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3 years ago

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Which of the following is a characteristic of the sun energy arrive on Earth A.it cannot travel through the air B.It cannot trav

el through space C.It is made up of rays of varying frequencies. D.It is made up of rays of the same wavelength

1 answer:

Svetradugi [14.3K]3 years ago

5 0

The answer to that is (C). Hope this helps

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The width of the central maxima, formed from light of wavelength 575 nm behind a single slit that has a width of 115 μm, is 1.15

Lady bird [3.3K]

Answer:

L = 1.15 m

Explanation:

The diffraction phenomenon is described by the equation

a sin θ = m λ

Where a is the width of the slit, λ the wavelength and m is an integer, the order of diffraction is left.

The diffraction measurements are made on a screen that is far from the slit, and the angles in the experiment are very small, let's use trigonometry

tan θ = y / L

tan θ = sint θ / cos θ≈ sin θ

We substitute in the first equation

a (y / L) = m λ

The first maximum occurs for m = 1

The distance is measured from the center point of maximum, which coincides with the center of the slit, in this case the distance is the total width of the central maximum, so the distance (y) measured from the center is

y = 1.15 / 2 = 0.575 cm

y = 0.575 10⁻² m

Let's clear the distance to the screen (L)

L = a y / λ

Let's calculate

L = 115 10⁻⁶ 0.575 10⁻² / 575 10⁻⁹

L = 1.15 m

3 0

3 years ago

The photon energies used in different types of medical x-ray imaging vary widely, depending upon the application. Single dental

pav-90 [236]

A) $5.0\cdot 10^{-11} m$

The energy of an x-ray photon used for single dental x-rays is

$E=25 keV = 25,000 eV \cdot (1.6\cdot 10^{-19} J/eV)=4\cdot 10^{-15} J$

The energy of a photon is related to its wavelength by the equation

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

where

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

$c=3\cdot 10^8 m/s$ is the speed of light

$\lambda$ is the wavelength

Re-arranging the equation for the wavelength, we find

$\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{4\cdot 10^{-15}J}=5.0\cdot 10^{-11} m$

B) $2.0\cdot 10^{-11} m$

The energy of an x-ray photon used in microtomography is 2.5 times greater than the energy of the photon used in part A), so its energy is

$E=2.5 \cdot (4\cdot 10^{-15}J)=1\cdot 10^{-14} J$

And so, by using the same formula we used in part A), we can calculate the corresponding wavelength:

$\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{1\cdot 10^{-14}J}=2.0\cdot 10^{-11} m$

4 0

3 years ago

Jill is getting ready to push little Frank in his swing. She pulls Frank back as high as she can and then releases the swing. Wh

makkiz [27]

The transfer of energy is potential energy to kinetic energy. The swing has potential energy when she pulls it back and once she lets go, allowing the swing to move, it then has kinetic energy.

5 0

3 years ago

ASAP pls answer right and I will mark brainiest

Ronch [10]

1. D:Asteroids have geological activity
2.A:Short-period comets circle around the sun in 200 years or less and originate in the Kuiper Belt
3.D:All of the above
4.A:It’s gravity is too weak to clear it’s orbit
5.B:Ceres
6.C:Haumea
7.A:Is bigger than Pluto

I hope this helps

6 0

3 years ago

If a car is accelerating downhill under a net force 3674 N , what additional force would cause the car to have a constant veloci

Semenov [28]

Answer:

For the car to move with constant velocity the additional force required is $F__{dg} }= -3674 \ N$

Explanation:

From the question we are told that

The net force of the car is $F_{net} = 3674 \ N$

Generally the total force acting on the car is the net force plus the force due to gravity acting in direction of the car (Let denote it as $F__{dg}$ )

So the total force acting on the car is mathematically represented as

$F_{net} + F__{dg}} = F$

Here this F representing the total force can be mathematically represented as

$F = ma$

Now for constant velocity to be attained, the acceleration of the car will be zero

So at constant velocity

$F = m * 0$

=> $F = 0 \ N$

So

$F_{net} + F__{dg}} = 0$

=> $F__{dg} }= -F_{net}$

=> $F__{dg} }= -3674 \ N$

6 0

3 years ago