You have an initial velocity of -3.0 m/s. You then experience an acceleration of 2.5 m/s2 for 9.0s; what

Suppose that the dipole moment associated with an iron atom of an iron bar is 2.8 × 10-23 J/T. Assume that all the atoms in the

masya89 [10]

To solve this exercise it is necessary to apply the equations related to the magnetic moment, that is, the amount of force that an image can exert on the electric currents and the torque that a magnetic field exerts on them.

The diple moment associated with an iron bar is given by,

$\mu = \alpha *N$

Where,

$\alpha =$ Dipole momento associated with an Atom

N = Number of atoms

$\alpha$ y previously given in the problem and its value is 2.8*10^{-23}J/T

$L = 5.8cm = 5.8*10^{-2}m$

$A = 1.5cm^2 = 1.5*10^{-4}m^2$

The number of the atoms N, can be calculated as,

$N = \frac{\rho AL}{M_{mass}}*A_n$

Where

$\rho =$ Density

$M_{mass} =$ Molar Mass

A = Area

L = Length

$A_n =$ Avogadro number

$N = \frac{(7.9g/cm^3)(1.5cm)(5.8cm^2)}{55.9g/mol}(6.022*10^{23}atoms/mol)$

$N = 7.4041*10^{23}atoms$

Then applying the equation about the dipole moment associated with an iron bar we have,

$\mu = \alpha *N$

$\mu = (2.8*10^{-23})*(7.4041*10^{23})$

$\mu = 20.72Am^2$

PART B) With the dipole moment we can now calculate the Torque in the system, which is

$\tau = \mu B sin(90)$

$\tau = (20.72)(2.2)$

$\tau = 45.584N.m$

<em>Note: The angle generated is perpendicular, so it takes 90 ° for the calculation made.</em>

3 0

3 years ago

A 100-W lightbulb is placed in a cylinder equipped with a moveable piston. The lightbulb is turned on for 0.010 hour, and the as

Taya2010 [7]

Answer:

$w = - 508.53$ joules

$q = - 3091.47$ joules

Explanation:

Let us convert the time in hours into seconds

$0.010* 3600\\= 36$

Change in internal energy

$\delta E = p * \delta t$

where E is the internal energy in Joules

p is the power in watts

and t is the time in seconds

$\delta E = - 100 * 36\\$

$\delta E = - 3600$ Joules

Amount of work done by the system

$w = - P * \delta V$

where P is the pressure and V is the volume

Substituting the given values in above equation, we get -

$w = - 1 * ( 5.92 -0.90)\\$

$w = -5.02$ liter-atmospheres

Work done in Joules

$- 5.02 * 101.3\\$ $= 508.53$ Joules

$q = \delta E - w\\$

Substituting the given values we get -

$q = - 3600 - (-508.53)\\q = - 3091.47$

Thus

$w = - 508.53$ joules

$q = - 3091.47$ joules

7 0

3 years ago

CClassify each characteristic of sound waves.

Alex787 [66]

Explanation : Explain each characteristic of sound waves.

Intensity : the intensity of the sound wave is understand as the power carry by sound wave per unit area in the direction perpendicular to that area.

Loudness : loudness is the quality of the loud and soft of the sound wave.

Frequency : Human normal hear sound frequency between 20 Hz to kHz.

Pitch : Pitch is the quality of low and high of sound wave . pitch relates to the frequency of the slowest vibration in the sound wave for simple sound.

4 0

3 years ago

Read 2 more answers

A laser is shone through a double slit and a particular interference pattern is observed on a screen some distance away. If the

Grace [21]

If the separation between the openings in a laser is increased, then the distance between the interference fringes decreases

<h3>What is Interference fringe ?</h3>

Interference fringe refers to bands caused by different lights which can be found in phase or not each other.

Distances between laser fringes are short which is due to light wavelength.
The interference fringes can be estimated by knowing slit separation and wavelength.

In conclusion, if the separation between the openings in a laser is increased, then the distance between the interference fringe decreases

Learn more about Interference fringe here:

brainly.com/question/14264436

#SPJ4

5 0

2 years ago

If this energy were used to vaporize water at 100.0 ∘C, how much water (in liters) could be vaporized? The enthalpy of vaporizat

Zanzabum

Answer:

0.429 L of water

Explanation:

First to all, you are not putting the value of the energy given to vaporize water, so, to explain better this problem, I will assume a value of energy that I took in a similar exercise before, which is 970 kJ.

Now, assuming that the water density is 1 g/mL, this is the same as saying that 1 g of water = 1 mL of water

If this is true, then, we can assume that 1 kg of water = 1 L of water.

Knowing this, we have to use the expression to get energy which is:

Q = m * ΔH

Solving for m:

m = Q / ΔH

Now "m" is the mass, but in this case, the mass of water is the same as the volume, so it's not neccesary to do a unit conversion.

Before we begin with the calculation, we need to put the enthalpy of vaporization in the correct units, which would be in grams. To do that, we need the molar mass of water:

MM = 18 g/mol

The enthalpy in mass:

ΔH = 40.7 kJ/mol / 18 g/mol = 2.261 kJ/g

Finally, solving for m:

m = 970 / 2.261 = 429 g

Converting this into volume:

429 g = 429 mL

429 / 1000 = 0.429 L of water

3 0

3 years ago