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3 years ago

Which of the following is an example of distance equaling displacement?

Physics

1 answer:

AleksandrR [38]3 years ago

3 0

Answer:

Explanation:

The student had displaced their in the class when she left. The phone is what's displaced and student leaving equals distance.

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X rays of wavelength 0.0169 nm are directed in the positive direction of an x axis onto a target containing loosely bound electr

mamaluj [8]

Answer:

a) 4.04*10^-12m

b) 0.0209nm

c) 0.253MeV

Explanation:

The formula for Compton's scattering is given by:

$\Delta \lambda=\lambda_f-\lambda_i=\frac{h}{m_oc}(1-cos\theta)$

where h is the Planck's constant, m is the mass of the electron and c is the speed of light.

a) by replacing in the formula you obtain the Compton shift:

$\Delta \lambda=\frac{6.62*10^{-34}Js}{(9.1*10^{-31}kg)(3*10^8m/s)}(1-cos132\°)=4.04*10^{-12}m$

b) The change in photon energy is given by:

$\Delta E=E_f-E_i=h\frac{c}{\lambda_f}-h\frac{c}{\lambda_i}=hc(\frac{1}{\lambda_f}-\frac{1}{\lambda_i})\\\\\lambda_f=4.04*10^{-12}m +\lambda_i=4.04*10^{-12}m+(0.0169*10^{-9}m)=2.09*10^{-11}m=0.0209nm$

c) The electron Compton wavelength is 2.43 × 10-12 m. Hence you can use the Broglie's relation to compute the momentum of the electron and then the kinetic energy.

$P=\frac{h}{\lambda_e}=\frac{6.62*10^{-34}Js}{2.43*10^{-12}m}=2.72*10^{-22}kgm\\$

$E_e=\frac{p^2}{2m_e}=\frac{(2.72*10^{-22}kgm)^2}{2(9.1*10^{-31}kg)}=4.06*10^{-14}J\\\\1J=6.242*10^{18}eV\\\\E_e=4.06*10^{-14}(6.242*10^{18}eV)=0.253MeV$

5 0

3 years ago

Help please I do not understand

ahrayia [7]

What don’t you understand? If you haven’t uploaded anything

3 0

3 years ago

The notes produced by a tuba range in frequency from approximately 45 Hz to 375 Hz. Find the possible range of wavelengths in ai

taurus [48]

Answer:

The possible range of wavelengths in air produced by the instrument is 7.62 m and 0.914 m respectively.

Explanation:

Given that,

The notes produced by a tuba range in frequency from approximately 45 Hz to 375 Hz.

The speed of sound in air is 343 m/s.

To find,

The wavelength range for the corresponding frequency.

Solution,

The speed of sound is given by the following relation as :

$v=f_1\lambda_1$

Wavelength for f = 45 Hz is,

$\lambda_1=\dfrac{v}{f_1}$

$\lambda_1=\dfrac{343}{45}=7.62\ m$

Wavelength for f = 375 Hz is,

$\lambda_2=\dfrac{v}{f_2}$

$\lambda_2=\dfrac{343}{375}=0.914\ m/s$

So, the possible range of wavelengths in air produced by the instrument is 7.62 m and 0.914 m respectively.

6 0

3 years ago

An office building has a 24-volt branch circuit installed for landscape lighting around the front of the building. The circuit w

Arturiano [62]

The circuit was installed in UF cable which requires a minimum burial depth of 6 inches for this circuit.

<h3>UF cable</h3>

UF cable is used as an underground feeder cable to distribute power from an existing building to outdoor equipment. UF cable can also be used as direct burial cable.

For the 24-volt branch circuit installed, the minimum burial depth will be 6 inches.

Thus, the circuit was installed in UF cable which requires a minimum burial depth of 6 inches for this circuit.

Learn more about UF cable here: brainly.com/question/8591560

5 0

2 years ago

What happened to the weight of an object when it is taken from Earth to the Moon? why?<br>

Sholpan [36]

Answer:

the weight of the object decreases when it is taken from the Earth to the Moon

Explanation:

The weight of an object is defined as the product of the mass of the object with the acceleration due to gravity of the Planet.

$W =mg$

where,

W = weight of the object

m = mass of the object

g = acceleration due to gravity on the planet

The mass of an object remains constant everywhere in the universe. Therefore, the weight is directly proportional to the value of acceleration due to gravity.

The value of acceleration due to gravity on the Moon is lesser than its value on the Earth.

<u>Hence, the weight of the object decreases when it is taken from the Earth to the Moon </u>

6 0

3 years ago