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Tems11 [23]
2 years ago
15

En una librería un libro de 0.6 kg colocado inicialmente en un estante a 40 cm del suelo fue movido a otro estante con una altur

a de 1.30 m. Determine la energía potencial en cada uno de esos lugares. ¿Cuál es el cambio en la energía potencial del libro?
Physics
1 answer:
jeyben [28]2 years ago
3 0

Considerando la definición de energía potencial:

  • la energía potencial del libro inicialmente es 2,3544 J y luego de ser movido a otro estante es 7,6518 J.
  • el cambio en la energía potencial del libro es 5,2974 J.
<h3 /><h3>Definición de energía potencial</h3>

La energía potencial es la energía que mide la capacidad que tiene un sistema para realizar un trabajo en función de su posición. En otras palabras, esta es la energía que tiene un cuerpo situado a una determinada altura sobre el suelo.

La energía potencial gravitatoria es la energía asociada con la fuerza gravitatoria. Esta dependerá de la altura relativa de un objeto a algún punto de referencia, la masa, y la fuerza de la gravedad.

Entonces para un objeto con masa m, en la altura h, la expresión aplicada a la energía gravitacional del objeto es:

Ep= m×g×h

Donde

  • Ep es la energía potencial en julios (J).
  • m es la masa en kilogramos (kg).
  • h es la altura en metros (m).
  • g es la aceleración de caída en m/s² (aproximadamente 9,81 m/s²).

<h3>Energía potencial en cada lugar</h3>

En primer lugar, se sabe inicialmente del libro:

  • m= 0,6 kg
  • g= 9,81 m/s²
  • h= 40 cm= 0,4 m (siendo 1 cm= 0,01 m)

Entonces, reemplazando en la definición de energía potencial:

Ep1= 0,6 kg× 9,81 m/s²× 0,4 m

Resolviendo:

<u><em>Ep1= 2,3544 J</em></u>

Por otro lado, se sabe los siguientes datos del libro luego de ser movido:

  • m= 0,6 kg
  • g= 9,81 m/s²
  • h= 1,30 m

Entonces, reemplazando en la definición de energía potencial:

Ep2= 0,6 kg× 9,81 m/s²× 1,30 m

Resolviendo:

<u><em>Ep2= 7,6518 J</em></u>

En resumen, la energía potencial del libro inicialmente es 2,3544 J y luego de ser movido a otro estante es 7,6518 J.

<h3>Cambio de energía potencial</h3>

El cambio en la energía potencial del libro será la diferencia entre la energía potencial luego de ser movido y la energía potencial inicial del libro:

ΔEp= Ep2 - Ep1

ΔEp= 7,6518 J - 2,3544 J

<u><em>ΔEp= 5,2974 J</em></u>

Finalmente, el cambio en la energía potencial del libro es 5,2974 J.

Aprende más sobre energía potencial:

brainly.com/question/21370800

brainly.com/question/25163940

brainly.com/question/25960490

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Answer:

It appears to rise and set because of the Earth's rotation on its axis. It makes one complete turn every 24 hours. It turns toward the east. As the Earth rotates toward the east, it looks like the sun is moving west.

Explanation:

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A 1.50 cm high diamond ring is placed 20.0 cm from a concave mirror with radius of curvature 30.00 cm. The magnification is ____
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Answer:

Magnification, m = -0.42

Explanation:

It is given that,

Height of diamond ring, h = 1.5 cm

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Radius of curvature of concave mirror, R = 30 cm

Focal length of mirror, f = R/2 = -15 cm (focal length is negative for concave mirror)

Using mirror's formula :

\dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{v}, f = focal length of the mirror

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\dfrac{1}{v}=\dfrac{1}{-15}+\dfrac{1}{-20}

v = -8.57 cm

The magnification of a mirror is given by,

m=\dfrac{-v}{u}

m=\dfrac{-(-8.57)}{-20}

m = -0.42

So, the magnification of the concave mirror is 0.42. Thew negative sign shows that the image is inverted.

5 0
3 years ago
7. 13 a turbine receives steam at 6 mpa, 600°c with an exit pressure of 600 kpa. Assume the turbine is adiabatic and neglect kin
AnnyKZ [126]

The work done by the turbine will be 708.2 kJ/kg. The work done by the turbine is the difference of the enthalpy at inlet and exit.

<h3 /><h3>What is temperature?</h3>

Temperature directs the hotness or coldness of a body. In clear terms, it is the method of finding the kinetic energy of particles within an entity. Faster the motion of particles, more the temperature.

If the given turbine is assumed to be reversible;

\rm P_I(Initial pressure)=60 mpa = 60 bar

\rm  T_i(Initial temperature)=600° C

\rm P_e (Exit pressure)=600 kpa=6 bar

The heat balance equation is;

\rm q-W_t=h_e-h_i\\\\ q=0\\\\\ W_t=h_i-h_e

The change in the entropy is;

\rm S_2-S_1=\frac{\delta q}{dt}

The work done by the turbine is;

\rm W_t = h_i-h_e\\\\ W_t =3658.4 -29.5019\\\\  W_t =708.2 \ kJ/kg

Hence,the work done by the turbine will be 708.2 kJ/kg.

To learn more about the temperature, refer to the link;

brainly.com/question/7510619

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6 0
2 years ago
How is thermal energy naturally transferred between objects?
Sladkaya [172]

The second law of thermodynamics establishes restrictions on the flow of thermal energy between two bodies. This law states that the energy does not flow spontaneously from a low temperature object T1, to another object that is at a high temperature T2.

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Finally we can say that the correct option is B: From the hotter object to the cooler object

5 0
3 years ago
X rays of wavelength 0.0169 nm are directed in the positive direction of an x axis onto a target containing loosely bound electr
mamaluj [8]

Answer:

a) 4.04*10^-12m

b) 0.0209nm

c) 0.253MeV

Explanation:

The formula for Compton's scattering is given by:

\Delta \lambda=\lambda_f-\lambda_i=\frac{h}{m_oc}(1-cos\theta)

where h is the Planck's constant, m is the mass of the electron and c is the speed of light.

a) by replacing in the formula you obtain the Compton shift:

\Delta \lambda=\frac{6.62*10^{-34}Js}{(9.1*10^{-31}kg)(3*10^8m/s)}(1-cos132\°)=4.04*10^{-12}m

b) The change in photon energy is given by:

\Delta E=E_f-E_i=h\frac{c}{\lambda_f}-h\frac{c}{\lambda_i}=hc(\frac{1}{\lambda_f}-\frac{1}{\lambda_i})\\\\\lambda_f=4.04*10^{-12}m +\lambda_i=4.04*10^{-12}m+(0.0169*10^{-9}m)=2.09*10^{-11}m=0.0209nm

c) The electron Compton wavelength is 2.43 × 10-12 m. Hence you can use the Broglie's relation to compute the momentum of the electron and then the kinetic energy.

P=\frac{h}{\lambda_e}=\frac{6.62*10^{-34}Js}{2.43*10^{-12}m}=2.72*10^{-22}kgm\\

E_e=\frac{p^2}{2m_e}=\frac{(2.72*10^{-22}kgm)^2}{2(9.1*10^{-31}kg)}=4.06*10^{-14}J\\\\1J=6.242*10^{18}eV\\\\E_e=4.06*10^{-14}(6.242*10^{18}eV)=0.253MeV

5 0
3 years ago
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