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oksian1 [2.3K]
4 years ago
12

A truck of mass 2500gm. moving at. 15 m. 's cellulose with a. car. of mass 1000. kg. moving. at. 5m/s. in. the. opposite. direct

ion. with. what. velocity. would the two move together
​
Physics
1 answer:
vlabodo [156]4 years ago
5 0

Answer:

23

Explanation:

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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

3 0
4 years ago
5) Find the length of guitar string that would vibrate at a fundamental frequency of 825 Hz if the string is stretched so that t
Mariulka [41]

Áp dụng công thức tính độ dài ạ

6 0
2 years ago
If a net force of 800n is acting on an object with a mass of 60kg. What would be the objects accelerations
aivan3 [116]

Answer: 13.33 m/s^2

Explanation:

Force is the product of mass of an object and its acceleration.

i.e Force = Mass x Acceleration

Since Force on object = 800N

Mass of object = 6Kg

Acceleration = ?

Then, Force = Mass x Acceleration

800N = 60Kg x Acceleration

Acceleration = (800N/60kg)

Acceleration = 13.33 m/s^2 (The unit of acceleration is metre per second square)

Thus, the objects accelerations is 13.33 m/s^2

5 0
4 years ago
A block of wood weighs 160 N and has a specific gravity of 0.60. To sink it in fresh water requires an additional downward force
AysviL [449]

Answer:

Your answer is: D) 110N

Hope this helped : )

Explanation:

In order for the wood to sink, it's specific gravity must equal 1 (or greater).

Specific gravity = .6 kg/m^3= Mass/volume

F (Newtons or kg x m/s^2) = Mass*9.8m/s^2.  

Therefore, Mass = 160 N / 9.8 m/s^2= 16.32 kg

Volume = Mass/Specific Gravity = 16.32kg/.6 kg/m^3 = 27.2 m^3

Desired Specific gravity = 1 = (Mass + additional force/9.8m/s^2)/Volume = (16.32 kg +X / 9.8m/s^2)/27.2m^3

Solving for the additional force: X = 106.66...N

The answer is D, 110 N will be required to sink the block of wood.

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3 years ago
A resting object exerts pressure equal to (i point!
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I think it’s the last one it’s weight divided by its contact area
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3 years ago
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