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2 years ago

What speed must an electron have if its momentum is to be the same as that of an x-ray photon with a wavelength of 0. 85 nm?

1 answer:

4 0

Speed must an electron have if its momentum is to be the same as that of an x-ray photon with a wavelength of 0. 85 nm is 8.6*10^5m/s.

To find the answer, we have to know about the energy of photon.

<h3>What is the speed of the electron here?</h3>

As we know that the momentum of an x-ray photon with a wavelength w,

$P=\frac{h}{w}$

where; h is the plank's constant.

Thus, the momentum will be,

$P=\frac{6.63*10^{-34}}{0.85*10^{-9}} =7.8*10^{-25}kgm/s.$

We have to find the speed of the electron, thus, we have the expression of linear momentum as,

$P=mv\\v=\frac{P}{m} =\frac{7.8*10^{-25}}{9.1*10^{-31}}=8.6*10^5m/s.$

Thus, we can conclude that, speed must an electron have if its momentum is to be the same as that of an x-ray photon with a wavelength of 0. 85 nm is 8.6*10^5m/s.

Learn more about the energy of photon here:

brainly.com/question/3584036

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There is a clever kitchen gadget for drying lettuce leaves after you wash them. It consists of a cylindrical container mounted s

lukranit [14]

Answer:

Option B

$a=19 m/s^{2}$

Explanation:

Given information

Radius of container, r=12cm=12/100=0.12m

Angular velocity= 2 rev/s, converted to rad/s we multiply by 2π

Angular velocity, $\omega=2*2\pi =12.56637061$

We know that speed, $v=r\omega$

Centripetal acceleration, $a=\frac {v^{2}}{r}$ and substituting $v=r\omega$ we obtain that

$a=r\omega^{2}$

Substituting \omega for 12.56637061 and r for 0.12

$a=0.12*(12.56637061)^{2}=18.94964045 m/s^{2}$

Rounded off, $a=19 m/s^{2}$

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3 years ago

1) What do (x) and (y) symbolize?

MArishka [77]

Answer:

x is vertical and y is horizontal

Explanation:

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3 years ago

Which combination of temperature and pressure correctly describes standard temperature and pressure

raketka [301]

Answer: A combination 0 degrees Celsius and 101.3 kPa or 1 atm correctly describes standard temperature and pressure.

Explanation:

The term standard temperature and pressure is also known as STP and it is most commonly used when we want to calculate the density of a gas.

The term standard temperature means $32^{o}$ Fahrenheit or $0^{o}C$ or 273 Kelvin. On the other hand, term standard pressure means 1 atmosheric pressure of a gas.

Thus, we can conclude that a combination 0 degrees Celsius and 101.3 kPa or 1 atm correctly describes standard temperature and pressure.

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3 years ago

How do the amplitudes of a 120 decibel sound and a 100-decibel sound compare?

NemiM [27]

The 120 decibel sound has more amplitude than the 100 decibel sound.

In Physics, the relation between amplitude and intensity is that the intensity of the wave is directly proportional to the square of its amplitude.

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2 years ago

A tennis ball connected to a string is spun around in a vertical, circular path at a uniform speed. The ball has a mass m = 0.15

Oksanka [162]

1) 5.5 N

When the ball is at the bottom of the circle, the equation of the forces is the following:

$T-mg = m\frac{v^2}{R}$

where

T is the tension in the string, which points upward

mg is the weight of the string, which points downward, with

m = 0.158 kg being the mass of the ball

g = 9.8 m/s^2 being the acceleration due to gravity

$m \frac{v^2}{R}$ is the centripetal force, which points upward, with

v = 5.22 m/s being the speed of the ball

R = 1.1 m being the radius of the circular trajectory

Substituting numbers and re-arranging the formula, we find T:

$T=mg+m\frac{v^2}{R}=(0.158 kg)(9.8 m/s^2)+(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=5.5 N$

2) 3.9 N

When the ball is at the side of the circle, the only force acting along the centripetal direction is the tension in the string, therefore the equation of the forces becomes:

$T=m\frac{v^2}{R}$

And by substituting the numerical values, we find

$T=(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=3.9 N$

3) 2.3 N

When the ball is at the top of the circle, both the tension and the weight of the ball point downward, in the same direction of the centripetal force. Therefore, the equation of the force is

$T+mg=m\frac{v^2}{R}$

And substituting the numerical values and re-arranging it, we find

$T=m\frac{v^2}{R}-mg=(0.158 kg)\frac{5.22 m/s)^2}{1.1 m}-(0.158 kg)(9.8 m/s^2)=2.3 N$

4) 3.3 m/s

The minimum velocity for the ball to keep the circular motion occurs when the centripetal force is equal to the weight of the ball, and the tension in the string is zero; therefore:

$T=0\\mg = m\frac{v^2}{R}$

and re-arranging the equation, we find

$v=\sqrt{gR}=\sqrt{(9.8 m/s^2)(1.1 m)}=3.3 m/s$

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3 years ago