Conventionally, the field strength around a charged object is the direction of the force acting on a .
2 answers:
Answer:
- <u>Field Strength: </u>
Any charged body has the capacity to effect any test charge that comes inside its field or region. It can be also defined as,
"The total amount or magnitude of force,F or intensity felt by any unit test charge when it enters an electromagnetic or electric field."
- For an electric field its unit will be, <u>volt per meter or simply V/m.</u>
Explanation:
<u>A unit test charge inside an electric field:</u>
When a unit test charge enters a given parameters or area set by the charged particle then it will surely experience a force,F equal to the magnitude of that charge body and it will be different as it continues to move closer or far inside the field.
- As, we have, "E=F/q",(where "F" is the field strength and "q" is the unit test charge placed inside the field).
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A) 320 count/min
B) 40 count/min
C) 80 count/min, 11400 years
Explanation:
A)
The activity of a radioactive sample is the number of decays per second in the sample.
The activity of a sample is therefore directly proportional to the number of nuclei in the sample:
where A is the activity and N the number of nuclei.
As a consequence, since the number of nuclei is proportional to the mass of the sample, the activity is also directly proportional to the mass of the sample:
where m is the mass of the sample.
In this problem:
- When the mass is , the activity is
- When the mass is , the activity is
So we can find A2 by using the rule of three:
B)
The equation describing the activity of a radioactive sample as a function of time is:
(1)
where
is the initial activity at time t = 0
t is the time
is the decay constant, which gives the probability of decay
The decay constant can be found using the equation
where is the half-life, which is the amount of time it takes for the radioactive sample to halve its activity.
In this problem, carbon-14 has half-life of
So its decay constant is
We also know that the tree died
t = 17,100 years ago
and that the initial activity was
(value calculated in part A, corresponding to a mass of 20 g)
So, substituting into eq(1), we find the new activity:
C)
We know that a sample of living wood has an activity of
per 1 g of mass.
Here we have 5 g of mass, therefore the activity of the sample when it was living was:
Moreover, here we have a sample of 5 g, with current activity of : it means that its activity per gram of mass is
We know that the activity halves after every half-life: Here the activity has became 1/4 of the original value, this means that 2 half-lives have passed, because:
- After 1 half-life, the activity drops from 16 count/min to 8 count/min
- After 2 half-lives, the activity dropd to 4 count/min
So the age of the wood is equal to 2 half-lives, which is: