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Ivan
3 years ago
11

A 12 N package of flour is suddenly placed on the pan of a grocer's scale. The pan is supported from below by a vertical spring

with a force constant of 325 N/m. If the pan has negligible weight and no energy is dissipated by friction, find the maximum distance the spring will be compressed.
Mathematics
1 answer:
Licemer1 [7]3 years ago
4 0

Answer:

  24/325 m ≈ 7.38 cm

Step-by-step explanation:

The equilibrium position of the package is the distance at which the package weight is balanced by the spring force. That distance is ...

  (12 N)/(325 N/m) = 12/325 m

Since the package is placed suddenly on the pan and there are no losses in the system, the position of the package will oscillate about this distance with an amplitude equal to this distance. That is, the maximum compression will be to a distance that is twice this amount, or ...

  24/325 m

__

This is the depth at which all of the potential energy of the package due to its height is transformed into the potential energy of the spring due to its compression.

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3 years ago
If: A + B = 8 [] C - A = 6 [] A + C = 13 [] then A=______ B=_______C=_______ (Hint:A,B, and C are not whole numbers!)
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Answer:

A=3.5

B=4.5

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7 0
3 years ago
If p is the incenter of triangle jkl, find each measure
Hunter-Best [27]

Answer:

PO = 7

PM = 7

MJ = 11

∠PJO = 32°

∠KJL = 64°

PL ≈ 18.385

OL = 17

∠PLO = 22°

∠NLO = 44°

∠JKL = 72°

∠MKP = 36°

∠NKP = 36°

∠PKN = 36°

KN = 10

PL ≈ 13.04

PK ≈ 12.21

JL = 28

JK = 21

LK = 27

Step-by-step explanation:

The given parameters are;

The point representing the incenter of the triangle = P

Therefore PO = PM = PN = 7

tan(32°) = PM/JM = 7/JM

∴JM = 7/(tan(32°)) ≈ 11.2

∠PJO = tan⁻¹(7/11)≈ 32.47°

∠PJO = ∠PJM = 32° similar triangles

∠KJL = ∠KJP + ∠PJO = 32 + 32 = 64°

∠KJL ≈ 64°

PL = √(7² + 17²) ≈ 18.385

OL = NL = 17 similar triangles

∠PLO = sin⁻¹(7/18.385) ≈ 22.380°

∠PLO = ∠PLN = 22°

∠NLO = ∠PNL + ∠OLP ≈ 22° + 22° ≈ 44°

∠NLO ≈ 44.380°

∠JKL = 180 - (∠KJL + ∠NLO)

∠JKL = 180° - (64° + 44°) ≈ 72°

∠JKL  ≈ 72°

∠MKP = ∠NKP = 72°/2 = 36°

∠MKP = 36°

∠NKP = 36°

∠PKN = ∠JKL - ∠MKP = 72° - 36° ≈ 36°

∠PKN  ≈ 36°

KN = KM = 10

MJ = OJ = 11

PL = √(7² + 11²) ≈ 13.04

PK = √(7² + 10²) ≈ 12.21

JL = JO + OL = 11 + 17 = 28

JK = JM + MK = 11 + 10 = 21

LK = LN + NK = 17 + 10 = 27

6 0
2 years ago
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