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2 years ago

Set local ground level to 700 ft, and record the inches of mercury.

1 answer:

6 0

The main way in which an altimeter measures the altitude of an object is by calculating the location's air pressure.

<h3>What is an Altimeter?</h3>

This refers to the instrument that is used to measure the altitude of an object when it is at a fixed level.

Hence, we can see that an altimeter should NOT be confused with a barometer, because although they both measure pressure, a barometer calculates the change in air pressure and elevation based on available weather.

Please note that your question is incomplete so I gave you a general overview to help you get a better understanding of the concept.

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What is the significant figure of 0.00000006

Maksim231197 [3]

<span>1 of them i think try it out</span>

3 0

3 years ago

Given two vectors A⃗ =4.00i^+7.00j^ and B⃗ =5.00i^−2.00j^ , find the vector product A⃗ ×B⃗ (expressed in unit vectors).

Reptile [31]

Answer:

$-43\hat{k}$

Explanation:

given,

$\vec{A} = 4 \hat{i} + 7 \hat{j}$

$\vec{B} = 5 \hat{i} - 2 \hat{j}$

vector product $\vec{A} \times \vec{B} = ?$

$\vec{A} \times \vec{B}$ = $\begin{bmatrix}i & j & k\\ 4 & 7 &0 \\ 5 & -2 & 0\end{bmatrix}$

now, expanding the vector

$\vec{A} \times \vec{B}= \hat{k}(-2\times 4 - 7\times 5)$

$\vec{A} \times \vec{B}= -43\hat{k}$

the vector product is equal to $-43\hat{k}$

4 0

4 years ago

Insert

nordsb [41]

A) 320 count/min

B) 40 count/min

C) 80 count/min, 11400 years

Explanation:

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:

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

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:

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

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:

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