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artcher [175]
2 years ago
6

The wavelength of a water wave is 54 m. It has a frequency of 0.03 Hz. What is the velocity of the wave?

Physics
1 answer:
dalvyx [7]2 years ago
6 0

Answer:

1.62 m/s

Explanation:

Wavelength of the water wave= 54 m

The frequency is 0.03 Hz

Therefore the velocity can be calculated as follows

Velocity= frequency × wavelength

= 0.03 × 54

= 1.62 m/s

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A long solenoid that has 1 200 turns uniformly distributed over a length of 0.420 m produces a magnetic field of magnitude 1.00
Fantom [35]

Answer:

<h2>The current required  winding is  2.65*10^-^2 mA</h2>

Explanation:

We can use the expression B=μ₀*n*I-------1 for the magnetic field that enters a coil  and

n= N/L (number of turns per unit length)

Given data

The number of turns n= 1200 turns

length L= 0.42 m

magnetic field B= 1*10^-4 T

μ₀= 4\pi*10^-^7 T.m/A

Applying the equation  B=μ₀*n*I

I= B/μ₀*n

I= B*L/μ₀*n

I= \frac{1*10^-^4*0.42}{4\pi*10^-^7*1.2*10^3 }

I= 2.65*10^-^2 mA

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If a boy believed that the only variable that affected the distance a ball traveled when thrown was the force with which it was
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Suppose Galileo dropped a lead ball (100 kilograms) and a glass ball (1 kilogram) from the Leaning Tower of Pisa. Which one hit
Reika [66]

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How many places are there for electrons in the innermost shell of an atom? A. 2 B. 4 C. 8 D. 12
Butoxors [25]

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5 0
3 years ago
A nonconducting sphere is made of two layers. The innermost section has a radius of 6.0 cm and a uniform charge density of −5.0C
Leno4ka [110]

Answer:

a) E =0, b)   E = 1,129 10¹⁰ N / C , c)    E = 3.33 10¹⁰ N / C

Explanation:

To solve this exercise we can use Gauss's law

        Ф = ∫ E. dA = q_{int} / ε₀

Where we must define a Gaussian surface that is this case is a sphere; the electric field lines are radial and parallel to the radii of the spheres, so the scalar product is reduced to the algebraic product.

           E A = q_{int} /ε₀

The area of ​​a sphere is

          A = 4π r²

         E = q_{int} / 4πε₀ r²

         k = 1 / 4πε₀

         E = k q_{int} / r²

To find the charge inside the surface we can use the concept of density

        ρ = q_{int} / V ’

         q_{int} = ρ V ’

         V ’= 4/3 π r’³

Where V ’is the volume of the sphere inside the Gaussian surface

 Let's apply this expression to our problem

a) The electric field in center r = 0

     Since there is no charge inside, the field must be zero

          E = 0

b) for the radius of r = 6.0 cm

In this case the charge inside corresponds to the inner sphere

        q_{int} = 5.0  4/3 π 0.06³

         q_{int} = 4.52 10⁻³ C

        E = 8.99 10⁹  4.52 10⁻³ / 0.06²

         E = 1,129 10¹⁰ N / C

c) The electric field for r = 12 cm = 0.12 m

In this case the two spheres have the charge inside the Gaussian surface, for which we must calculate the net charge.

     The charge of the inner sphere is q₁ = - 4.52 10⁻³ C

The charge for the outermost sphere is

       q₂ =  ρ 4/3 π r₂³

       q₂ = 8.0 4/3 π 0.12³

       q₂ = 5.79 10⁻² C

The net charge is

     q_{int} = q₁ + q₂

     q_{int} = -4.52 10⁻³ + 5.79 10⁻²

     q_{int} = 0.05338 C

The electric field is

        E = 8.99 10⁹ 0.05338 / 0.12²

        E = 3.33 10¹⁰ N / C

8 0
2 years ago
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