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wlad13 [49]
2 years ago
5

Complete the equation to show the radioactive decay of carbon-14 to nitrogen-14

Physics
1 answer:
blsea [12.9K]2 years ago
7 0

Answer:

The beta decay takes place.

Explanation:

The reaction of radioactivity of carbon 14 to nitrogen 14 is

There is a beta decay.  

The reaction is

C_{6}^{14}\rightarrow N_{7}^{14}+\beta _{-1}^{0}+ energy

Here some energy is released in form of neutrino.

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1. I think a drum would be the easiest instrument to explain. You can vary the pitch and loudness by hitting it with different intensities or at different points or angles.

PS. I'm not talking about a drumset. I'm talking about an ordinary drum.

2. Different types of instruments sound different even though they play the same notes because they use different mechanisms for producing sound.

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3 years ago
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(b) Find a point between the two charges on the horizontal line where the electric potential is zero. (Enter your answer as meas
aleksley [76]

Complete Question: A charge q1 = 2.2 uC is at a distance d= 1.63m from a second charge q2= -5.67 uC. (b) Find a point between the two charges on the horizontal line where the electric potential is zero. (Enter your answer as measured from q1.)

Answer:

d= 0.46 m

Explanation:

The electric potential is defined as the work needed, per unit charge, to bring a positive test charge from infinity to the point of interest.

For a point charge, the electric potential, at a distance r from it, according to Coulomb´s Law and the definition of potential, can be expressed as follows:

V = \frac{k*q}{r}

We have two charges, q₁ and q₂, and we need to find a point between them, where the electric potential due to them, be zero.

If we call x to the distance from q₁, the distance from q₂, will be the distance between both charges, minus x.

So, we can find the value of x, adding the potentials due to q₁ and q₂, in such a way that both add to zero:

V = \frac{k*q1}{x} +\frac{k*q2}{(1.63m-x)} = 0

⇒k*q1* (1.63m - x) = -k*q2*x:

Replacing by the values of q1, q2, and k, and solving for x, we get:

⇒ x = (2.22 μC* 1.63 m) / 7.89 μC = 0.46 m from q1.

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2 years ago
On Mars, where air resistance is negligible, an astronaut drops a rock from a cliff and notes that the rock falls about d meters
dimulka [17.4K]

Answer:

d_1 = 16 d

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v_i = 0

also the acceleration due to gravity on Mars is g

so we have

d = v_i t + \frac{1}{2}gt^2

now we have

d = 0 + \frac{1}{2}g t^2

now if the same is dropped for 4t seconds of time

then again we will use above equation

d_1 = 0 + \frac{1}{2}g(4t)^2

d_1 = 16(\frac{1}{2}gt^2)

d_1 = 16 d

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LiRa [457]

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