Which statement is not true about radiation

Runner x starts 50 feet in front of runner z and moves with a velocity of 8.0 m/s. At what speed would runner z have to run to c

Mariulka [41]

Answer:

15.625

Explanation:

8 0

3 years ago

Insert

nordsb [41]

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:

$A\propto N$

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:

$A\propto m$

where m is the mass of the sample.

In this problem:

- When the mass is $m_1 = 1 g$ , the activity is $A_1=16 count/min$

- When the mass is $m_2=20 g$ , the activity is $A_2$

So we can find A2 by using the rule of three:

$\frac{A_1}{m_1}=\frac{A_2}{m_2}\\A_2=A_1 \frac{m_2}{m_1}=(16)\frac{20}{1}=320 count/min$

B)

The equation describing the activity of a radioactive sample as a function of time is:

$A(t)= A_0 e^{-\lambda t}$ (1)

where

$A_0$ is the initial activity at time t = 0

t is the time

$\lambda$ is the decay constant, which gives the probability of decay

The decay constant can be found using the equation

$\lambda = \frac{ln2}{t_{1/2}}$

where $t_{1/2}$ 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

$t_{1/2}=5700 y$

So its decay constant is

$\lambda=\frac{ln2}{5700}=1.22\cdot 10^{-4} y^{-1}$

We also know that the tree died

t = 17,100 years ago

and that the initial activity was

$A_0 = 320 count/min$ (value calculated in part A, corresponding to a mass of 20 g)

So, substituting into eq(1), we find the new activity:

$A(17,100) = (320)e^{-(1.22\cdot 10^{-4})(17,100)}=40 count/min$

C)

We know that a sample of living wood has an activity of

$A=16 count/min$ per 1 g of mass.

Here we have 5 g of mass, therefore the activity of the sample when it was living was:

$A_0 = A\cdot 5 = (16)(5)=80 count/min$

Moreover, here we have a sample of 5 g, with current activity of $A=20 count/min$ : it means that its activity per gram of mass is

$A'=\frac{20}{5}=4 count/min$

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:

$t=2t_{1/2}=2(5700)=11,400 y$

3 0

4 years ago

Which of the following is a difference between sound and light?

inn [45]

Answer:

4

Explanation:

Humans can detect light and sound, also light and sound travel in waves. Light is faster than sound but sound waves cannot travel through space because there is no medium for the sound waves to travel through.

3 0

3 years ago

Read 2 more answers

Two drag cars race. They line up at the starting line at rest. The winning car accelerates at a constant rate a and reaches the

Mila [183]

Answer: $d=\frac{v^2}{8a}$

Explanation:

Given

winning car accelerates with a and its final velocity is v

considering they both start from rest

time taken by winning car is

v=u+at

where u=initial velocity

a=acceleration

t=time

$v=at$

$t=\frac{v}{a}$

Now loosing car is accelerating with $\frac{a}{4}$

Distance traveled by loosing car in time t

$s_1=ut+\frac{at^2}{2\cdot 4}$

$s_1=0+\frac{a}{8}\times (\frac{v}{a})^2$

$s_1=\frac{v^2}{8a}$

Thus distance d traveled by loosing car is given by $d=\frac{v^2}{8a}$

5 0

4 years ago

What might happen if water molecules did not have a slight negative charge on one end and a slight positive charge on the other?

Katena32 [7]

Answer:

It would not be possible the cohesion among water molecules by the polar covalent bonding.

Well, to understand this in a better way, let's begin by explaining that water is special due to its properties, which makes this fluid useful for many purposes and for the existence of life.

In this sense, one of the main properties of water is cohesion (molecular cohesion), which is the attraction of molecules to others of the same type. So, water molecule ( $H_{2}O$ ) has 2 hydrogen atoms attached to 1 oxygen atom and can stick to itself through hydrogen bonds.

How is this possible?

By the polar covalent bonding, a process in which electrons are shared unequally between atoms, due to the unequal distribution of electrons between atoms of different elements. In other words: slightly positive and slightly negative charges appear in different parts of the molecule.

Now, it can be said that a water molecule has a negative side (oxygen) and a positive side (hydrogen). This is how the oxygen atom tends to monopolize more electrons and keeps them away from hydrogen. Thanks to this polarity, water molecules can stick together.

5 0

4 years ago