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yKpoI14uk [10]
2 years ago
15

1. Calcular la masa de mercurio que pasó de 30 °C hasta 120 °C y absorbió 4400 cal. Calor específico del

Physics
1 answer:
timofeeve [1]2 years ago
7 0

Answer:

Masa, m = 0.088 kg

Explanation:

Given the following data;

Temperatura inicial = 30°C

Temperatura final = 120°C

Capacidad calorífica específica = 138J/kg.K

Calor absorbido, Q = 4400 cal.

Para encontrar la masa;

La capacidad calorífica viene dada por la fórmula;

Q = mct

Dónde;

Q representa la capacidad calorífica o la cantidad de calor.

m representa la masa de un objeto.

c representa la capacidad calorífica específica del agua.

dt representa el cambio de temperatura.

dt = T2 - T1

dt = 120 - 30

dt = 90°C to kelvin = 273 + 90 = 363K

Sustituyendo en la fórmula, tenemos;

4400 = m*138*363

4400 = 50094m

m = \frac {4400}{50094}

Masa, m = 0.088 kg

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A 2 eV electron encounters a barrier 5.0 eV high and width a. What is the probability b) 0.5 that it will tunnel through the bar
denis23 [38]

Answer:

The tunnel probability for 0.5 nm and 1.00 nm are  5.45\times10^{-4} and 7.74\times10^{-8} respectively.

Explanation:

Given that,

Energy E = 2 eV

Barrier V₀= 5.0 eV

Width = 1.00 nm

We need to calculate the value of \beta

Using formula of \beta

\beta=\sqrt{\dfrac{2m}{\dfrac{h}{2\pi}}(v_{0}-E)}

Put the value into the formula

\beta = \sqrt{\dfrac{2\times9.1\times10^{-31}}{(1.055\times10^{-34})^2}(5.0-2)\times1.6\times10^{-19}}

\beta=8.86\times10^{9}

(a). We need to calculate the tunnel probability for width 0.5 nm

Using formula of tunnel barrier

T=\dfrac{16E(V_{0}-E)}{V_{0}^2}e^{-2\beta a}

Put the value into the formula

T=\dfrac{16\times 2(5.0-2.0)}{5.0^2}e^{-2\times8.86\times10^{9}\times0.5\times10^{-9}}

T=5.45\times10^{-4}

(b). We need to calculate the tunnel probability for width 1.00 nm

T=\dfrac{16\times 2(5.0-2.0)}{5.0^2}e^{-2\times8.86\times10^{9}\times1.00\times10^{-9}}

T=7.74\times10^{-8}

Hence, The tunnel probability for 0.5 nm and 1.00 nm are  5.45\times10^{-4} and 7.74\times10^{-8} respectively.

6 0
3 years ago
Which of these is the best example of translational motion?
Tanya [424]
B, air blowing from across the field is as a bullet fired from a rifle
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3 years ago
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Mimas, a moon of Saturn, has an orbital radius of 1.62 × 108 m and an orbital period of about 23.21 h. Use Newton’s version of K
Drupady [299]

Answer:

3.60432\times 10^{26}\ kg

Explanation:

a = Orbital radius = 1.62\times 10^8\ m

T = Orbital period = 23.21 hours

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

From Kepler's third law we get

M=\frac{4\pi^2a^3}{GT^2}\\\Rightarrow M=\frac{4\pi^2\times (1.62\times 10^8)^3}{6.67\times 10^{-11}\times (23.21\times 3600)^2}\\\Rightarrow M=3.60432\times 10^{26}\ kg

From the given data the mass of Saturn is 3.60432\times 10^{26}\ kg

8 0
3 years ago
When light with a wavelength of 238 nm is incident on a certain metal surface, electrons are ejected with a maximum kinetic ener
erastova [34]

Answer:

The wavelength is 173 nm.

Explanation:

This kind of phenomenon is known as photoelectric effect, it occurs when photons of light inside the metal surface and if they have the right amount of energy electrons absorb it and got expelled from the metal as photo electrons. The maximum kinetic energy of that photo electrons is given by the expression:

K_{max} =E_{photon} - \Phi (1)

With E the energy of the photon and Φ the work function of the material. The work function is a value characteristic of each material and is related with how much the electron is attached to the material, the energy of the photon is the Planck's constant (h=6.63\times10^{-34}) times the frequency of light (\nu) , then (1) is:

K_{max} =h\nu - \Phi (2)

The frequency of an electromagnetic wave is related with the wavelength (\lambda) by:

\nu=\frac{c}{\lambda} (3)

with c the velocity of light (c=3.0\times10^{8})

Using (3) on (2):

K_{max} =\frac{hc}{\lambda} - \Phi

Solving for \Phi:

\Phi=\frac{hc}{\lambda}-K_max=\frac{(6.63\times10^{-34})(3.0\times10^{8})}{238\times10^{-9}}-3.13\times10^{-19}

\Phi=5.23\times10^{-19} J

That's the work function of the metal we're dealing. So now if we want to know the wavelength to obtain the double of the kinetic energy we use:

2K_{max} =\frac{hc}{\lambda} - \Phi

Solving for \lambda:

\lambda = \frac{hc}{2K_{max}+\Phi}=\frac{(6.63\times10^{-34})(3.0\times10^{8})}{2(3.13\times10^{-19})+5.23\times10^{-19}}=1.73\times10^{-7}

\lambda=173 nm

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