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mixer [17]
3 years ago
11

Hospital X-ray generators emit X-rays with wavelength of about 15.0 nanometers (nm), where 1nm=10−9m. What is the energy of a ph

oton of the X-rays? Express your answer to three significant figures and include the appropriate units.
Physics
1 answer:
Sergio [31]3 years ago
4 0

Answer:

E=1.32\times 10^{-17}\ J

Explanation:

Given that,

The wavelength of x ray, \lambda=15\ nm=15\times 10^{-9}\ m

We need to find the energy if a photon on the x-rays. The formula for the energy of the photon is given by :

E=\dfrac{hc}{\lambda}

E=\dfrac{6.63\times 10^{-34}\times 3\times 10^8}{15\times 10^{-9}}

E=1.32\times 10^{-17}\ J

So, the energy of a photon of the X-rays is 1.32\times 10^{-17}\ J. Hence, this is the required solution.

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An object travels with a constant speed in a circular path. The net force on the object is
Pepsi [2]

Answer:

toward the center

Explanation:

Before answering, let's remind the first two Newton Laws:

1) An object at rest tends to stay at rest and an object moving at constant velocity tends to continue its motion at constant velocity, unless acted upon a net force

2) An object acted upon a net force F experiences an acceleration a according to the equation

F=ma

where m is the mass of the object.

In this problem, we have an object travelling at constant speed in a circular path. The fact that the trajectory of the object is circular means that the direction of motion of the object is constantly changing: this means that its velocity is changing, so it has an acceleration. And therefore, a net force is acting on it. The force that keeps the object travelling in the circular path is called centripetal force, and it is directed towards the center of the circle (because it prevents the object from continuing its motion straight away).

So, the correct answer is

toward the center

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3 years ago
When a voltage difference is applied to a piece of metal wire, a 5.0 mA current flows through it. If this metal wire is now repl
zheka24 [161]

Answer:

I = 21.13 mA ≈ 21 mA

Explanation:

If

I₁ = 5 mA

L₁ = L₂ = L

V₁ = V₂ = V

ρ₁ = 1.68*10⁻⁸ Ohm-m

ρ₂ = 1.59*10⁻⁸ Ohm-m

D₁ = D

D₂ = 2D

S₁ = 0.25*π*D²

S₂ = 0.25*π*(2*D)² = π*D²

If we apply the equation

R = ρ*L / S

where (using Ohm's Law):

R = V / I

we have

V / I = ρ*L / S

If V and L are the same

V / L =  ρ*I / S

then

(V / L)₁ = (V / L)₂  ⇒     ρ₁*I₁ / S₁ = ρ₂*I₂ / S₂

If

S₁ = 0.25*π*D²   and

S₂ = 0.25*π*(2*D)² = π*D²

we have

ρ₁*I₁ / (0.25*π*D²) = ρ₂*I₂ / (π*D²)

⇒    I₂ = 4*ρ₁*I₁ / ρ₂

⇒     I₂ = 4*1.68*10⁻⁸ Ohm-m*5 mA / 1.59*10⁻⁸ Ohm-m

⇒     I₂ = 21.13 mA

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