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lubasha [3.4K]
3 years ago
6

Please help question in the picture ​

Physics
1 answer:
mr Goodwill [35]3 years ago
8 0
B because it accelerates by 40m/s same as from 0 to 40
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An arachnid has eight legs.
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Three bulbs are connected by tubing, and the tubing is evacuated. The volume of the tubing is 45.0 mL. The first bulb has a volu
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Answer:

The final pressure of the whole system is 34.80 atm.

Explanation:

Given that,

Volume = 45.0 ml

Volume of first bulb = 77.0 mL

Pressure  = 8.89 atm

Volume of second  bulb = 250 mL

Pressure = 2.82 atm

Volume of third  bulb = 21.0 mL

Pressure = 8.42 atm

We need to calculate the final pressure of the whole system

Using formula of pressure

P_{1}V_{1}+P_{2}V_{2}+P_{3}V_{3}+P_{t}V_{t}=P_{f}V_{f}

Where, P_{1}= pressure of first bulb

P_{2}= pressure of second bulb

P_{3}= pressure of third bulb

P_{4}= initial pressure of tube

V_{1}= Volume of first bulb

V_{2}=Volume of second bulb

V_{3}= Volume of third bulb

V_{4}= Initial volume of tube

Put the value into the formula

8.89\times77.0+250\times2.82+21.0\times8.42+0=P_{f}\times45

P_{f}=\dfrac{1566.35}{45}

P_{f}=34.80\ atm

Hence, The final pressure of the whole system is 34.80 atm.

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3 years ago
A car begins at rest (0 velocity), 5 seconds later it is travelling at 20 meters/per second. What was the acceleration of the ve
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use the formula

v= u+ at

v is final velocity , u is initial velocity , a is acceleration and t is time

put the values

20 = 0+ a×5

a = 4 m/s²

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Read 2 more answers
Development is best described as:
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Answer:

A. Change that normally occurs in people throughout their lifetimes.

Explanation:

APEX verified

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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
wlad13 [49]

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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4 years ago
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