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Inessa05 [86]
3 years ago
13

ASAPP!!!!!!

Physics
1 answer:
lord [1]3 years ago
5 0
I would say the plastic grip because glass, wood, and plastic are all good conductors of electricity
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A tree falls in a forest. How many years must pass before the 14C activity in 1.03 g of the tree's carbon drops to 1.02 decay pe
Illusion [34]

Answer:

t = 5.59x10⁴ y

Explanation:

To calculate the time for the ¹⁴C drops to 1.02 decays/h, we need to use the next equation:

A_{t} = A_{0}\cdot e^{- \lambda t}    (1)

<em>where A_{t}: is the number of decays with time, A₀: is the initial activity, λ: is the decay constant and t: is the time.</em>

To find A₀ we can use the following equation:  

A_{0} = N_{0} \lambda   (2)

<em>where N₀: is the initial number of particles of ¹⁴C in the 1.03g of the trees carbon </em>

From equation (2), the N₀ of the ¹⁴C in the trees carbon can be calculated as follows:        

N_{0} = \frac{m_{T} \cdot N_{A} \cdot abundance}{m_{^{12}C}}

<em>where m_{T}: is the tree's carbon mass, N_{A}: is the Avogadro's number and m_{^{12}C}: is the ¹²C mass.  </em>

N_{0} = \frac{1.03g \cdot 6.022\cdot 10^{23} \cdot 1.3\cdot 10^{-12}}{12} = 6.72 \cdot 10^{10} atoms ^{14}C    

Similarly, from equation (2) λ is:

\lambda = \frac{Ln(2)}{t_{1/2}}

<em>where t 1/2: is the half-life of ¹⁴C= 5700 years </em>

\lambda = \frac{Ln(2)}{5700y} = 1.22 \cdot 10^{-4} y^{-1}

So, the initial activity A₀ is:  

A_{0} = 6.72 \cdot 10^{10} \cdot 1.22 \cdot 10^{-4} = 8.20 \cdot 10^{6} decays/y    

Finally, we can calculate the time from equation (1):

t = - \frac{Ln(A_{t}/A_{0})}{\lambda} = - \frac {Ln(\frac{1.02decays \cdot 24h \cdot 365d}{1h\cdot 1d \cdot 1y \cdot 8.20 \cdot 10^{6} decays/y})}{1.22 \cdot 10^{-4} y^{-1}} = 5.59 \cdot 10^{4} y              

I hope it helps you!

4 0
3 years ago
According to the law of conservation of momentum, objects will either stick together or bounce off of each other when they colli
Tanzania [10]
F - False.

The law of conservation of momentum states that the total momentum is conserved.
3 0
3 years ago
Why is it not necessary for radio telescope surfaces to be as smooth as a mirror?
Law Incorporation [45]
It doesn't on account of radio waves are longer than optical waves. Radio waves are a sort of electromagnetic radiation with wavelengths in the electromagnetic range longer than infrared light. These long waves are in the radio locale of the electromagnetic range.
8 0
3 years ago
What is the speed of a wave that has a frequency of 125 Hz and a wavelength of 1.25 meters?
kirill115 [55]
Using the "v = f. λ" <span>equation...
 
Your "v" or </span>velocity = 156.25 meters/second
5 0
3 years ago
Reception devices pick up the variation in the electric field vector of the electromagnetic wave sent out by the satellite. Give
Keith_Richards [23]

Complete Question

A satellite in geostationary orbit is used to transmit data via electromagnetic radiation. The satellite is at a height of 35,000 km above the surface of the earth, and we assume it has an isotropic power output of 1 kW (although, in practice, satellite antennas transmit signals that are less powerful but more directional).

Reception devices pick up the variation in the electric field vector of the electromagnetic wave sent out by the satellite. Given the satellite specifications listed in the problem introduction, what is the amplitude E0 of the electric field vector of the satellite broadcast as measured at the surface of the earth? Use ϵ0=8.85×10^−12C/(V⋅m) for the permittivity of space and c=3.00×10^8m/s for the speed of light.

Answer:

The electric field vector of the satellite broadcast as measured at the surface of the earth is  E_o = 6.995 *10^{-6} \ V/m

Explanation:

From the question we are told that

     The height of the satellite is  r  = 35000 \ km  =  3.5*10^{7} \ m

      The power output of the satellite is P  = 1 \ KW  =  1000 \ W

       

Generally the intensity of the electromagnetic radiation of the satellite at the surface of the earth is  mathematically represented as  

     I  =  \frac{P}{4 \pi r^2}

substituting values

      I  =  \frac{1000}{4 * 3.142 (3.5*10^{7})^2}

      I  = 6.495*10^{-14} \  W/m^2

This intensity of the electromagnetic radiation of the satellite at the surface of the earth can also be   mathematically represented as  

          I  =  c * \epsilon_o * E_o^2

Where E_o is the amplitude of the electric field vector of the satellite broadcast so

         E_o =  \sqrt{\frac{2 * I}{c * \epsilon _o} }

substituting values

          E_o =  \sqrt{\frac{2 * 6.495 *10^{-14}}{3.0 *10^{8} * 8.85*10^{-12}} }

           E_o = 6.995 *10^{-6} \ V/m

 

   

4 0
3 years ago
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