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2 years ago

A stone is thrown upward with an initial upward velocity of 15 m/s from a

Physics

1 answer:

tensa zangetsu [6.8K]2 years ago

7 0

Answer

for every meters it will go up 15 so if it was 2 secoonds it woudl be 30 and if it was 3 seconds it would be 45 meters

Explanation:

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X-rays with an energy of 300 keV undergo Compton scattering from a target. If the scattered rays are detected at 30 relative to

lys-0071 [83]

Answer:

a) $\Delta \lambda = \lambda' -\lambda_o = \frac{h}{m_e c} (1-cos \theta)$

For this case we know the following values:

$h = 6.63 x10^{-34} Js$

$m_e = 9.109 x10^{-31} kg$

$c = 3x10^8 m/s$

$\theta = 37$

So then if we replace we got:

$\Delta \lambda = \frac{6.63x10^{-34} Js}{9.109 x10^{-31} kg *3x10^8 m/s} (1-cos 37) = 4.885x10^{-13} m * \frac{1m}{1x10^{-15} m}= 488.54 fm$

b) $\lambda_0 = \frac{hc}{E_0}$

With $E_0 = 300 k eV= 300000 eV$

And replacing we have:

$\lambda_0 = \frac{1240 x10^{-9} eV m}{300000eV}=4.13 x10^{-12}m = 4.12 pm$

And then the scattered wavelength is given by:

$\lambda ' = \lambda_0 + \Delta \lambda = 4.13 + 0.489 pm = 4.619 pm$

And the energy of the scattered photon is given by:

$E' = \frac{hc}{\lambda'}= \frac{1240x10^{-9} eVm}{4.619x10^{-12} m}=268456.37 eV - 268.46 keV$

c) $E_f = E_0 -E' = 300 -268.456 kev = 31.544 keV$

Explanation

Part a

For this case we can use the Compton shift equation given by:

$\Delta \lambda = \lambda' -\lambda_0 = \frac{h}{m_e c} (1-cos \theta)$

For this case we know the following values:

$h = 6.63 x10^{-34} Js$

$m_e = 9.109 x10^{-31} kg$

$c = 3x10^8 m/s$

$\theta = 37$

So then if we replace we got:

$\Delta \lambda = \frac{6.63x10^{-34} Js}{9.109 x10^{-31} kg *3x10^8 m/s} (1-cos 37) = 4.885x10^{-13} m * \frac{1m}{1x10^{-15} m}= 488.54 fm$

Part b

For this cas we can calculate the wavelength of the phton with this formula:

$\lambda_0 = \frac{hc}{E_0}$

With $E_0 = 300 k eV= 300000 eV$

And replacing we have:

$\lambda_0 = \frac{1240 x10^{-9} eV m}{300000eV}=4.13 x10^{-12}m = 4.12 pm$

And then the scattered wavelength is given by:

$\lambda ' = \lambda_0 + \Delta \lambda = 4.13 + 0.489 pm = 4.619 pm$

And the energy of the scattered photon is given by:

$E' = \frac{hc}{\lambda'}= \frac{1240x10^{-9} eVm}{4.619x10^{-12} m}=268456.37 eV - 268.46 keV$

Part c

For this case we know that all the neergy lost by the photon neds to go into the recoiling electron so then we have this:

$E_f = E_0 -E' = 300 -268.456 kev = 31.544 keV$

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2 years ago

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Red, green, and blue light rays each enter a drop of water from the same direction.Which light ray's path through the drop will

KengaRu [80]

Blue will be the most and red the least

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3 years ago

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How would you measure the potential difference across a resistor?

coldgirl [10]

You would need to connect a potential difference across a resistor in parallel

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2 years ago

Home heating systems usually work in the following way: When the furnace warms a house above the set temperature of the thermost

Paraphin [41]

Home heating systems work through negative feedback by regulating the temperature through thermostat.

<h3>How home heating systems work through negative feedback?</h3>

In thermostat, when the temperature of the room environment increases by the negative feedback system the furnace gets turn off, allowing the room to get to its normal temperature and when the temperature lower down it turns on again and maintain the temperature in equilibrium state.

So we can conclude that home heating systems work through negative feedback by regulating the temperature through thermostat.

Learn more about temperature here: brainly.com/question/24746268

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2 years ago

A solid disk has a total kinetic energy k. what is its rotational kinetic energy krot if it's rolling without slipping? express

Dominik [7]

the total energy is the sum of the linear and rotational energy: <span>
K = K_rot + K_lin

first, we find the rotational kinetic energy of a rotating disc with an angular velocity of w. see the references for the moment of inertia of a disc.
K_rot = (1/2)(I)(w^2)
I = (1/2)(m)(r^2)
K_rot = (1/4)(m)(r^2)(w^2)

next, we find the linear kinetic energy of a rolling disc:
K_lin = (1/2)(m)(v^2)
v = angular velocity * circumference
= w * (pi * 2 * r)
K_lin = (1/2)(m)(w*2*pi*r)^2
= (2*pi^2)(m)(r^2)(w^2)

we find the total kinetic energy:
K = K_rot + K_lin
= (1/4)(m)(r^2)(w^2) + (2*pi^2)(m)(r^2)(w^2)

and find the rotational contribution:
K_rot = K * [K_rot/K]
K_rot = K * [K_rot/(K_rot+K_lin)]
K_rot = K * (1/4) / [(1/4) + (2*pi^2)]
</span>K_rot = K / (8*pi^2 + 1)

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3 years ago