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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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ohaa [14]

I believe your answer is correct, because 8.7*10^-7 is equal to 0.00000085347.

Hope you do well!

4 0
3 years ago
What are the main parts of metabolism
meriva

The main parts of metabolism includes the two components which are: <em>Catholicism</em>, and <em>Anabolism</em>.

7 0
3 years ago
I need help!!!! I don't understand the question
Rudiy27
Mgraphs are proportional ofrelationships
8 0
4 years ago
Professional baseball pitchers deliver pitches that can reach the blazing speed of 100 mph (miles per hour). A local team has dr
Wittaler [7]

Answer:

A. ) K =126. 7 J

B. ) h= 91.1 m.

Explanation:

A)

  • Assuming no air resistance, once released by the pitcher, the speed must keep constant through all the trajectory, so the kinetic energy of the ball can be expressed as follows:

       K = \frac{1}{2}*m*v^{2}  =  \frac{1}{2}*0.142 kg*(42.24m/s)^{2} = 126.7 J (1)

B)

  • Neglecting air resistance, total mechanical energy must be the same at any point, so, if we choose the ground level as the zero reference level for the gravitational potential energy, and assuming that the ball attains this kinetic energy just before striking ground, this value must be equal to the gravitational potential energy just before be dropped, so we can write the following equality:

        U_{o} = K_{f} = 126. 7 J (2)

        ⇒ m*g*h = 126. 7 J

  • Solving for h, we get:

       h = \frac{K_{f}}{m*g} = \frac{126.7J}{0.1420kg*9.8m/s2} = 91.1 m (3)

4 0
3 years ago
Two planets have the same surface gravity, but planet b has twice the radius of planet
nevsk [136]
The formula for the acceleration due to gravity is:

a = Gm/r²
where
G is the universal gravitational constant = 6.6726 x 10⁻¹¹ N-m²/kg²
m is the mass of planet
r is the radius of planet

So, if they have the same a:

m₁/r₁² = m₂/r₂²
So, if m₁ = m and r₂ = 2r₁,
m/r₁² = m₂/(2r₁)²
m₂ = 4m

<em>Thus, the answer is D.</em>
8 0
4 years ago
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