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3 years ago

HURRY!!!!

Physics

2 answers:

nevsk [136]3 years ago

7 0

Answer:

light strikes the grooves → different wavelengths of light bend at different angles → diffracted wavelength reach the eyes → the eyes see different colors

Explanation:

A light is a combination of colors which can be separated appropriately by the use of diffraction grating. Diffraction grating is an optical device that splits and diffracts light into several beams and colors with respect to their wavelength. The size of the grooves and wavelength determine the direction of dispersion.

Bingel [31]3 years ago

4 0

I'm not exactly sure but I'm thinking that it's the last one. Sorry if I'm wrong

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Gold has a higher density than tin? True or false

Ganezh [65]

Answer:

True. Gold does have a higher density than tin

3 0

3 years ago

Answer the amplitude part of the questions in 7, 8, & 9

erastova [34]

I actually believe for the first question, it would be complete destructive interference as the amplitude and the approximate wavelength for each are the same and will completely or entirely cancel out, rather than simply decreasing or lowering the amplitude as in the bottom question.

The amplitude for the first will be 0, as the 2 waves will cancel each other out. The amplitude of the second, will be 3x, I believe, assuming the amplitude of the first is 2x and the second is 1x, in a constructive interference, I believe the amplitudes would add up.

Likewise for the bottom, I believe you would be subtracting the supposed amplitude of the first which is 2x from 1x which would be 1x.

5 0

3 years ago

What is the time period of vibration?

Y_Kistochka [10]

Answer:

A time period is denoted by 'T' . It is the time to complete one cycle of vibration. As the frequency of a wave increases, the time period of the wave decreases. The unit for time period is 'seconds

HOPE IT HELPS :)

PLEASE MARK IT THE BRAINLIEST!

4 0

3 years ago

What is the shortest-wavelength x-ray radiation in m that can be generated in an x-ray tube with an applied voltage of 93.3 kV?

VikaD [51]

(a) $1.33\cdot 10^{-11} m$

The x-rays in the tube are emitted as a result of the collisions of electrons (accelerated through the potential difference applied) on the metal target. Therefore, all the energy of the accelerated electron is converted into energy of the emitted photon:

$e \Delta V = \frac{hc}{\lambda}$

where the term on the left is the electric potential energy given by the electron, and the term on the right is the energy of the emitted photon, and where:

$e=1.6\cdot 10^{-19}C$ is the electron's charge

$\Delta V = 93.3 kV = 93300 V$ is the potential difference

$h=6.63\cdot 10^{-34} Js$ is the Planck constant

$c=3.00\cdot 10^8 m/s$ is the speed of light

$\lambda$ is the wavelength of the emitted photon

Solving the formula for $\lambda$ , we find:

$\lambda=\frac{hc}{e\Delta V}=\frac{(6.63\cdot 10^{-34})(3\cdot 10^8)}{(1.6\cdot 10^{-19})(93300)}=1.33\cdot 10^{-11} m$

(b) 93300 eV (93.3 keV)

The energy of the emitted photon is given by:

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

where

h is Planck constant

c is the speed of light

$\lambda=1.33\cdot 10^{-11} m$ is the wavelength of the photon, calculated previously

Substituting,

$E=\frac{(6.63\cdot 10^{-34})(3\cdot 10^8)}{1.33\cdot 10^{-11}}=1.50\cdot 10^{-14} J$

Now if we want to convert into electronvolts, we have to divide by the charge of the electron:

$E=\frac{1.50\cdot 10^{-14} J}{1.6\cdot 10^{-19} J/eV}=93300 eV$

(c) The following statements are correct:

The maximum photon energy is just the applied voltage times the electron charge. (1)

The value of the voltage in volts equals the value of the maximum photon energy in electron volts.

In fact, we see that statement (1) corresponds to the equation that we wrote in part (a):

$e \Delta V = \frac{hc}{\lambda}$

While statement (2) is also true, since in part (b) we found that the photon energy is 93.3 keV, while the voltage was 93.3 kV.

3 0

3 years ago

A straight wire in a magnetic field experiences a force of 0.023 N when the current in the wire is 2.5 A. The current in the wir

Tatiana [17]

Answer:

The new current in the wire is 6.3 A.

Explanation:

Given that,

Force in a straight wire, F = 0.023 N

Current in the wire, I = 2.5 A

If new force is, F' = 0.058 N, we need to find the new current.

We know that, magnetic force is given by :

$F=ilB\sin\theta$

It is clear that force is directly proportional to electric current. So, we can say:

$\dfrac{F}{F'}=\dfrac{I}{I'}$

I' is new current

$I'=\dfrac{IF'}{F}\\\\I'=\dfrac{2.5\times 0.058}{0.023}\\\\I'=6.3\ A$

So, the new current in the wire is 6.3 A. Hence, this is the required solution.

8 0

3 years ago