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Yuliya22 [10]
3 years ago
5

A pair of opposite electric charges of equal magnitude is called a(n)

Physics
2 answers:
ladessa [460]3 years ago
8 0

A pair of opposite electric charges of equal magnitude is called a(n) A) Dipole

Kryger [21]3 years ago
6 0

Answer:

A) dipole.

Explanation:

As we know that when two equal and opposite charges are placed at some distance from each other then this combination of two equal and opposite charges are known as electric dipole.

Here electric dipole has dipole moment which is a vector quantity which is measured as the product of charge and distance between the two charges

it is given as

P = qd

here the direction of dipole moment is from negative charge to positive charge.

so correct answer is given as

A) dipole.

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What are the characteristics of the radiation emitted by a blackbody? According to Wien's Law, how many times hotter is an objec
jasenka [17]

Answer:

a) What are the characteristics of the radiation emitted by a blackbody?

The total emitted energy per unit of time and per unit of area depends in its temperature (Stefan-Boltzmann law).

The peak of emission for the spectrum will be displaced to shorter wavelengths as the temperature increase (Wien’s displacement law).

The spectral density energy is related with the temperature and the wavelength (Planck’s law).

b) According to Wien's Law, how many times hotter is an object whose blackbody emission spectrum peaks in the blue, at a wave length of 450 nm, than a object whose spectrum peaks in the red, at 700 nm?

The object with the blackbody emission spectrum peak in the blue is 1.55 times hotter than the object with the blackbody emission spectrum peak in the red.

Explanation:

A blackbody is an ideal body that absorbs all the thermal radiation that hits its surface, thus becoming an excellent emitter, as these bodies express themselves without light radiation, and therefore they look black.

The radiation of a blackbody depends only on its temperature, thus being independent of its shape, material and internal constitution.

If it is study the behavior of the total energy emitted from a blackbody at different temperatures, it can be seen how as the temperature increases the energy will also increase, this energy emitted by the blackbody is known as spectral radiance and the result of the behavior described previously is Stefan's law:

E = \sigma T^{4}  (1)

Where \sigma is the Stefan-Boltzmann constant and T is the temperature.

The Wien’s displacement law establish how the peak of emission of the spectrum will be displace to shorter wavelengths as the temperature increase (inversely proportional):

\lambda max = \frac{2.898x10^{-3} m. K}{T}   (2)

Planck’s law relate the temperature with the spectral energy density (shape) of the spectrum:

E_{\lambda} = {{8 \pi h c}\over{{\lambda}^5}{(e^{({hc}/{\lambda \kappa T})}-1)}}}  (3)

b) According to Wien's Law, how many times hotter is an object whose blackbody emission spectrum peaks in the blue, at a wavelength of 450 nm, than a object whose spectrum peaks in the red, at 700 nm?

It is need it to known the temperature of both objects before doing the comparison. That can be done by means of the Wien’s displacement law.

Equation (2) can be rewrite in terms of T:

T = \frac{2.898x10^{-3} m. K}{\lambda max}   (4)

Case for the object with the blackbody emission spectrum peak in the blue:

Before replacing all the values in equation (4), \lambda max (450 nm) will be express in meters:

450 nm . \frac{1m}{1x10^{9} nm}  ⇒ 4.5x10^{-7}m

T = \frac{2.898x10^{-3} m. K}{4.5x10^{-7}m}

T = 6440 K

Case for the object with the blackbody emission spectrum peak in the red:

Following the same approach above:

700 nm . \frac{1m}{1x10^{9} nm}  ⇒ 7x10^{-7}m

T = \frac{2.898x10^{-3} m. K}{7x10^{-7}m}

T = 4140 K

Comparison:

\frac{6440 K}{4140 K} = 1.55

The object with the blackbody emission spectrum peak in the blue is 1.55 times hotter than the object with the blackbody emission spectrum peak in the red.

4 0
3 years ago
A solid cylinder of mass M = 45 kg, radius R = 0.44 m and uniform density is pivoted on a frictionless axle coaxial with its sym
user100 [1]

Answer:

w_f = 1.0345 rad/s

Explanation:

Given:

- The mass of the solid cylinder M = 45 kg

- Radius of the cylinder R = 0.44 m

- The mass of the particle m = 3.6 kg

- The initial speed of cylinder w_i = 0 rad/s

- The initial speed of particle V_pi = 3.3 m/s

- Mass moment of inertia of cylinder I_c = 0.5*M*R^2

- Mass moment of inertia of a particle around an axis I_p = mR^2

Find:

- What is the magnitude of its angular velocity after the collision?

Solution:

- Consider the mass and the cylinder as a system. We will apply the conservation of angular momentum on the system.

                                     L_i = L_f

- Initially, the particle is at edge at a distance R from center of cylinder axis with a velocity V_pi = 3.3 m/s contributing to the initial angular momentum of the system by:

                                    L_(p,i) = m*V_pi*R

                                    L_(p,i) = 3.6*3.3*0.44

                                    L_(p,i) = 5.2272 kgm^2 /s

- While the cylinder was initially stationary w_i = 0:

                                    L_(c,i) = I*w_i

                                    L_(c,i) = 0.5*M*R^2*0

                                    L_(c,i) = 0 kgm^2 /s

The initial momentum of the system is L_i:

                                    L_i = L_(p,i) + L_(c,i)

                                    L_i = 5.2272 + 0

                                    L_i = 5.2272 kg-m^2/s

- After, the particle attaches itself to the cylinder, the mass and its distribution around the axis has been disturbed - requires an equivalent Inertia for the entire one body I_equivalent. The final angular momentum of the particle is as follows:

                                   L_(p,f) = I_p*w_f

- Similarly, for the cylinder:

                                   L_(c,f) = I_c*w_f

- Note, the final angular velocity w_f are same for both particle and cylinder. Every particle on a singular incompressible (rigid) body rotates at the same angular velocity around a fixed axis.

                                  L_f = L_(p,f) + L_(c,f)

                                  L_f = I_p*w_f + I_c*w_f

                                  L_f = w_f*(I_p + I_c)

-Where, I_p + I_c is the new inertia for the entire body = I_equivalent that we discussed above. This could have been determined by the superposition principle as long as the axis of rotations are same for individual bodies or parallel axis theorem would have been applied for dissimilar axes.

                                  L_i = L_f

                                  5.2272 = w_f*(I_p + I_c)

                                  w_f =  5.2272/ R^2*(m + 0.5M)

Plug in values:

                                  w_f =  5.2272/ 0.44^2*(3.6 + 0.5*45)

                                  w_f =  5.2272/ 5.05296

                                  w_f = 1.0345 rad/s

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