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3 years ago

What is the frequency of an xray with a wavelength of 2x10-10m?

Physics

1 answer:

Fudgin [204]3 years ago

7 0

Answer:

1.5 x 10¹⁸hz

Explanation:

Given parameters:

Wavelength = 2 x 10⁻¹⁰m

Unknown:

Frequency = ?

Solution:

To find the frequency, use the expression below;

V = f x wavelength

V is the speed of light = 3 x 10⁸m/s

f is the frequency

Now;

Insert the parameters

3 x 10⁸ = 2 x 10⁻¹⁰ x frequency

Wavelength = $\frac{3 x 10^{8} }{2 x 10^{-10} }$ = 1.5 x 10¹⁸hz

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3 years ago

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calcula la potencia por hora de un radiador, sabiendo que esta conectado a un contacto común 110 v. y requiere 20 Amp.

Ira Lisetskai [31]

calculate the power per hour of a radiator, knowing that it is connected to a common 110 v contact. and requires 20 Amp.

Answer:

2.2kWh

Explanation:

Given parameters:

Potential difference = 110v

Current = 20A

Unknown:

Power = ?

Solution:

To solve this problem, we use the expression below:

Power = IV

Power = 110 x 20 = 2200W

This is therefore 2.2kW

Power per hour = 2.2kWh

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3 years ago

If the axes of the two cylinders are parallel, but displaced from each other by a distance d, determine the resulting electric f

Novay_Z [31]

Answer:

E = ρ ( R1²) / 2 ∈o R

Explanation:

Given data

two cylinders are parallel

distance = d

radial distance = R

d < (R2−R1)

to find out

Express answer in terms of the variables ρE, R1, R2, R3, d, R, and constants

solution

we have two parallel cylinders

so area is 2 $\pi$ R × l

and we apply here gauss law that is

EA = Q(enclosed) / ∈o ......1

so first we find Q(enclosed) = ρ Volume

Q(enclosed) = ρ ( $\pi$ R1² × l )

so put all value in equation 1

we get

EA = Q(enclosed) / ∈o

E(2 $\pi$ R × l) = ρ ( $\pi$ R1² × l ) / ∈o

E = ρ ( R1²) / 2 ∈o R

6 0

3 years ago

A wire is oriented along the x-axis. It is connected to two batteries, and a conventional current of 2.4 A runs through the wire

Deffense [45]

Answer:

$\vec{F}=0.40176 N \hat{k}$

Explanation:

To calculate the force we need to use this equation

$\vec{F}=\int\limits^L_0 {i(\vec{dl}\times\vec{B})}$

where L is the total length of the wire

So in this case the small element of current is

$\vec{dl} = dx \hat{i}$

Because x is the direction of the current flow.

As is said in the problem B is such that

$\vec{B} = B \hat{j} = 0.62\hat{j} [ T]$

so to use the equation above we first calculate the following cross product:

$\vec{dl}\times\vec{B}=dx \hat{i}\times B \hat{j} = Bdx\hat{k}$

so the force:

$F = \int\limits^L_0 {i(\vec{dl}\times\vec{B})}=\int\limits^L_0{iBdx\hat{k}}$

So here we use the fact that B=0 in any point of the x axis that is not $x^{'}=0.27 [m]$ , that means that we only need to do the integration between a very short distant behind the point $x^{'}=0.27 [m]$ and a very short distant after that point, meaning: