Ask question

Ask question

All categories

2 years ago

10

Un caballo parte del reposo y alcanza una velocidad de 15m/s en un tiempo de 8s. ¿ Cuál fue su aceleración y que distancia recor

rió?

1 answer:

Lemur [1.5K]2 years ago

7 0

Answer: A 120 metros por segundo

Explanation: multiplicas la velocidad por el tiempo

You might be interested in

All of the following equations are statements of the ideal gas law except

elena55 [62]

Answer:

a. P = nRTV

Explanation:

The question is incomplete. Here is the complete question.

"All of the following equations are statements of the ideal gas law except a. P = nRTV b. PV/T = nR c. P/n = RT/v d. R = PV/nT"

Ideal gas equation is an equation that describes the nature of an ideal gas. The molecule of an ideal gas moves at a particular velocity depending on the temperature. This gases collides with one another elastically. The collision that an ideal gas experience is a perfectly elastic collision.

The ideal gas equation is expressed as shown:

PV = nRT where:

P is the pressure of the gas

V is the volume

n is the number of moles

R is the ideal gas constant

T is the temperature.

Based on the formula given for an ideal gas, it can be inferred that the equation. P = nRTV is not a statement of an ideal gas equation.

The remaining option will results to an ideal gas equation if they are cross multipled.

7 0

3 years ago

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

Read 2 more answers

Use the work–energy theorem to solve each of these problems. You can use Newton’s laws to check your answers.A skier moving at 5

erastovalidia [21]

Answer:

Yo. what is the answer.

Explanation:

8 0

3 years ago

Magnetic pole reversal worksheet

Likurg_2 [28]

??????????????????????

8 0

3 years ago

What do you do if you approach a railroad crossing with a crossbuck sign that has no lights or gates

miv72 [106K]

Proceed with caution before crossing

3 0

3 years ago