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2 years ago

Iron is a 1. Compound 2.an element 3.a mixture

Physics

2 answers:

Tcecarenko [31]2 years ago

8 0

Answer:

Iron is a element

Explanation:

a substance that cannot be broken down into simpler substances by chemical means, and is characterized by its atomic number, , which represents the number of positively charged particles within that element's nucleus

ziro4ka [17]2 years ago

4 0

Answer:

compound

Explanation:

Iron is a mineral that our bodies need for many functions.

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Two geological field teams are working in a remote area. A global positioning system (GPS) tracker at their base camp shows the

Masja [62]

Answer:

distance of 2nd team from 1st team will be: 58.2

Direction of 2nd team from 1st team will be: 14.90 deg North of east

Explanation:

ASSUME Vector is R and makes angle A with +x-axis,

therefore component of vector R is

$R_x = Rcos A$

$R_y = Rsin A$

From above relation

Assuming base camp as the origin, location of 1st team is

$R_1 = 37 km$ away at 21 deg North of west (North of west is in 2nd quadrant, So x is -ve and y is positive)

$R_{1x} = -R_1*cos A_1 = -37*cos 21 deg = -34.54 km$

$R_{1y} = R_1*sin A_1 = 37*sin 21 deg = 13.25 km$

location of 2nd team is at

$R_2 = 32 km$ , at 38 deg East of North = 32 km, at 58 deg North of east (North of east is in 1st quadrant, So x and y both are +ve)

$R_{2x} = R_2*cos A_2 = 32*cos 58 deg = 16.95 km$

$R_{2y} = R_2*sin A_2 = 32*sin 58 deg = 27.13 km$

Now position of 2nd team with respect to 1st team will be given by:

$R_3 = R_2 - R_1$

$R_3 = (R_{2x} - R_{1x}) i + (R_{2y} - R_{1y}) j$

Using above values:

$R_3 = (16.95 - (-34.54)) i + (27.13 - 13.42) j$

$R_3 = 51.49 i + 13.71 j$

distance of 2nd team from 1st team will be:

$\left | R_3 \right | = \sqrt (51.49^2 +13.71^2)$

$\left | R_3 \right | = 53.28 km = 58.2 km$

Direction of 2nd team from 1st team will be:

$Direction = tan^{-1} \frac{R_{3y}}{R_{3x}} = tan^{-1}[ \frac{13.71}{51.49}]$

Direction = 14.90 deg North of east

6 0

2 years ago

Arm ab has a constant angular velocity of 16 rad/s counterclockwise. At the instant when theta = 60

geniusboy [140]

The <em>linear</em> acceleration of collar D when <em>θ = 60°</em> is - 693.867 inches per square second.

<h3>How to determine the angular velocity of a collar</h3>

In this question we have a system formed by three elements, the element AB experiments a <em>pure</em> rotation at <em>constant</em> velocity, the element BD has a <em>general plane</em> motion, which is a combination of rotation and traslation, and the ruff experiments a <em>pure</em> translation.

To determine the <em>linear</em> acceleration of the collar ( $a_{D}$ ), in inches per square second, we need to determine first all <em>linear</em> and <em>angular</em> velocities ( $v_{D}$ , $\omega_{BD}$ ), in inches per second and radians per second, respectively, and later all <em>linear</em> and <em>angular</em> accelerations ( $a_{D}$ , $\alpha_{BD}$ ), the latter in radians per square second.

By definitions of <em>relative</em> velocity and <em>relative</em> acceleration we build the following two systems of <em>linear</em> equations:

<h3>Velocities</h3>

$v_{D} + \omega_{BD}\cdot r_{BD}\cdot \sin \gamma = -\omega_{AB}\cdot r_{AB}\cdot \sin \theta$ (1)

$\omega_{BD}\cdot r_{BD}\cdot \cos \gamma = -\omega_{AB}\cdot r_{AB}\cdot \cos \theta$ (2)

<h3>Accelerations</h3>

$a_{D}+\alpha_{BD}\cdot \sin \gamma = -\omega_{AB}^{2}\cdot r_{AB}\cdot \cos \theta -\alpha_{AB}\cdot r_{AB}\cdot \sin \theta - \omega_{BD}^{2}\cdot r_{BD}\cdot \cos \gamma$ (3)

