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What is the frequency of a photon with an energy of 4. 56 x 10^-19 j

Sauron [17]

The frequency of a photon with an energy of 4.56 x 10⁻¹⁹ J is 6.88×10¹⁴ s⁻¹.

<h3>What is a frequency?</h3>

The number of waves that travel through a particular point in a given length of time is described by frequency. So, if a wave takes half a second to pass, the frequency is 2 per second.

Given that the energy of the photon is 4.56 x 10⁻¹⁹ J. Therefore, the frequency of the photon can be written as,

$\rm \gamma = \dfrac{E}{h} = \dfrac{4.56x10^{-19} J}{6.626 \times 10^{-34}\ Jsec^{-1}}\\\\\\\gamma = 6.88 \times 10^{14}\ s^{-1}$

Hence, the frequency of a photon with an energy of 4.56 x 10⁻¹⁹ J is 6.88×10¹⁴ s⁻¹.

Learn more about Frequency:

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5 0

2 years ago

Read 2 more answers

When u write on a piece of glass sheet with a piece of chalk , the writing is not clear explain .

svetlana [45]

Becuse your weighting with chalk that has pigment

6 0

3 years ago

As a new electrical technician, you are designing a large solenoid to produce a uniform 0.170 T magnetic field near the center o

MrRa [10]

Answer:

18.6012339739 A

Explanation:

$\mu_0$ = Vacuum permeability = $4\pi \times 10^{-7}\ H/m$

L = Length of wire = 55 cm

N = Number of turns = 4000

I = Current

Magnetic field is given by

$B=\dfrac{\mu_0NI}{L}\\\Rightarrow I=\dfrac{BL}{\mu_0N}\\\Rightarrow I=\dfrac{0.17\times 0.55}{4\pi \times 10^{-7}\times 4000}\\\Rightarrow I=18.6012339739\ A$

The current necessary to produce this field is 18.6012339739 A

7 0

3 years ago

You are working as a summer intern for the Illinois Department of Natural Resources (DNR) and are assigned to some initial work

serious [3.7K]

Answer:

Check the explanation

Explanation:

1) Pressure acting on the plug = Patm + P

Pressure = Patm + rho*g*h (Here h = D2)

Pressure = 101325 + 1000*9.8*7

Pressure = 169925 Pa

so, Force = PA

Force = 169925*pi*0.0152

Force = 120.1 N

7 0

2 years ago

Un puente de acero de 100 m de largo a 8° C aumenta su temperatura a 24°C ¿Cuánto medirá su longitud? Valor del coeficiente de d

BartSMP [9]

La longitud final del puente de acero es 100.018 metros.

Asumamos que la dilatación térmica experimentada por el puente de acero es pequeña, de modo que podemos emplear la siguiente aproximación lineal para determinar la longitud final del puente de acero ( $L$ ), en metros:

$L = L_{o}\cdot [1+\alpha\cdot (T_{f}-T_{o})]$ (1)

Donde:

$L_{o}$ - Longitud inicial del puente, en metros.
$\alpha$ - Coeficiente de dilatación, sin unidad.
$T_{o}$ - Temperatura inicial, en grados Celsius.
$T_{f}$ - Temperatura final, en grados Celsius.

Si tenemos que $L_{o} = 100\,m$ , $\alpha = 11.5\times 10^{-6}$ , $T_{o} = 8\,^{\circ}C$ y $T_{f} = 24\,^{\circ}C$ , entonces la longitud final del puente de acero es:

$L = (100\,m)\cdot [1+(11.5\times 10^{-6})\cdot (24\,^{\circ}C - 8\,^{\circ}C)]$

$L = 100.018\,m$

La longitud final del puente de acero es 100.018 metros.

Para aprender más sobre dilatación térmica, invitamos cordialmente a ver esta pregunta verificada: brainly.com/question/24953416

5 0

2 years ago