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3 years ago

A particle moves with a velocity v in a circle of radius r, then its angular velocity is equal

Physics

1 answer:

saul85 [17]3 years ago

3 0

Answer:

Given that,

Speed = v

Radius = r

We have to ascertain the precise speed

Utilizing equation of speed

Where, v = velocity/speed

r = radius

= precise speed/ angular velocity

Angular Velocity :

The precise speed is characterized by the speed of turn.

The precise speed is straightforwardly corresponding to the direct speed and contrarily relative to the range of the molecule.

Subsequently, The precise speed is v/r

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Itzel travels south 4 miles and then goes east 2 miles and then north 10 miles. What is her displacement? Round your answer to t

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The displacement of Itzel according to the question is 6.3 miles SW

Displacement is defined as the distance moved by a body in a specified direction

Find the diagram attached

From the diagram given, we can see that AB is the displacement

To get the length AB, we will have to use the Pythagoras theorem:

$AB^2=2^2+ 6^2\\AB^2 ^2=4+36\\AB^2=40\\AB=\sqrt{40}\\AB= 6.3 miles\\$

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2 years ago

two masses are separated by 1 m. suppose the masses are moved so they are 2 m apart how will the gravitational force change

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3 years ago

What are (a) the energy of a photon corresponding to wavelength 6.0 nm, (b) the kinetic energy of an electron with de Broglie wa

dlinn [17]

Explanation:

Given that,

Wavelength = 6.0 nm

de Broglie wavelength = 6.0 nm

(a). We need to calculate the energy of photon

Using formula of energy

$E = \dfrac{hc}{\lambda}$

$E=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{6.0\times10^{-9}}$

$E=3.315\times10^{-17}\ J$

(b). We need to calculate the kinetic energy of an electron

Using formula of kinetic energy

$\lambda=\dfrac{h}{\sqrt{2mE}}$

$E=\dfrac{h^2}{2m\lambda^2}$

Put the value into the formula

$E=\dfrac{(6.63\times10^{-34})^2}{2\times9.1\times10^{-31}\times(6.0\times10^{-9})^2}$

$E=6.709\times10^{-21}\ J$

(c). We need to calculate the energy of photon

Using formula of energy

$E = \dfrac{hc}{\lambda}$

$E=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{6.0\times10^{-15}}$

$E=3.315\times10^{-11}\ J$

(d). We need to calculate the kinetic energy of an electron

Using formula of kinetic energy

$\lambda=\dfrac{h}{\sqrt{2mE}}$

$E=\dfrac{h^2}{2m\lambda^2}$

Put the value into the formula

$E=\dfrac{(6.63\times10^{-34})^2}{2\times9.1\times10^{-31}\times(6.0\times10^{-15})^2}$

$E=6.709\times10^{-9}\ J$

Hence, This is the required solution.

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3 years ago