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3 years ago

Determine the specific heat of iron if 6.1 j of energy are needed to warm 1.50 g of iron from 20.0◦c to 29.0◦c?

1 answer:

8 0

As the iron core isprogressively laid down, the mechanism of iron oxidation changes from a protein dominated process with H202 being the primary product of O2 reduction to a mineral surface dominated process where H20 is the primary product. These results emphasize the importance of the apoferritin shell in facilitating iron oxidation in the early stage of iron deposition prior to significant development of the polynuclear iron core.

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As the temperature of an object increases, the wavelength of the brightest light emitted _______

alexira [117]

Answer:

option (B) decreases

Explanation:

According to the Wein's displacement law, the minimum wavelength of the radiated emission is inversely proportional to the absolute temperature of the body which emits radiation.

$\lambda_{m}\alpha \frac{1}{T}$

Where, T is the absolute temperature of the body and λm is the minimum wavelength of heat radiated.

Here, as the temperature increases, the wavelength decreases.

4 0

4 years ago

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

Suppose you hit a steel nail with a 0.500-kg hammer, initially moving at 15.0 m/s and brought to rest in 2.80 mm. How much is th

katrin [286]

Complete Question

Suppose you hit a steel nail with a 0.500-kg hammer, initially moving at 15.0 m/s and brought to rest in 2.80 mm. How much is the nail compressed if it is 2.50 mm in diameter and 6.00-cm long.What Average force is excreted on the Nail

Answer:

$F=2*10^{4}N$

Explanation:

From the question we are told that:

Mass $m=0.500kg$

Initial Velocity $V=15.0m/s$

Distance $x=2.80mm=>0.00280m$

Diameter $d=2.50mm=>0.00250m$

Length $l=6.00cm=>0.6m$

Generally the equation for Force is mathematically given by

$F=\frac{mv^2}{2d}$

$F=\frac{0.500*15^2}{2.80*10^{-3}}$

$F=2*10^{4}N$

6 0

3 years ago

A lightweight vertical spring of force constant k has its lower end mounted on a table. You compress the spring by a distance d,

shusha [124]

Answer:

$v=d\sqrt{\frac{k}{m}}$

Explanation:

In order to solve this problem, we can do an analysis of the energies involved in the system. Basically the addition of the initial potential energy of the spring and the kinetic energy of the mass should be the same as the addition of the final potential energy of the spring and the kinetic energy of the block. So we get the following equation:

$U_{0}+K_{0}=U_{f}+K_{f}$

In this case, since the block is moving from rest, the initial kinetic energy is zero. When the block loses contact with the spring, the final potential energy of the spring will be zero, so the equation simplifies to:

$U_{0}=K_{f}$