$-\alpha_{BD}\cdot r_{BD}\cdot \cos \gamma = - \omega_{AB}^{2}\cdot r_{AB}\cdot \sin \theta + \alpha_{AB}\cdot r_{AB}\cdot \cos \theta - \omega_{BD}^{2}\cdot r_{BD}\cdot \sin \gamma$ (4)

If we know that $\theta = 60^{\circ}$ , $\gamma = 19.889^{\circ}$ , $r_{BD} = 10\,in$ , $\omega_{AB} = 16\,\frac{rad}{s}$ , $r_{AB} = 3\,in$ and $\alpha_{AB} = 0\,\frac{rad}{s^{2}}$ , then the solution of the systems of linear equations are, respectively:

<h3>Velocities</h3>

$v_{D}+3.402\cdot \omega_{BD} = -41.569$ (1)

$9.404\cdot \omega_{BD} = -24$ (2)

$v_{D} = -32.887\,\frac{in}{s}$ , $\omega_{BD} = -2.552\,\frac{rad}{s}$

<h3>Accelerations</h3>

$a_{D}+3.402\cdot \alpha_{BD} = -445.242$ (3)

$-9.404\cdot \alpha_{BD} = -687.264$ (4)

$a_{D} = -693.867\,\frac{in}{s^{2}}$ , $\alpha_{BD} = 73.082\,\frac{rad}{s^{2}}$

The <em>linear</em> acceleration of collar D when <em>θ = 60°</em> is - 693.867 inches per square second. $\blacksquare$

<h3>Remark</h3>

The statement is incomplete and figure is missing, complete form is introduced below:

<em>Arm AB has a constant angular velocity of 16 radians per second counterclockwise. At the instant when θ = 60°, determine the acceleration of collar D.</em>

To learn more on kinematics, we kindly invite to check this verified question: brainly.com/question/27126557

5 0

1 year ago

You place ammonium nitrate crystals in water and stirred. As you do so, the container becomes cold to the touch. This an example

ankoles [38]

The process of flask becoming cold is due to endothermic reaction.

Answer: Option B

<u>Explanation:</u>

So two kinds of heat transfer can be possible in any chemical reaction. If the sample is considered as system and the sample container is considered as the surrounding, then heat transfer can occur between them.

If the heat is transferred from the surrounding to the system , then it is an endothermic reaction. And in those cases, the sample holder will be becoming colder. This is because the heat from the surrounding that is the container will be utilized to complete the reaction.

While when there is transfer of heat from the system to surrounding , it will be exothermic reaction and the beaker will be getting hot in this process. So in the present case, the container is becoming cold because of occurrence of endothermic process.

8 0

3 years ago

A thermos bottle works well because:

baherus [9]

Answer:

A thermos bottle works well because:

A) Its glass walls are thin

7 0

3 years ago

Read 2 more answers

What is the energy stored between 2 Carbon nuclei that are 1.00 nm apart from each other? HINT: Carbon nuclei have 6 protons and

Andrei [34K]

Answer:

$A. 8.29\times 10^{-18}\ J$

Explanation:

Given that:

p = magnitude of charge on a proton = $1.6\times 10^{-19}\ C$

k = Boltzmann constant = $9\times 10^{9}\ Nm^2/C^2$

r = distance between the two carbon nuclei = 1.00 nm = $1.00\times 10^{-9}\ m$

Since a carbon nucleus contains 6 protons.

So, charge on a carbon nucleus is $q = 6p=6\times 1.6\times 10^{-19}\ C=9.6\times 10^{-19}\ C$

We know that the electric potential energy between two charges q and Q separated by a distance r is given by:

$U = \dfrac{kQq}{r}$

So, the potential energy between the two nuclei of carbon is as below:

$U= \dfrac{kqq}{r}\\\Rightarrow U = \dfrac{kq^2}{r}\\\Rightarrow U = \dfrac{9\times10^9\times (9.6\times 10^{-19})^2}{1.0\times 10^{-9}}\\\Rightarrow U =8.29\times 10^{-18}\ J$

Hence, the energy stored between two nuclei of carbon is $8.29\times 10^{-18}\ J$ .

8 0

2 years ago