What statement most accurately classifies this cell?

6) K2O Use IUPAC nomenclature rules to properly identify this compound. A) potassium oxide B) dipotassium oxide C) potassium (I)

tino4ka555 [31]

NO,,The answer is A) potassium oxide

4 0

3 years ago

1. Ignore friction and determine an expression for the distance d the boxes travel before coming to rest. Choose the floor as th

Nadusha1986 [10]

Newton's second law allows us to find the results for the displacement and work on box 1 are:

1) The displacement is $x = \frac{v_o^2 m_2}{2(m_1+m_2)} g$

2) The work is $W= \frac{1}{2} m_1v_o^2$

1) Newton's second law says that force is directly proportional to the mass and acceleration of bodies.

F = ma

Where the bold letters indicate vectors, F is the force, m the mass and the acceleration of the bodies

The reference system is a coordinate system with respect to which the decompositions of the forces are carried out and all the measurements are made, in this case we take a system where the x axis is horizontal, the y axis is vertical and the zero of the system is located at soil

In the attached we can see a free body diagram of the system where the external forces are indicated, let's apply Newton's second law to body 1

body 1

x-axis

-T = m₁ a

body 2

y-axis

T - W₂ = m₂ a

W₂ = m₂ g

let's solve the system

- m₂ g = (m₁ + m₂) a

a = $- \frac{m_2}{m_1+m_2} \ g$ - m2 / m1 + m2 g

Kinematics studies the movement of bodies, let's use the expression

v² = $v_o^2$ - 2 a x

When the body stops the velocity is zero

x = $\frac{v_o^2}{2a}$

We substitute

$x = \frac{v_o^2}{2} \frac{m_2}{m_1+m_2} \ g$

Since the two boxes are connected by a rope, they both travel the same distance

2) They ask for Tension work on box 1

Work is defined by the scalar product of force and displacement

W = F. d

Where the bold letters indicate vectors, W is the work that is a scalar, F the force and d the displacement

We can write this expression by developing the dot product

W = F d cos θ

Where θ is the angle between force and displacement.

In this case the force is the tension of the rope, from the attached graph we see that the force is directed to the right and the displacement is to the left, therefore the angle is 180º and the cos 180 is equal to -1

We look for the tension from Newton's second law for box 1

T = - m₁ a

Let's substitute

T = $- m_1 \ ( - \frac{m_2}{m_1+m_2} ) \ g$

T = $\frac{m_1m_2}{m_1+m_2} \ g$

We calculate the work

$W = - \frac{m_1m_2}{m_1+m_2 } \ g ( \frac{v_o^2 (m1+m_2)}{2 m_2 g} )$

W = - ½ m₁ v₀²

In conclusion using Newton's second law we can find the results for the displacement and work on box 1 are:

1) The displacement is x= $\frac{v_o^2 m_2}{2(m_1+m_2)} g$

2) The work is W = - ½ m₁ v₀²

Learn more here: brainly.com/question/17290735

3 0

2 years ago

An ore sample weighs 17.50 N in air. When the sample is suspended by a light cord and totally immersed in water, the tension in

valkas [14]

Answer:

Volume of the sample: approximately $\rm 0.6422 \; L = 6.422 \times 10^{-4} \; m^{3}$ .

Average density of the sample: approximately $\rm 2.77\; g \cdot cm^{3} = 2.778 \times 10^{3}\; kg \cdot m^{3}$ .

Assumption:

$\rm g = 9.81\; N \cdot kg^{-1}$ .
$\rho(\text{water}) = \rm 1.000\times 10^{3}\; kg \cdot m^{-3}$ .
Volume of the cord is negligible.

Explanation:

<h3>Total volume of the sample</h3>

The size of the buoyant force is equal to $\rm 17.50 - 11.20 = 6.30\; N$ .

That's also equal to the weight (weight, $m \cdot g$ ) of water that the object displaces. To find the mass of water displaced from its weight, divide weight with $g$ .

$\displaystyle m = \frac{m\cdot g}{g} = \rm \frac{6.30\; N}{9.81\; N \cdot kg^{-1}} \approx 0.642\; kg$ .

Assume that the density of water is $\rho(\text{water}) = \rm 1.000\times 10^{3}\; kg \cdot m^{-3}$ . To the volume of water displaced from its mass, divide mass with density $\rho(\text{water})$ .

$\displaystyle V(\text{water displaced}) = \frac{m}{\rho} = \rm \frac{0.642\; kg}{1.000\times 10^{3}\; kg \cdot m^{-3}} \approx 6.42201 \times 10^{-4}\; m^{3}$ .

Assume that the volume of the cord is negligible. Since the sample is fully-immersed in water, its volume should be the same as the volume of water it displaces.

$V(\text{sample}) = V(\text{water displaced}) \approx \rm 6.422\times 10^{-4}\; m^{3}$ .

<h3>Average Density of the sample</h3>

Average density is equal to mass over volume.

To find the mass of the sample from its weight, divide with $g$ .

$\displaystyle m = \frac{m \cdot g}{g} = \rm \frac{17.50\; N}{9.81\; N \cdot kg^{-1}} \approx 1.78389 \; kg$ .

The volume of the sample is found in the previous part.

Divide mass with volume to find the average density.

$\displaystyle \rho(\text{sample, average}) = \frac{m}{V} = \rm \frac{1.78389\; kg}{6.42201 \times 10^{-4}\; m^{3}} \approx 2.778\; kg \cdot m^{-3}$ .

3 0

4 years ago

Hey plz help me<br>suggest a situations where you might need to use ear defenders

xenn [34]

You would need to ear defenders (I'm assuming these are ear protectors for use with loud sounds) while firing at a gun range. You could use it for loud construction areas.

4 0

3 years ago

What is the mass of an object that weigh 150 N on earth

aev [14]

450 g =0.9921 lb I hope this helps you out.

8 0

3 years ago