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3 years ago

What is the term for substances that have several unpaired electrons and are strongly magnetic

Physics

1 answer:

Murljashka [212]3 years ago

8 0

Answer:

Scientific definitions for ferromagnetic

The property of being strongly attracted to either pole of a magnet. Ferromagnetic materials, such as iron, contain unpaired electrons, each with a small magnetic field of its own, that align readily with each other in response to an external magnetic field.

Explanation:

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Question 11

notka56 [123]

The volume of the gas once it reaches the surface of water is 2 liters.

The volume of the air in balloon at depth of 100ft (30m) is 500ml.

The pressure at this point is 4 atm.

Assuming that the balloon have no compression by rubber of balloon the volume of air at the water surface is V.

The pressure at the surface of water is 1 atms.

As we know, from the ideal gas equation,

PV = nRT

Where,

P is the Pressure of gas,

V is the volume of the gas,

n is the number of moles,

R is the gas constant whose value is 0.082057 L atm mol-1 K-1,

T is the temperature.

Assuming that the temperature is constant,

We know,

PV = nRT

All quantities on the right side are constants,

So, we can write,

P₁V₁ = P₂V₂

Putting all the the values,

4(0.5) = 1V₂

V₂ = 2 Liters.

The volume of the air at the surface is 2 liters.

1 liter = 0.001 m

Hence,

2 liters = 0.002 m³

So the volume of air at the surface is 0.002m³.

To know more about Ideal gas Equation, visit,

brainly.com/question/20348074

#SPJ9

3 0

1 year ago

Suppose a 49-N sled is resting on packed snow. The coefficient of kinetic friction is 0.11. If a person weighing 585 N sits on t

Andre45 [30]

Answer:

69.74 N

Explanation:

We are given that

Weight of sled=49 N

Coefficient of kinetic friction $,\mu_k=0.11$

Weight of person=585 N

Total weight==mg=49+585=634 N

We know that

Force needed to pull the sled across the snow at constant speed,F=Kinetic friction

$F=\mu_k N$

Where N= Normal=mg

$F=0.11\times 634=69.74 N$

Hence, the force is needed to pull the sled across the snow at constant speed=69.74 N

7 0

3 years ago

He is the chemical symbol for which element? Select one: a. hydrogen b. hassium c. helium d. hafnium

Ludmilka [50]

The answer is C . Helium

7 0

3 years ago

Does the voltage of a battery affect the strength of an electromagnet?

swat32

I'm trying to make an electromagnet that's strength is constantly getting incremented by small amounts every second. I need to know, which would have a greater effect on the electromagnet's strength, amps or volts? (I know increasing the turns and/or density of the magnet wire will increase the strength, but I am looking for answers other than that particular one.)

7 0

3 years ago

A cold beverage can be kept cold even a warm day if it is slipped into a porous ceramic container that has been soaked in water.

Arisa [49]

Answer:

The rate at which the container is losing water is 0.0006418 g/s.

Explanation:

Under the assumption that the can is a closed system, the conservation law applied to the system would be: $E_{in}-E_{out}=E_{change}$ , where $E_{in}$ is all energy entering the system, $E_{out}$ is the total energy leaving the system and, $E_{change}$ is the change of energy of the system.
As the purpose is to kept the beverage can at constant temperature, the change of energy ( $E_{change}$ ) would be 0.
The energy that goes into the system, is the heat transfer by radiation from the environment to the top and side surfaces of the can. This kind of transfer is described by: $Q=\varepsilon*\sigma*A_S*(T_{\infty}^4-T_S^4)$ where $\varepsilon$ is the emissivity of the surface, $\sigma=5.67*10^{-8}\frac{W}{m^2K}$ known as the Stefan–Boltzmann constant, $A_S$ is the total area of the exposed surface, $T_S$ is the temperature of the surface in Kelvin, $T_{\infty}$ is the environment temperature in Kelvin.
For the can the surface area would be ta sum of the top and the sides. The area of the top would be $A_{top}=\pi* r^2=\pi(0.0252m)^2=0.001995m^2$ , the area of the sides would be $A_{sides}=2*\pi*r*L=2*\pi*(0.0252m)*(0.09m)=0.01425m^2$ . Then the total area would be $A_{total}=A_{top}+A_{sides}=0.01624m^2$
Then the radiation heat transferred to the can would be $Q=\varepsilon*\sigma*A_S*(T_{\infty}^4-T_S^4)=1*5.67*10^{-8}\frac{W}{m^2K}*0.01624m^2*((32+273K)^4-(17+273K)^4)=1.456W$ .
The can would lost heat evaporating water, in this case would be $Q_{out}=\frac{dm}{dt}*h_{fg}$ , where $\frac{dm}{dt}$ is the rate of mass of water evaporated and, $h_{fg}$ is the heat of vaporization of the water ( $2257\frac{J}{g}$ ).
Then in the conservation balance: $Q_{in}-Q_{out}=Q_{change}$ , it would be $1.45W-\frac{dm}{dt}*2257\frac{j}{g}=0$ .
Recall that $1W=1\frac{J}{s}$ , then solving for $\frac{dm}{dt}$ : $\frac{dm}{dt}=\frac{1.45\frac{J}{s} }{2257\frac{J}{g} }=0.0006452\frac{g}{s}$

5 0

3 years ago