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just olya [345]
3 years ago
15

Phương trình chuyển động thẳng đều của một chất điểm có dạng: x = 2t – 10. (x: km, t: h). Quãng đường đi được của chất điểm sau

2h là bao nhiêu?
Physics
2 answers:
siniylev [52]3 years ago
8 0

Answer:

Distance cover in 2  hour = 6 kilometer

Explanation:

Given equation:

x = 2t - 10

where

x = kilometer

t = hour

Find:

Distance cover in 2  hour

Computation:

T = 2

So,

x = 2t - 10

x = 2(2) - 10

x = 4 - 10

x = -6

Distance cover in 2  hour = 6 kilometer

vodomira [7]3 years ago
6 0

Answer:

The distance is 6 km.

Explanation:

The equation of uniform linear motion of a particle has the form: x = 2t – 10. (x: km, t: h). What is the distance traveled by the particle after 2 hours?

x = 2t - 10

distance traveled after t = 2 hours

Substitute t = 2 in the given expression

x = 2 x 2 - 10

x = 4 - 10

x = - 6 km

So, the distance is 6 km.

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Consider two ideal gases, A & B, at the same temperature. The rms speed of the molecules of gas A is twice that of gas B. Ho
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option (d)

Explanation:

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So for two gases

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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
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Correct order, from lowest potential energy to highest potential energy:

E - C - D - B - A

Explanation:

The gravitational potential energy of the car is given by:

U=mgh

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h is the height of the car relative to the ground

In the formula, we see that m and g are constant, so the potential energy of the car depends only on its height above the ground, h. The higher the car from the ground, the larger its potential energy. Therefore, the position with least potential energy will be E, since the height is the minimum. Then, C will have more potential energy, because the car is at higher position, and so on: the position with greatest potential energy is A, because the height of the car is maximum.

5 0
3 years ago
Read 2 more answers
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