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GrogVix [38]
4 years ago
5

Si te dijera que la energía es una cualidad o capacidad que tienen los cuerpos materiales ¿para qué sirve o emplean los cuerpos

esa cualidad o capacidad?
Physics
1 answer:
Vlada [557]4 years ago
7 0

Answer:

La energía es una cualidad o capacidad que tienen los cuerpos materiales y posee diversas funciones. Hay distintos tipos de energía, pero ésta permite que un cuerpo se mueva por ejemplo, cambie de temperatura o de estado.

Explanation:

La Energía de los cuerpos puede ser definida de distintas maneras según la ciencia en la cual estemos trabajando. <em>Para la física, la energía es la capacidad para realizar un trabajo. </em>Esa energía puede ser cinética, térmica, química, y cada una de ellas define distintas propiedades de los cuerpos, que hacen que puedan cambiar de posición, temperatura, estado.

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If the car’s speed decreases at a constant rate from 64 mi/h to 30 mi/h in 3.0 s, what is the magnitude of its acceleration, ass
mixas84 [53]

Answer:3.874 m/s^2

Explanation:

Given

Car speed decreases at a constant rate from 64 mi/h to 30 mi/h

in 3 sec

60mi/h \approx 26.8224m/s

34mi/h \approx 15.1994 m/s

we know acceleration is given by =\frac{velocity}{Time}

a=\frac{15.1994-26.8224}{3}

a=-3.874 m/s^2

negative indicates that it is stopping the car

Distance traveled

v^2-u^2=2as

\left ( 15.1994\right )^2-\left ( 26.8224\right )^2=2\left ( -3.874\right )s

s=\frac{488.419}{2\times 3.874}

s=63.038 m

7 0
3 years ago
A police car chases a speeder along a straight road towards a cliff both vehicles move at 160km/h the siren on the police car pr
natta225 [31]

Answer:

f ’= 97.0 Hz

Explanation:

This is an exercise of the doppler effect use the frequency change due to the relative movement of the fort and the observer

in this case the source is the police cases that go to vs = 160 km / h

and the observer is vo = 120 km / h

the relationship of the doppler effect is

          f ’= f₀ (v + v₀ / v- v_{s})

let's reduce the magnitude to the SI system

            v_{s} = 160 km / h (1000 m / 1km) (1h / 3600s) = 44.44 m / s

            v₀ = 120 km / h (1000m / 1km) (1h / 3600s) = 33.33 m / s

we substitute in the equation of the Doppler effect

          f ‘= 100 (330+ 33.33 / 330-44.44)

          f ’= 97.0 Hz

4 0
3 years ago
Ultraviolet light of wavelength 270 nm strikes a metal whose work function is 2.3 eV.What is the shortest de Broglie wavelength
soldier1979 [14.2K]

Answer:

The shortest de Broglie wavelength for the electrons that are produced as photoelectrons is 0.81 nm

Explanation:

Given;

wavelength of ultraviolet light, λ = 270 nm

work function of the metal, φ = 2.3 eV = 2.3 x 1.602 x 10⁻¹⁹ J = 3.685 x 10⁻¹⁹ J

The energy of the ultraviolet light is given by;

E = \frac{hc}{\lambda}\\\\E = \frac{(6.626*10^{-34} )(3*10^{8}) }{270*10^{-9} }\\\\E = 7.362 * 10^{-19} \ J

The energy of the incident light is related to kinetic energy of the electron and work function of the metal by the following equation;

E = φ  + K.E

K.E = E - φ

K.E = (7.362 x 10⁻¹⁹ J) - (3.685 x 10⁻¹⁹ J )

K.E = 3.677 x 10⁻¹⁹ J

K.E = ¹/₂mv²

mv² = 2K.E

velocity of the electron is given by;

V = \sqrt{\frac{2K.E}{m} }\\\\V =  \sqrt{\frac{2(3.677*10^{-19}) }{9.1*10^{-31} } }\\\\V = 8.99*10^{5}  \ m/s

the shortest de Broglie wavelength for the electrons is given by;

\lambda = \frac{h}{mv}\\ \\\lambda = \frac{6.626*10^{-34} }{(9.1*10^{-31})( 8.99*10^{5} )}\\\\\lambda = 8.10*10^{-10} \ m\\\\\lambda = 0.81 \ nm

Therefore, the shortest de Broglie wavelength for the electrons that are produced as photoelectrons is 0.81 nm

7 0
4 years ago
A battery-operated car utilizes a 120.0 V battery with negligible internal resistance. Find the charge, in coulombs, the batteri
fgiga [73]

Answer:

4.29×10⁵ C

Explanation:

From the question,

The energy stored in the battery = Kinetic energy of the car+ Energy needed to make the car climbed the hill+Energy required to exert a force.

E = 1/2mv²+mgh+Fd.................... Equation 1

Where E = Energy stored in the battery, m = mass of the car, v = velocity of the car, h = height of the hill, F = force exerted on the car, d = distance traveled by the car.

But,

d = vt.................... Equation 2

Where v = velocity, t = time.

Substitute equation 2 into equation 1

E = 1/2mv²+mgh+F(vt)................... Equation 3

Given: m = 770 kg, v = 26 m/s, h = 2.15×10² m = 215 m, F = 5.3×10² N = 530 N, t = 1 hour = 3600 s, g = 9.8 m/s²

Substitute into equation 1

E = 1/2(770)(26²)+(770)(9.8)(215)+(530)(26)(3600)

E = 260260+1622390+49608000

E = 51490650 J

Using,

E = qV................. Equation 4

Where q = charge of the battery, V = Voltage.

make q the subject of the equation

q = E/V............... Equation 5

Given: E = 51490650 J, V = 120 V

Substitute into equation 5

q = 51490650/120

q = 429088.75 C

q = 4.29×10⁵ C

7 0
4 years ago
If a student flicks a stationary 0.1 kg ball with 5N of force for 0.1 seconds. What is its final
Alisiya [41]

5m/s

Explanation:

Given parameters:

Mass of ball = 0.1kg

Force on the ball = 5N

time taken = 0.1s

Unknown:

final speed of the ball = ?

Solution:

According to newton's second law "the net force on a body is the product of its mass and acceleration".

  Force = mass x acceleration      equation 1

Acceleration =

  V is the final velocity

  U is the initial velocity

  T is the time taken

 U = O since it is a stationary body;

      a = \frac{V}{T}

Input "a" into equation 1

  F = m x \frac{V}{T}

 5 = 0.1 x \frac{V}{0.1}

 V = 5m/s

learn more:

Newton's laws brainly.com/question/11411375

#learnwithBrainly

3 0
3 years ago
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