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melomori [17]
2 years ago
12

What is the frequency of radiation whose wavelength is 11.5 a0 ?

Physics
1 answer:
irakobra [83]2 years ago
4 0

Answer:

The frequency of radiation is 2.61 \times 10^{17} s^{-1}

Explanation:

Given:

Wavelength \lambda = 11.5 \times 10^{-10} m

Speed of light c = 3 \times 10^{8} \frac{m}{s}

For finding the frequency of radiation,

  c = f \lambda

  f = \frac{c}{\lambda}

  f = \frac{3 \times 10^{8} }{11.5 \times 10^{-10} }

  f = 2.61 \times 10^{17} s^{-1}

Therefore, the frequency of radiation is 2.61 \times 10^{17} s^{-1}

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A sphere of radius 5.00 cm carries charge 3.00 nC. Calculate the electric-field magnitude at a distance 4.00 cm from the center
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Answer:

a)   E = 8.63 10³ N /C,  E = 7.49 10³ N/C

b)   E= 0 N/C,  E = 7.49 10³ N/C  

Explanation:

a)  For this exercise we can use Gauss's law

         Ф = ∫ E. dA = q_{int} /ε₀

We must take a Gaussian surface in a spherical shape. In this way the line of the electric field and the radi of the sphere are parallel by which the scalar product is reduced to the algebraic product

The area of ​​a sphere is

        A = 4π r²

 

if we use the concept of density

        ρ = q_{int} / V

        q_{int} = ρ V

the volume of the sphere is

      V = 4/3 π r³

         

we substitute

         E 4π r² = ρ (4/3 π r³) /ε₀

         E = ρ r / 3ε₀

the density is

         ρ = Q / V

         V = 4/3 π a³

         E = Q 3 / (4π a³) r / 3ε₀

         k = 1 / 4π ε₀

         E = k Q r / a³

 

let's calculate

for r = 4.00cm = 0.04m

        E = 8.99 10⁹ 3.00 10⁻⁹ 0.04 / 0.05³

        E = 8.63 10³ N / c

for r = 6.00 cm

in this case the gaussine surface is outside the sphere, so all the charge is inside

         E (4π r²) = Q /ε₀

         E = k q / r²

let's calculate

         E = 8.99 10⁹ 3 10⁻⁹ / 0.06²

          E = 7.49 10³ N/C

b) We repeat in calculation for a conducting sphere.

For r = 4 cm

In this case, all the charge eta on the surface of the sphere, due to the mutual repulsion between the mobile charges, so since there is no charge inside the Gaussian surface, therefore the field is zero.

         E = 0

In the case of r = 0.06 m, in this case, all the load is inside the Gaussian surface, therefore the field is

        E = k q / r²

      E = 7.49 10³ N / C

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A 5.0-kg crate is resting on a horizontal plank. The coefficient of static friction is 0.50 and the coefficient of kinetic frict
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Answer:

The mass of the crate is 5kg.

We know that the force of friction can be obtained by:

F = N*k

where k is the coefficient of friction, where we use the static one if the object is at rest, and the kinetic one if the object os moving. N is the normal force

If we tilt the base making an angle of 30° with the horizontal, now the normal force against the plank will be equal to the fraction of the weight in the direction normal to the surface of the plank.

Knowing that the angle is 30°, then the fraction of the weight that pushes against the normal is Cos(30°)*W = cos(30°)*5kg*9.8m/s^2 = 42.4N

The fraction of the force in the parallel direction to the plank (the force that would accelerate the crate downwards) is:

F = sin(30°)*5k*9,8m/s = 24.5N

now, the statical friction force is:

Fs = 42.4N*0.5 = 21.2N

The statical force is less than the 24.5N, so the crate will move downwards, then the force that acts on the crate is the kinetic force of friction:

Fk = 42.4N*0.4 = 16.96N

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Answer:

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Since the body moves at a constant speed, there is no acceleration so the sum of forces on the X-axis must be equal to zero.

The frictional force is defined as the product of the coefficient of friction by the normal force. In this way, we can calculate the coefficient of friction.

The process of solving this problem can be seen in the attached image.

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