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Genrish500 [490]
3 years ago
11

A sledgehammer strikes an anvil with a velocity of 50 ft/sec (Fig. 2.59). The hammer and the anvil weigh 12 lb and 100 lb, respe

ctively. The anvil is supported on four springs, each of stiffness Find the resulting motion of the anvil (a) if the hammer remains in contact with the anvil and (b) if the hammer does not remain in contact with the anvil after the initial impact
Physics
1 answer:
saveliy_v [14]3 years ago
4 0
Hard question thx for the points give me brainlest points plz
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The force required to compress a non-standard spring as a function of displacement from equilibrium x is given by the equation f
Hoochie [10]
<span>Answer: F(x) = ax^2 - bx or F(x) = ax² - bx F(x) = 30x² - 6x â«F(x)dx = â«(30x² - 6x)dx as this is evaluated from zero to x W = 10xÂł - 3x² <===ANS W = 10(0.42Âł) - 3(0.42²) - [10(0Âł) - 3(0²)] W = 0.212 J <===ANS W = 10(0.72Âł) - 3(0.72²) - [10(0.42Âł) - 3(0.42²)] W = 1.966 J <===ANS</span>
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3 years ago
If there are no other changes, explain what effect reducing the mass of the car will have on its acceleration when starting to m
densk [106]

Answer:

when the mass of an object is decreased, the acceleration will increase

when mass is increased, acceleration decreases

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3 years ago
How has the mix of energy sources changed in the last 50 years in production and consumption?
Tcecarenko [31]

As new energy sources have been developed, the energy consumption in the history has changed significantly.

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3 years ago
If 2.2 lbs = 1.0 kg, and Megan Progress weighs 130 lbs, what is her weight in newtons? W = N (Round your answer to nearest whole
lord [1]

as it is given that

2.2 lbs = 1 kg

here we know that Megan Progress weighs 130 lbs

so its mass in kg is given as

m = 130 * \frac{1}{2.2} = 59.1 kg

now to find the weight in Newton unit we can say

W = mg

W = 59.1* 9.8

W = 579 N

so 130 lbs weighs as 579 N in SI units

5 0
3 years ago
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
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