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3 years ago

Calculate the energy in electron volts of X-rays that have a frequency of 3.00 x 100 Hz.

Physics

1 answer:

inn [45]3 years ago

8 0

Answer:

E = 124 eV

Explanation:

Given,

The frequency of the X-rays, ν = 3.0 x 10¹⁰ Hz

The formula for calculating the energy of the electromagnetic waves of a single photon of having frequency 'ν' is given by the relation

E = hν joules

Where,

h - Planck's constant (6.626 x 10⁻³⁴ Js)

Substituting the values in the above equation

E = 6.626 x 10⁻³⁴ Js x 3.0 x 10¹⁰ Hz

= 1.9878 x 10⁻²³ J

Converting it into eV

E = 1.9878 x 10⁻²³ x 6.242 x 10¹⁸ eV

= 124 eV

Therefore, the energy in electron volts of X-rays is, E = 124 eV

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Explanation:

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3 years ago

The 1.53-kg uniform slender bar rotates freely about a horizontal axis through O. The system is released from rest when it is in

OlgaM077 [116]

Answer:

The spring constant = 104.82 N/m

The angular velocity of the bar when θ = 32° is 1.70 rad/s

Explanation:

From the diagram attached below; we use the conservation of energy to determine the spring constant by using to formula:

$T_1+V_1=T_2+V_2$

$0+0 = \frac{1}{2} k \delta^2 - \frac{mg (a+b) sin \ \theta }{2} \\ \\ k \delta^2 = mg (a+b) sin \ \theta \\ \\ k = \frac{mg(a+b) sin \ \theta }{\delta^2}$

Also;

$\delta = \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2}$

Thus;

$k = \frac{mg(a+b) sin \ \theta }{( \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2})^2}$

where;

$\delta$ = deflection in the spring

k = spring constant

b = remaining length in the rod

m = mass of the slender bar

g = acceleration due to gravity

$k = \frac{(1.53*9.8)(0.6+0.2) sin \ 64 }{( \sqrt{0.6^2 +0.6^2 +2*0.6*0.6 sin \ 64} - \sqrt{0.6^2 +0.6^2})^2}$

$k = 104.82\ \ N/m$

Thus; the spring constant = 104.82 N/m

The angular velocity can be calculated by also using the conservation of energy;

$T_1+V_1 = T_3 +V_3 \\ \\ 0+0 = \frac{1}{2}I_o \omega_3^2+\frac{1}{2}k \delta^2 - \frac{mg(a+b)sin \theta }{2} \\ \\ \frac{1}{2} \frac{m(a+b)^2}{3} \omega_3^2 + \frac{1}{2} k \delta^2 - \frac{mg(a+b)sin \ \theta }{2} =0$