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ExtremeBDS [4]
3 years ago
13

The distance between planets varies as they move in their orbits. The minimum distance from the Earth to Mars is about 0.35 AU.

About how many kilometers is that?
Physics
2 answers:
Natalija [7]3 years ago
6 0

Answer:

c on edge

Explanation:

Alina [70]3 years ago
5 0

Answer:

52,360,000km

Explanation:

To solve this problem you use a conversion factor.

By taking into account that 1UA = 1.496*10^{8}km you obtain:

0.35UA*\frac{1.496*10^{8}km}{1UA}=52,360,000km

hence, 0.35UA is about 52,360,000km. This is the least distance between Mars and Earth

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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
Read 2 more answers
Scientists have organized all forms of energy into two types—kinetic and potential. Kinetic energy is the energy of motion, and
mihalych1998 [28]

Answer:

Potential energy only

Explanation:

at the top of its swing the pendulum stops moving , (therefore it has no KINETIC energy)  thus all of the energy is stored as potential energy.

7 0
3 years ago
ONE sound wave travels through two containers of different gasses. Wave through container 1 has a wavelength of 1.2 m. Wave thro
Zarrin [17]
I’m not 100% but the answer is b
7 0
2 years ago
The average specific heat of the human body is 3.6 kJ/kg·°C. If the body temperature of a(n) 96-kg man rises from 37°C to 39°C d
d1i1m1o1n [39]

Answer:

691200 J

Explanation:

From specific heat capacity,

ΔQ = cmΔt.................. Equation 1

Where ΔQ = increase in thermal energy, c = specific heat capacity of the body, m = mass of the man, Δt = rise in temperature.

Given: c = 3.6 kJ/kg.°C = 3600 J/kg.°C, m = 96 kg, Δt = 39-37 = 2 °C.

Substitute into equation 1

ΔQ = 3600×96×2

ΔQ = 691200 J.

Hence the change in the thermal energy of the body = 691200 J

5 0
3 years ago
Two blocks are connected by a massless cable which goes through the center of a rotating turntable. The blocks have masses M1 =
Airida [17]

Answer:

If the final question is; at what velocity will the first block start to move outward in m/s?

v = 3.5596 \frac{m}{s}

Explanation:

The motion have the velocity that will make the block move using:

F_{1}*F_{r}+ F_{2}= Ec \\Ec= \frac{M*v^{2} }{r}

m_{1} = 1.2 Kg

m_{2} = 2.8 Kg

r = 0,45 m

μs= 0,54

Resolving:

\frac{m_{1}*v^{2}  }{r} = us*m_{1}* g + m_{2} * g

v^{2} = \frac{((us *m_{1} + m_{2})*g )* r}{m_{1} }

v^{2} = \frac{((0.54 *1.2 + 2.8)*9.8 )* 0.45}{1.2}

v^{2} = 12.6714 \frac{m^{2} }{s^{2} }

v = 3.559 \frac{m}{s}

4 0
3 years ago
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