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Lesechka [4]
3 years ago
8

60m27Co → 6027Co Predict the type of radioactive emission produced from the decay of metastable cobalt-60 to cobalt-60. Describe

this type of emission and its reaction to an electric field. A) During the radioactive decay, alpha particles are released. These positive particles are attracted to the negative plate in the electric field and represent a ground energy level. B) Beta particles are released during the radioactive decay. These negative particles are attracted to the positive plate in the electric field and represent an excited energy state. C) Radioactive gamma decay is produced by the reaction. This neutral electromagnetic radiation allows the isotope to return to its ground state and is not attracted to the electric field. D) Both types of radioactive emissions, particles and electromagnetic radiation, are produced during this decay. None of these are attracted to the electric field and both present an intermediate level. Eliminate
Physics
1 answer:
lara [203]3 years ago
4 0
Some one already asked this question and you can copy paste and google it but I believe it is c you may want to double check
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What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1
notka56 [123]

Complete Question

Part of the question is shown on the first uploaded image

The rest of the question

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1 and q2 at x3 = -1.220 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures.

Answer:

The net force exerted on the third charge is  F_{net}=  3.22*10^{-5} \ J

Explanation:

From the question we are told that

    The third charge is  q_3 =  55 nC =  55 *10^{-9} C

    The position of the third charge is  x = -1.220 \ m

     The first charge is q_1 =  -16 nC  =  -16 *10^{-9} \ C

     The position of the first charge is x_1 =  -1.650m

      The second charge is  q_2 =  32 nC  =  32 *10^{-9} C

      The position of the second charge is  x_2 =   0  \ m  

The distance between the first and the third charge is

      d_{1-3} =  -1.650 -(-1.220)

     d_{1-3} = -0.43 \ m

The force exerted on the third charge by the first is  

     F_{1-3} =  \frac{k  q_1 q_3}{d_{1-3}^2}

Where k is the coulomb's constant with a value  9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.

substituting values

      F_{1-3} =  \frac{9*10^{9}* 16 *10^{-9} * (55*10^{-9})}{(-0.43)^2}

       F_{1-3} = 4.28 *10^{-5} \ N

 The distance between the second and the third charge is      

  d_{2-3} =  0- (-1.22)

   d_{2-3} =1.220 \ m

The force exerted on the third charge by the first is mathematically evaluated as

       F_{2-3} =  \frac{k  q_2 q_3}{d_{2-3}^2}

substituting values

       F_{2-3} =  \frac{9*10^{9} * (32*10^{-9}) *(55*10^{-9})}{(1.220)^2}

       F_{2-3} =  1.06*10^{-5} N

The net force is

      F_{net} =  F_{1-3} -F_{2-3}

substituting values

    F_{net} = 4.28 *10^{-5} - 1.06*10^{-5}

    F_{net}=  3.22*10^{-5} \ J

6 0
2 years ago
Consider a 2.54-cm-diameter power line for which the potential difference from the ground, 19.6 m below, to the power line is 11
tiny-mole [99]

Answer:

The line charge density is 1.59\times10^{-4}\ C/m

Explanation:

Given that,

Diameter = 2.54 cm

Distance = 19.6 m

Potential difference = 115 kV

We need to calculate the line charge density

Using formula of potential difference

V=EA

V=\dfrac{\lambda}{2\pi\epsilon_{0}r}\times\pi r^2

\lambda=\dfrac{V\times2\epsilon_{0}}{r}

Where, r = radius

V = potential difference

Put the value into the formula

\lambda=\dfrac{115\times10^{3}\times2\times8.8\times10^{-12}}{1.27\times10^{-2}}

\lambda=1.59\times10^{-4}\ C/m

Hence, The line charge density is 1.59\times10^{-4}\ C/m

4 0
3 years ago
Manipulate the equation "v=d/t" to find the answers to these problems using
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GarryVolchara [31]
High energy waves have Gamma rays
3 0
2 years ago
Read 2 more answers
Why might a major volcanic eruption lead to cooler global temperatures? Explain your answer.
irina [24]

Answer:

Volcanic eruptions cool down  the planet

Explanation:

Volcanic eruptions actually cool the planet because the particles ejected from volcanoes shade incoming solar radiation. ... The small ash and aerosol particles decrease the amount of sunlight reaching the surface of the Earth and lower average global temperatures.

Hope this helps!!! :D

4 0
2 years ago
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