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olga2289 [7]
3 years ago
5

Two asteroids begin to gravitationally attract one another. If one asteroid has twice the mass of the other, which one experienc

es the greater force
Physics
1 answer:
asambeis [7]3 years ago
4 0

Answer:

F = G  \frac{2 m^{2} }{r^{2} }

action and reaction forces with the same magnitude, but in the opposite direction, one applied to each body.

Explanation:

The force between the asteroids is given by the law of universal gravitation

           F = G m₁ m₂ / r²

in this case they give us the mass of each asteroid

           m₁ = m

           m₂ = 2m

we substitute

           F = G m 2m / r²

           F = G 2 m² / r²

            F = G  \frac{2 m^{2} }{r^{2} }

Let's analyze this expression, this is the force that one asteroid exerts on the other, therefore there are two forces, one applied to each asteroid, as the order of the products does not affect the result, the magnitude of the force is the same on each asteroid.

These are action and reaction forces with the same magnitude, but in the opposite direction, one applied to each body.

Consequently the two asteroid experiences the same magnitude of force

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Using the formula:


a = (Vf - Vi) / t


Our initial velocity is 0 m/s, and our final velocity is 8.15 m/s, with a time period of 5 seconds:


a = (8.15 - 0.0) / 5

a = 1.63 m/s^2


If you know the acceleration due to gravity on the Moon, you can confirm this answer. The recorded gravitational acceleration on the Moon is 1.62 m/s^2.

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What mRNA sequence would result from the following DNA sequence?
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Answer:

UAC CUG AGG AUC

Explanation:

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<u>According to Chargaff's base pairing rule, the purine bases always pair with pyrimidine bases. Specifically, Adenine base must pair with Thymine base while Guanine base must pair with Cytosine base. In RNA, Thymine base is replaced with Uracil base.</u>

Hence:

ATG   GAC   TCC   TAG will pair with

UAC   CUG   AGG   AUC

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A machinist turns the power on to a grinding wheel, at rest at tome t=0 s. the wheel accelerates uniformly for 10 s and reaches
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Define orbital velocity
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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
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