A 5 kg mass is moving at 3 ms-1. Calculate its kinetic energy.

Write one to two sentences explaining whether this passage from the story is exposition, rising action, climax, or resolution: “

fiasKO [112]

Answer:

This passage is part of the resolution because it shows what happens after the climax. It wraps up the conflict, and then the story is over.

Explanation: cause I am smart thats why

4 0

3 years ago

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The wattage of a bulb is 24 W when it is connected to a 12 V battery. Calculate its effective wattage if it operates on a 6 V ba

tresset_1 [31]

It would be 12W because: 6v is half of 12v so half of 24w would be 12w

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3 years ago

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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:

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

3 0

3 years ago

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Help with speed problems #1-3

-Dominant- [34]

1. Each plot represents the meters traveled by both the Hare and the Tortoise over a certain period of time (minutes).

2. The Tortoise lines show it lines is steadily increasing over a period of time. So as more time elapses the faster the tortoise becomes it travels more meters. The Tortoise line shows steady acceleration.

3. The Hare in the first 5 minutes had a rapid fast advancement up to 40 meters. But for the 5-20 mins. period the Hare did not move at all. Its speed stayed at the same place. But towards the end 20-25 mins. marks the Hare started moving again. At the end the Hare at first had a rapid acceleration but stopped for a long time then it sped up briefly.

8 0

3 years ago

If the PLATE SEPARATION of an isolated charged parallel-plate capacitor is doubled: A. the electric field is doubled

mash [69]

Explanation:

The electric field of an isolated charged parallel-plate capacitor is given by :

$E=\dfrac{q}{A\epsilon_o}$ ........(1)

Where

q is the electric charge

A is the area of cross section of parallel plate

It is clear from equation (1) that the electric field of a parallel plate capacitor is directly proportional to the charge on the plate and inversely proportional to the area of cross section of a plate.

So, the correct option is (E) i.e. "none of the above".

5 0

3 years ago