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3 years ago

a water wav e has frequency of 2hz and there are 3 meters between each crest on the wave. how fast is the wave moving

Physics

1 answer:

Dafna11 [192]3 years ago

6 0

Answer:

6 m/s

Explanation:

The velocity of waves, v is a product of frequency and wavelength expressed as v=fw where v is the velocity, f is frequency and w is wavelength.

Wavelength is defined as distance between two successive crests or troughs. In this case, the wavelength is 3m

Substituting 2 Hz for frequency and 3 m for wavelength then the speed of waves will be v=2*3=6 m/s

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How much kinetic energy does a proton gain if it is accelerated, with no friction, through a potential difference of 1.00 V? The

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Answer:

If energy is conserved, then the sum of the potential energy and the kinetic energy is a constant.

Assuming the proton starts from rest, so it's kineitc energy is zero, but it has a potential energy, PE equal to:

PE = qV

where q =1.6 x 10^-19 C

and V = 1.00 V

Assuming the proton no longer experiences the potential energy and it is all converted to kinetic energy then:

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3 years ago

Two electrons are at rest and separated by a distance of 4.32 × 10-10 m. When they are released they accelerate away from each o

sasho [114]

Answer:

Speed of electron when their separation increased by a factor of 4.10 is 9.41 x 10⁵ m/s .

Explanation:

The electric potential energy is given by the relation :

$U = \frac{kq_{1}q_{2} }{r}$

Here q₁ and q₂ are the two charge particles and r is the distance between them and k is electric constant.

In this case, there are two electrons which are separated by the distance 4.32 x 10⁻¹⁰ m.

Let e be the electron charge and r₁ be the distance between them. Then, the initial electric potential energy is :

$U_{1} = \frac{ke^{2} }{r_{1} }$

Now, the distance between the electrons increases by the factor of 4.10. Let r₂ be the new distance between them i.e. r₂ = 4.10 r₁.

Thus, the new electric potential energy is :

$U_{2} = \frac{ke^{2} }{r_{2} }=\frac{ke^{2} }{4.10r_{1} }$

Applying law of conservation of energy :

ΔU = ΔK

Here ΔU is change in electric potential energy and ΔK is change in kinetic energy.

( U₁ - U₂ ) = ( K₂ - K₁ )

Here K₂ and K₁ are initial and final kinetic energy of electron.

Since, the electron initially is at rest, so its initial kinetic energy is zero. Thus, the above equation becomes:

K₂ = U₁ - U₂

$\frac{1}{2}mv^{2}=\frac{ke^{2} }{r_{1} }- \frac{ke^{2} }{4.10r_{1} }$

Here m and v are the mass and final speed of electron respectively.

$v^{2}=\frac{2}{m} \frac{ke^{2} }{r_{1} }(1- \frac{1 }{4.10 })$

Substitute 9.1 x 10⁻³¹ kg for m, 9 x 10⁹ N m² C⁻² for k, 1.6 x 10⁻¹⁹ C for e and 4.32 x 10⁻¹⁰ m for r₁ in the above equation.

$v^{2}=\frac{2}{9.1\times10^{-31} } \frac{9\times10^{9}\times(1.6\times10^{-19})^{2} }{4.32\times10^{-10} }(1- \frac{1 }{4.10 })$

$v^{2}=8.86\times10^{11}$

v = 9.41 x 10⁵ m/s

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