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yanalaym [24]
3 years ago
15

Naturally occurring element X exists in three isotopic forms: X-28 (27.977 amu, 92.23% abundance), X-29 (28.976 amu, 4.67% abund

ance), and X-30 (29.974 amu, 3.10% abundance). Calculate the atomic weight of X. A. 86.93 amu B. 48.63 amu C. 28.09 amu D. 28.97 amu E. 27.16 amu
Physics
1 answer:
vova2212 [387]3 years ago
6 0

Answer:

C. 28.09 amu

Explanation:

The natural occurring element exist in 3 isotopic forms: namely X-28 (27.977 amu, 92.23% abundance),  X-29 (28.976 amu, 4.67% abundance) and  X-30 (29.974 amu, 3.10% abundance).

The atomic weight of elements depends on the isotopic abundance. If you know the fractional abundance and the mass of the isotopes the atomic weight can be computed.

The atomic weight is computed as follows:

atomic weight = mass of X-28 × fractional abundance + mass of X-29 × fractional abundance + mass of  X-30 × fractional abundance

atomic weight = 27.977 × 0.9223 + 28.976 × 0.0467 + 29.974 ×  0.0310

atomic weight = 25.8031871 + 1.3531792 + 0.929194

atomic weight = 28.0855603 amu

To 2 decimal place atomic weight = 28.09 amu

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A 3.00-kg object undergoes an acceleration given by a = (2.00 i + 5.00 j) m/s^2. Find (a) the resultant force acting on the obje
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(a): The resultant force acting on the object are F= (5.99 i + 14.98 j).

(b): The magnitude of the resultant force are F= 16.4 N < 68.19º .

Explanation:

m= 3kg

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3 years ago
A thin rod of length 0.75 m and mass 0.42 kg is suspended
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Answer:

a)  K = 0.63 J, b)  h = 0.153 m

Explanation:

a) In this exercise we have a physical pendulum since the rod is a material object, the angular velocity is

         w² = \frac{m g d}{I}

where d is the distance from the pivot point to the center of mass and I is the moment of inertia.

The rod is a homogeneous body so its center of mass is at the geometric center of the rod.

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the moment of inertia of the rod is the moment of a rod supported at one end

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an oscillatory system is described by the expression

              θ = θ₀ cos (wt + Φ)

the angular velocity is

             w = dθ /dt

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In this exercise, the kinetic energy is requested in the lowest position, in this position the energy is maximum. For this expression to be maximum, the sine function must be equal to ±1

In the exercise it is indicated that at the lowest point the angular velocity is

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the kinetic energy is

           K = ½ I w²

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b) for this part let's use conservation of energy

starting point. Lowest point

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final point. Highest point

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energy is conserved

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             h = 1/6 L² w² / g

             h = 1/6 0.75² 4.0² / 9.8

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