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NeTakaya
3 years ago
6

Dejamos caer un objeto desde lo alto de una torre y medimos el tiempo que tarda en llegar al suelo que resulta ser de 0,02 minut

os. calcula :a) la altura de la torre, b) la velocidad al impactar al suelo.
Physics
1 answer:
Harman [31]3 years ago
4 0

Answer: a) 11.76 m/s  b) 7.056 m

Explanation:

The described situation is as follows:

An object is dropped from the top of a tower and when measuring the time it takes to reach the ground that turns out to be 0.02 minutes.

This situation is related to free fall, this also means we have constant acceleration, hence the equations we will use are:

V_{f}=V_{o}+at (1)  

{V_{f}}^{2}={V_{o}}^{2}+2ad (2)  

Where:  

V_{f} Is the final velocity of the object

V_{o}=0 Is the initial velocity of the object (it was dropped)

a=9.8 m/s^{2} is the acceleration due gravity

d is the height of the tower

t=0.02min=1.2 s is the time it takes to the object to reach the ground

b) Begining with (1):

V_{f}=0+at (3)  

V_{f}=at=(9.8 m/s^{2})(1.2 s) (4)  

V_{f}=11.76 m/s (5)  This is the final velocity of the object

a) Substituting (5) in (2):

(11.76 m/s)^{2}=0+2(9.8 m/s^{2})d (6)  

Clearing d:

d=\frac{(11.76 m/s)^{2}}{2(9.8 m/s^{2})} (7)  

d=7.056 m (8)  This is the height of the tower

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You want to find out how many atoms of the isotope 65Cu are in a small sample of material. You bombard the sample with neutrons
serious [3.7K]

Answer:

a) number of copper atoms 65 (⁶⁵Cu)  is 7.692 10⁶ atoms

b) m_total Cu = 1.585 10⁹ u = 2.632 10⁻¹⁸ kg

Explanation:

a) For this exercise let's start by using the radioactive decay ratio

           N = N₀  e^{- \lambda t}o e - lambda t

The half-life time is defined as the time it takes for half of the radioactive (activated) atoms to decay, therefore after two half-lives there are

            N = ½ (½ N₀) = ¼ N₀

            N₀ = 4 N

in each decay a photon is emitted so we can use a direct rule of proportions. If an atom emits a photon it has Eo = 1,04 Mev, how many photons it has energy E = 10,000 MeV

          # _atoms = 1 atom (photon) (E / Eo)

          # _atoms = 1 10000 / 1.04

          # _atoms = 9615,4 atoms

          N₀ = 4 #_atoms

          N₀ = 4 9615,4

          N₀=  38461.6  atoms

in the exercise indicates that half of the atoms decay in this way and the other half decays directly to the base state of Zinc, so the total number of activated atoms

          N_activated = 2 # _atoms

          N_activated = 2 38461.6

          N_activated = 76923.2

also indicates that 1% = 0.01 of the nuclei is activated by neutron bombardment

          N_activated = 0.01 N_total

          N_total = N_activated / 0.01

          N_total = 76923.2 / 100

          N_total = 7.692 10⁶ atoms

so the number of copper atoms 65 (⁶⁵Cu)  is 7.692 10⁶

b) the natural abundance of copper is

  ⁶³Cu     69.17%

  ⁶⁵Cu    30.83%

Let's use a direct proportion rule. If there are 7.692 10⁶  ⁶⁵Cu that represents 30.83, how much ⁶³Cu is there that represents 69.17%

                # _63Cu = 69.17%  (7.692 10⁶    / 30.83%)

                # _63Cu = 17.258 10⁶  atom  ⁶³Cu

the total amount of comatose is

              #_total Cu = #_ 65Cu + # _63Cu

              #_total Cu = (7.692 + 17.258) 10⁶

              #_total Cu = 24.95 10⁶

the atomic mass of copper is m_Cu = 63.546 u

          m_total = #_totalCu m_Cu

          m_total = 24.95 10⁶ 63,546 u

          m_total = 1.585 10⁹ u

let's reduce to kg

           m_total Cu = 1.585 10⁹ u (1,66054 10⁻²⁷ kg / 1 u)

           m_total Cu = 2.632 10⁻¹⁸ kg

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the average distance to the moon, d_m=238,900\ mi

The factor by which the distance to moon is greater than the width of Atlantic ocean:

f=\frac{d_m}{d_i}

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the distance to the moon greater than the distance across the Atlantic ocean by an order of 1.

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The balloon will collapse

Explanation:

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3 years ago
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