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AveGali [126]
4 years ago
14

What is the electric potential 16.0 cm from a 4.00 μC point charge?

Physics
1 answer:
AveGali [126]4 years ago
7 0

Answer:

the potential at a distance of 16 cm from the charge 4\mu C will be 225000 volt

Explanation:

We have given charge q=4\mu C=4\times 10^{-6}C

Distance between the charge r = 16 cm = 0.16 m

Electric potential is given by V=\frac{1}{4\pi \varepsilon _0}\frac{q}{r}=\frac{Kq}{r}, here K  is constant which value is 9\times 10^9Nm^2/C^2

So potential V=\frac{9\times 10^9\times 4\times 10^{-6}}{0.16}=225000volt

So the potential at a distance of 16 cm from the charge 4\mu C will be 225000 volt

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The energy change in an endothermic reaction is: A. Internal B. External C. Negative D. Positive
77julia77 [94]

Answer:

Positive

Explanation:

In an endothermic reaction, the products are at a higher energy than the reactants. This means that the enthalpy change of the reaction (∆H) is positive

4 0
3 years ago
10 points
MariettaO [177]

Answer:

true

Explanation:

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8 0
3 years ago
a uniform rod is hung at onen end and is partially submerged in water. If the density of the rod is 5/9 than of wter, find the f
34kurt

Answer:

    \frac{h_{liquid} }{ h_{body} } = 5/9

Explanation:

This is an exercise that we can solve using Archimedes' principle which states that the thrust is equal to the weight of the desalted liquid.

         B = ρ_liquid  g V_liquid

let's write the translational equilibrium condition

         B - W = 0

let's use the definition of density

        ρ_body = m / V_body

        m = ρ_body  V_body

        W = ρ_body  V_body  g

we substitute

          ρ_liquid  g  V_liquid = ρ_body  g  V_body

          \frac{\rho_{body}   }{\rho_{liquid} } } =  \frac{V_{liquid}   }{V_{body} } }

In the problem they indicate that the ratio of densities is 5/9, we write the volume of the bar

          V = A h_bogy

Thus

          \frac{V_{liquid} }{V_{1body} } = \frac{ h_{liquid} }{h_{body} }

we substitute

           5/9 = \frac{h_{liquid} }{ h_{body} }

8 0
3 years ago
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
5. The volume of physical activity attained by an individual who exercises at a level of 7 METs for 35 min · day-1, 4 days · wk-
Airida [17]
D. 980, this is the best answer because 35 x 7 is 980 :)

3 0
3 years ago
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