Answer:
1. 18.25 m/s
2. 0 m/s
Explanation:
1.So the centripetal acceleration of the ball at this lowest point must be, taking gravity into account
The speed at this point would then be
2. Similarly, if T = mg, then the centripetal acceleration must be
As the ball has no centripetal acceleration, its speed must also be 0 as well.
Answer:
the energy when it reaches the ground is equal to the energy when the spring is compressed.
Explanation:
For this comparison let's use the conservation of energy theorem.
Starting point. Compressed spring
Em₀ = K_e = ½ k x²
Final point. When the box hits the ground
Em_f = K = ½ m v²
since friction is zero, energy is conserved
Em₀ = Em_f
1 / 2k x² = ½ m v²
v = x
Therefore, the energy when it reaches the ground is equal to the energy when the spring is compressed.
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: