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Ksenya-84 [330]
3 years ago
15

The wave frequency is where 1 hertz equals 1 wave passed a fixed point in 1 second.

Physics
1 answer:
ahrayia [7]3 years ago
3 0

                     (10 waves) / (5 sec) 

                 =  (10/5) (wave/sec)

                 =      2      per sec

                 =      2        Hz .
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Inside a vacuum tube, an electron is in the presence of a uniform electric field with a magnitude of 320 N/C. (a) What is the ma
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(a) The magnitude of the acceleration of the electron is 5.62 x 10¹³ m/s².

(b) The speed of the electron after the given time is  4.78 x 10⁵ m/s.

<h3>Acceleration of the electron</h3>

The acceleration of the electron is calculated as follows;

F = qE

ma = qE

a = qE/m

a = (1.6 x 10⁻¹⁹ x 320)/(9.11 x 10⁻³¹)

a = 5.62 x 10¹³ m/s²

<h3>Speed of the electron</h3>

v = at

v = 5.62 x 10¹³ m/s² x  8.50 x 10⁻⁹ s

v = 4.78 x 10⁵ m/s

Learn more about speed here: brainly.com/question/4931057

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2 years ago
How does the currently accepted model of the nucleus provide evidence of the existence of the strong nuclear force?
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Protons and neutrons are all attracted to each other as a result - the strong nuclear force. This is an attractive force that only has an effect over a very short range in the nucleus.

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3 years ago
A state patrol officer saw a car start from rest at a highway​ on-ramp. She radioed ahead to another officer 20 mi along the hig
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Answer:

We know from the basic speed distance relation that

Speed=\frac{Distance}{Time}

Since the car started from rest and it covered the distance between the 2 officer's in 19 minutes we have speed of the car

Speed=\frac{Distance}{Time}\\\\Speed=\frac{20}{\frac{19}{60}}=63.16mph

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3 years ago
A pillow is thrown downward
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Answer:

Is that the question?

Explanation:

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(a) If a proton with a kinetic energy of 6.2 MeV is traveling in a particle accelerator in a circular orbit with a radius of 0.5
Tju [1.3M]

Answer:

The fraction of its energy that it radiates every second is 3.02\times10^{-11}.

Explanation:

Suppose Electromagnetic radiation is emitted by accelerating charges. The rate at which energy is emitted from an accelerating charge that has charge q and acceleration a is given by

\dfrac{dE}{dt}=\dfrac{q^2a^2}{6\pi\epsilon_{0}c^3}

Given that,

Kinetic energy = 6.2 MeV

Radius = 0.500 m

We need to calculate the acceleration

Using formula of acceleration

a=\dfrac{v^2}{r}

Put the value into the formula

a=\dfrac{\dfrac{1}{2}mv^2}{\dfrac{1}{2}mr}

Put the value into the formula

a=\dfrac{6.2\times10^{6}\times1.6\times10^{-19}}{\dfrac{1}{2}\times1.67\times10^{-27}\times0.51}

a=2.32\times10^{15}\ m/s^2

We need to calculate the rate at which it emits energy because of its acceleration is

\dfrac{dE}{dt}=\dfrac{q^2a^2}{6\pi\epsilon_{0}c^3}

Put the value into the formula

\dfrac{dE}{dt}=\dfrac{(1.6\times10^{-19})^2\times(2.3\times10^{15})^2}{6\pi\times8.85\times10^{-12}\times(3\times10^{8})^3}

\dfrac{dE}{dt}=3.00\times10^{-23}\ J/s

The energy in ev/s

\dfrac{dE}{dt}=\dfrac{3.00\times10^{-23}}{1.6\times10^{-19}}\ J/s

\dfrac{dE}{dt}=1.875\times10^{-4}\ ev/s

We need to calculate the fraction of its energy that it radiates every second

\dfrac{\dfrac{dE}{dt}}{E}=\dfrac{1.875\times10^{-4}}{6.2\times10^{6}}

\dfrac{\dfrac{dE}{dt}}{E}=3.02\times10^{-11}

Hence, The fraction of its energy that it radiates every second is 3.02\times10^{-11}.

5 0
3 years ago
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