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Elena-2011 [213]
3 years ago
8

How many coulombs of positive charge are there in 47.0 gm of plutonium, given its atomic mass is 244 and that each plutonium ato

m has 96 protons?
(enter answer to three significant figures and in the exponent form: Example,

1.38e5)
Physics
1 answer:
enot [183]3 years ago
7 0

Answer:

  • 1.78×10⁶ C

Explanation:

Using the atomic mass of pluonium atoms (244 g/mol), you can calculate the number of atoms in 47.0 g. Then, knowing that each plutonium atom has 96 protons, you calculate the number of protons in the 47.0 g sample. Finally, using the positive charge of one proton, you calculate the total positive charge in the 47.0 g of plutonium.

<u>1. Number of atoms of plutonium in 47.0 g</u>

  • Number of moles = mass / atomic mass = 47.0 g / 244 = 0.1926 moles

  • Number of atoms = number of moles × 6.022 × 10²³ atoms/mol

  • Number of atoms = 0.1926 mol × 6.022 × 10²³ atoms/mol = 1.15998×10²³ atoms

<u>2. Number of protons</u>

  • Number of protons = 1.15998×10²³ atoms × 96 protons/atom = 1.11385×10²⁵ protons

<u>3. Charge</u>

<u />

  • Charge = charge of one proton × number of protons

  • Charge = 1.602×10⁻¹⁹ C/proton × 1.11385×10²⁵ protons = 1.78×10⁶C
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<em>W</em> = - (1/2 <em>kx</em> ² - 1/2 <em>k</em> (0.0400 m)²)

(note that <em>x</em> > 4.00 cm, and the restoring force of the spring opposes its elongation, so the total work is negative)

By the work-energy theorem, the total work is equal to the change in the glider's kinetic energy as it moves from 4.00 cm from equilibrium to <em>x</em>, so

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Solve for <em>x</em> :

- (1/2 (160 N/m) <em>x</em> ² - 1/2 (160 N/m) (0.0400 m)²) = -1/2 (0.190 kg) (0.835 m/s)²

==>   <em>x</em> ≈ 0.0493 m ≈ 4.93 cm

(b) The glider attains its maximum speed at the equilibrium point. The work done by the spring as it is stretched away from equilibrium to the 4.00 cm position is

<em>W</em> = - 1/2 <em>k</em> (0.0400 m)²

If <em>v</em> is the glider's maximum speed, then by the work-energy theorem,

<em>W</em> = ∆<em>K</em> = 1/2 <em>m</em> (0.835 m/s)² - 1/2 <em>mv</em> ²

Solve for <em>v</em> :

- 1/2 (160 N/m) (0.0400 m)² = 1/2 (0.190 kg) (0.835 m/s)² - 1/2 (0.190 kg) <em>v</em> ²

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the amplitude of the wave is which of the following A. Resonance B. Reflection C. Energy D. Magnitude
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D.Magnitude is the answer
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Answer:

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   From the question we are told that

           The height is h_s = 30 \ cm

            The angle of the slope is \theta = 15^o

According to the law of conservation of energy

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Where I is the moment of inertia which is mathematically represented as this for  a sphere

                    I = \frac{2}{5} mr^2

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                         w = \frac{v}{r}

So the equation for conservation of energy becomes

               mgh_s = \frac{1}{2} [\frac{2}{5} mr^2 ][\frac{v}{r} ]^2 + \frac{1}{2}mv^2

              mgh_s = \frac{1}{2} mv^2 [\frac{2}{5} +1 ]

             mgh_s = \frac{1}{2} mv^2 [\frac{7}{5} ]

            gh_s =[\frac{7}{10} ] v^2

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Substituting this into the conservation equation above

              mgh_c = \frac{1}{2} (mr^2)[\frac{v}{r} ] ^2 + \frac{1}{2} mv^2

Where h_c is the height where the circular hoop would be released to equal the speed of the sphere at the bottom

                 mgh_c  = mv^2

                     gh_c = v^2

                     h_c = \frac{v^2}{g}

Recall that   v^2 = \frac{10gh_s}{7}

                    h_c= \frac{\frac{10gh_s}{7} }{g}

                      = \frac{10h_s}{7}

            Substituting values

                   h_c = \frac{10(30)}{7}

                       h_c = 42.86 \ cm    

       

     

                         

8 0
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