For the board to remain stable, the sum of moments about either edge of the chair must tend to rotate the board toward the chair. This sum will depend on the locations of the sitters relative to each edge of the chair, so there is significant freedom in choosing locations. To make the problem tractable, we have made some specific assumptions about where the board is and what the locations of the sitters might be. YMMV

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A charge +q is located at the origin, while an identical charge is located on the x axis at x = +0.57 m. A third charge of +2q i
Jlenok [28]

Answer:

The third charge placed is 0.80 m.

Explanation:

Given that,

Distance = 0.57 m

First charge = q

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Using formula of electrostatic force

F_{21}=\dfrac{kq_{1}q_{2}}{r_{1}^2}

When placed another charge q₃ at certain distance from origin, then the net force on charge q₁ due to both charges is

F_{net}=F_{21}+F_{31}

The net electrostatic force on the charge at the origin doubles.

2F_{21}=F_{21}+F_{31}

F_{31}=F_{21}

\dfrac{kq_{3}q_{1}}{r_{2}^2}=\dfrac{kq_{2}q_{1}}{r_{1}^2}

\dfrac{q_{3}}{r_{2}^2}=\dfrac{q_{2}}{r_{1}^2}

r_{2}^2=\dfrac{q_{3}}{q_{2}\timesr_{1}^2}

r_{2}=\sqrt{\dfrac{q_{3}}{q_{2}}}r_{1}

Put the value into the formula

r_{2}=\sqrt{\dfrac{2q}{q}}\times0.57

r_{2}=\sqrt{2}\times0.57

r_{2}=0.80\ m

Hence, The third charge placed is 0.80 m from origin in x-axis.

4 0
3 years ago
If you place 48 cm object 15 m in front of a convex mirror of focal length 5 m, what is the magnification of the image?
Arisa [49]

Answer:

m = 0.25

Explanation:

Given that,

Object distance, u = -15cm

Height of the object, h = 48

Focal length, f = cm

We need to find the magnification of the image.

Let v is the image distance. Using mirror's equation.

\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}\\\\\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}\\\\=\dfrac{1}{5}-\dfrac{1}{(-15)}\\\\=3.75\ cm

Magnification,

m=\dfrac{-v}{u}\\\\=\dfrac{-3.75}{-15}\\\\=0.25

Hence, the magnification of the image is 0.25.

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Electron groups could be considered which of the following?
PSYCHO15rus [73]

Electron groups could be considered as  Lone pair electrons and bonded pairs of electrons.

Answer: Option D & B

<u>Explanation:</u>

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If a set of displacement vectors laid head to tail make a closed polygon, what is the resultant vector?
Papessa [141]

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Explanation:

When you add displacement vectors using the head to tail method, you follow this procedure:

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.. and so on.

The resultant of the all vectors will be the vector connecting the tail of the 1st vector to the head of the last vector.

In this problem, the vectors make a closed polygon: this means that the head of the last vector coincides with the tail of the first vector.

Therefore, this means that the length of the resultant vector is zero.

Learn more about vector addition:

brainly.com/question/11220787

brainly.com/question/2892784

#LearnwithBrainly

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