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Gnesinka [82]
3 years ago
12

An x-ray has a wavelength of 4.18 Å. Calculate the energy (in J) of one photon of this radiation. Enter your answer in scientifi

c notation
Physics
1 answer:
Damm [24]3 years ago
4 0

Answer:

E = 4.75 x 10⁻¹⁶ J

Explanation:

given,

wavelength of the x-ray , λ = 4.18 Å

Energy of photon = ?

we know

E = \dfrac{hc}{\lambda}

where h is the planks constant

          c is the speed of light

h = 6.626 x 10⁻³⁴ m² kg / s

c = 3 x 10⁸ m/s

now,

E = \dfrac{6.626\times 10^{-34}\times 3 \times 10^8}{4.18\times 10^{-10}}

E = 4.75 x 10⁻¹⁶ J

hence, the energy of the photon is equal to E = 4.75 x 10⁻¹⁶ J

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Explanation:

Meters per second

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The velocity of the ball when it strikes the ground, given the data is 21.56 m/s

<h3>Data obtained from the question</h3>

From the question given above, the following data were obtained:

  • Time to reach ground from maximum height (t) = 2.2 s
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  • Acceleration due to gravity (g) = 9.8 m/s²
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The velocity of the ball when it strikes the ground can be obtained as illustrated below:

v = u + gt

v = 0 + (9.8 × 2.2)

v = 0 + 21.56

v = 21.56 m/s

Thus, the velocity of the ball when it strikes the ground is 21.56 m/s

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brainly.com/question/22719691

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5 0
2 years ago
An object of mass m = 5.0 kg hangs from a cord around a light pulley: The length of the cord between the oscillator and the pull
puteri [66]

Answer:

\mu=0.0049Kg/m

Explanation:

When a standing wave is formed with six loops means the normal mode of the wave is n=6, the frequency of the normal mode is given by the expression:

f_n=\frac{nv}{2L}

Where L is the length of the string and v the velocity of propagation. Use this expression to find the value of v.

f_6=\frac{6v}{2L}\\(150)=\frac{6v}{2(2)} \\150=\frac{3v}{2} \\3v=150(2)\\ v=\frac{300}{3} \\v=100m/s

The velocity of propagation is given by the expression:

v=\sqrt{\frac{T}{\mu }

Where \mu is the desirable variable of the problem, the linear mass density, and T is the tension of the cord. The tension is equal to the weight of the mass hanging from the cord:

T=W=mg=(5)(9.81)=49.05N

With the value of the tension and the velocity you can find the mass density:

v=\sqrt{\frac{T}{\mu}

v^2=\frac{T}{\mu}\\ \mu=\frac{T}{v^2} =\frac{49.05}{(100)^2} =\frac{49.05}{10000} =0.0049Kg/m

6 0
3 years ago
The primary gases of the atmosphere are _____. nitrogen and carbon dioxide nitrogen, oxygen, and argon nitrogen and oxygen nitro
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<span>The primary gases of the atmosphere are _____:
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A car that experiences a deceleration of -41.62 m/s² and comes to a stop after 10.99 m has an initial velocity of 30.60 m/s.

A car experiences a deceleration (a) of -41.62 m/s² and comes to a stop (final velocity = v = 0 m/s) after 10.99 m (s).

We can calculate the initial velocity of the car (u) using the following kinematic equation.

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A car that experiences a deceleration of -41.62 m/s² and comes to a stop after 10.99 m has an initial velocity of 30.60 m/s.

Learn more: brainly.com/question/14851168

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