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Vlada [557]
3 years ago
5

If an ocean has a wavelength of 2 m and a frequency of 1 wave/s then it's speed is ___ m/s.

Physics
1 answer:
scoundrel [369]3 years ago
7 0
V = f x lamda

Wave speed = frequency x wavelength
2x1 = 2 so wave speed = 2m/s

Let me know if you have any questions
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To practice Problem-Solving Strategy 23.2 for continuous charge distribution problems. A straight wire of length L has a positiv
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Answer:

             E = k Q / [d(d+L)]

Explanation:

As the charge distribution is continuous we must use integrals to solve the problem, using the equation of the elective field

       E = k ∫ dq/ r² r^

"k" is the Coulomb constant 8.9875 10 9 N / m2 C2, "r" is the distance from the load to the calculation point, "dq" is the charge element  and "r^" is a unit ventor from the load element to the point.

Suppose the rod is along the x-axis, let's look for the charge density per unit length, which is constant

         λ = Q / L

If we derive from the length we have

        λ = dq/dx       ⇒    dq = L dx

We have the variation of the cgarge per unit length, now let's calculate the magnitude of the electric field produced by this small segment of charge

        dE = k dq / x²2

        dE = k λ dx / x²

Let us write the integral limits, the lower is the distance from the point to the nearest end of the rod "d" and the upper is this value plus the length of the rod "del" since with these limits we have all the chosen charge consider

        E = k \int\limits^{d+L}_d {\lambda/x^{2}} \, dx

We take out the constant magnitudes and perform the integral

        E = k λ (-1/x){(-1/x)}^{d+L} _{d}

   

Evaluating

        E = k λ [ 1/d  - 1/ (d+L)]

Using   λ = Q/L

        E = k Q/L [ 1/d  - 1/ (d+L)]

 

let's use a bit of arithmetic to simplify the expression

     [ 1/d  - 1/ (d+L)]   = L /[d(d+L)]

The final result is

     E = k Q / [d(d+L)]

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Six artificial satellites complete one circular orbit around a space station in the same amount of time. Each satellite has mass
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<h3>What is a good illustration of gravity?</h3>

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An english scientist created the first periodic table of elements
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Very large forces are produced in joints when a person jumps from some height to the ground. Calculate the force produced if an
krok68 [10]

Answer:

A) 31 kJ

B)  1.92 KJ

C) 40 ,  2.48

Explanation:

weight of person ( m ) = 79 kg

height of jump ( h ) = 0.510 m

Compression of joint material ( d ) = 1.30 cm ≈  0.013 m

A) calculate the force

Fd = mgh

F = mgh / d

W = mg

F(net) = W + F  = mg ( 1 + \frac{h}{d} )

          =  79 * 9.81 ( 1 + (0.51 / 0.013) )

          = 774.99 ( 40.231 ) ≈ 31 KJ

B) calculate the force when the stopping distance = 0.345 m

d = 0.345 m

Fd = mgh  hence  F = mgh / d

F(net) = W + F  = mg ( 1 + \frac{h}{d} )

          =  79 * 9.81 ( 1 + (0.51 / 0.345) )

          = 774.99 ( 2.478 ) = 1.92 KJ

C) Ratio of force in part a with weight of person

=  31000 / ( 79 * 9.81 ) = 31000 / 774.99 = 40

  Ratio of force in part b with weight of person

= 1920 / 774.99 = 2.48

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3 years ago
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