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Free_Kalibri [48]
3 years ago
7

Consult Conceptual Example 9 in preparation for this problem. Interactive LearningWare 6.3 also provides useful background. The

drawing shows a person who, starting from rest at the top of a cliff, swings down at the end of a rope, releases it, and falls into the water below. There are two paths by which the person can enter the water. Suppose he enters the water at a speed of 13.4 m/s via path 1. How fast is he moving on path 2 when he releases the rope at a height of 2.34 m above the water? Ignore the effects of air resistance.
Physics
1 answer:
Alex73 [517]3 years ago
3 0

Answer:

11.56066 m/s

Explanation:

m = Mass of person

v = Velocity of person = 13.4 m/s

g = Acceleration due to gravity = 9.81 m/s²

v' = Velocity of the person in the second

The kinetic and potential energy will balance each other at the surface

\dfrac{1}{2}mv^2=mgh\\\Rightarrow h=\dfrac{v^2}{2g}\\\Rightarrow h=\dfrac{13.4^2}{2\times 9.81}\\\Rightarrow h=9.15188\ m

Height of the cliff is 9.15188 m

Let height of the fall be h' = 2.34 m

\dfrac{1}{2}mv'^2+mgh'=mgh\\\Rightarrow v'=\sqrt{2g(h-h')}\\\Rightarrow v'=\sqrt{2\times 9.81(9.15188-2.34)}\\\Rightarrow v'=11.56066\ m/s

The speed of the person is 11.56066 m/s

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Polaris is used to identify North direction. Since, the Earth rotates on its axis which is along North-south, Polaris never seems to rise and set from the Northern hemisphere. This is because Polaris lies above north pole. Thus, in the given diagram, Polaris is above the North pole on the axis represented by top-most red circle.

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Molybdenum has the BCC crystal structure, has a density of 10.22 g cm-3 and an atomic mass of 95.94 g mol-1. What is the atomic
puteri [66]

Answer:

atomic concentration = 2 atoms/unit cell

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atomic radius: r= 1.39 x 10⁻¹⁰m

Explanation:

The atomic concentration is the number of atoms that can fit into a unit cell. It is a known number for each unit cell crystal structure. For a BCC (body-centered cube) crystal structure, atomic concentration is 2 atoms/unit cell because there are a 1/8 part of an atom in each corner of the cube (1/8 x 8= 1 atom) and 1 central atom in the central position of the cube ⇒ n= 1 atom + 1 atom= 2 atoms/unit cell

In order to calculate the lattice parameter a, we introduce the atomic mass 95.94 g/mol and the density 10.22 g/cm³ in the expression for the volume of the cube:

Vc= a³= \frac{(95.94 g/mol) x (2 atoms/unit cell)}{(10.2 g/cm^{3}) x (6.023 x 10^{23} atoms/mol)  }

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Once we know the lattice parameter a, we can calculate the atomic radius r by using the expression of a for a BCC structure:

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3 years ago
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