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Usimov [2.4K]
3 years ago
13

The earth travels around the sun once a year in an approximately circular orbit whose radius is 1.50x10^11 m. From this data det

ermine (a) the orbital speed of the earth and (b) the mass of the sun.
Physics
1 answer:
seraphim [82]3 years ago
8 0
(a) Determine the circumference of the Earth through the equation,
            C = 2πr
Substituting the known values, 
           C = 2π(1.50 x 10¹¹ m)
             C = 9.424 x 10¹¹ m

Then, divide the answer by time which is given to a year which is equal to 31536000 s. 
          orbital speed = (9.424 x 10¹¹ m)/31536000 s

               orbital speed = 29883.307 m/s

Hence, the orbital speed of the Earth is ~29883.307 m/s.

(b) The mass of the sun is ~1.9891 x 10³⁰ kg. 
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Does this marble have potential energy?<br> 1) no<br> 2)yes
jek_recluse [69]

Answer:

yes

Explanation:

because it has the potential to move

5 0
3 years ago
Find the kinetic energy of an electron whose de broglie wavelength is 34.0 nm.
dezoksy [38]
The De Broglie wavelength of the electron is
\lambda=34.0 nm=34 \cdot 10^{-9} m
And we can use De Broglie's relationship to find its momentum:
p= \frac{h}{\lambda}= \frac{6.6 \cdot 10^{-34} Js}{34 \cdot 10^{-9} m}=1.94 \cdot 10^{-26} kg m/s

Given p=mv, with m being the electron mass and v its velocity, we can find the electron's velocity:
v= \frac{p}{m}= \frac{1.94 \cdot 10^{-26} kgm/s}{9.1 \cdot 10^{-31} kg}=  2.13 \cdot 10^4 m/s

This velocity is quite small compared to the speed of light, so the electron is non-relativistic and we can find its kinetic energy by using the non-relativistic formula:
K= \frac{1}{2}mv^2= \frac{1}{2}(9.1 \cdot 10^{-31} kg)(2.13 \cdot 10^4 m/s)^2=2.06 \cdot 10^{-22} J
3 0
3 years ago
If these gas molecules were compressed further, how would the average kinetic energy of the molecules be affected? A. It would i
Advocard [28]
I believe the correct response would be B. It would decrease.
8 0
3 years ago
Which, if any, of the following statements about electric field lines is/are true? A.The electric field is always perpendicular
jonny [76]
The electric field is always perpendicular to the surface outside of a conductor. TRUE

<span> If an electron were placed on an electric field line, it would move in a direction perpendicular to the field. FALSE, it would move in an anti-parallel direction because its charge is negative </span>
 
<span>Electric field lines originate on positive charge and terminate on negative charge. TRUE ; but they can also go to infinity </span>
 
It is possible for two electric field lines to cross each other.
<span> Usually FALSE; though technically possible at special points where field is zero. </span>
 
If an electron and a positron were in the presence of a very strong electric field, they would move away from each other.
<span> TRUE; one is positive, and one is negative. If the field is strong enough, the action of the field will overcome the mutual attraction between them </span>
 
It is not possible for the electric field to ever be zero. FALSE: it IS possible, inside a conductor for instance
   
If a proton were placed on an electric field line, it would move in a direction anti-parallel to the field.
<span> FALSE: being positive, it would move in the SAME direction as the field</span>ic 
8 0
3 years ago
a stone with a mass of 2.40 kg is moving with velocity (6.60î − 2.40ĵ) m/s. find the net work (in j) on the stone if its velocit
ch4aika [34]

By the work energy theorem, the total work done on the stone is given by its change in kinetic energy,

W = \Delta K = \dfrac m2 ({v_2}^2 - {v_1}^2)

We have

\vec v_1 = (6.60\,\vec\imath - 2.40\,\vec\jmath)\dfrac{\rm m}{\rm s} \implies {v_1}^2 = \|\vec v_1\|^2 = 49.32 \dfrac{\rm m^2}{\rm s^2}

\vec v_2 = (8.00\,\vec\imath + 4.00\,\vec\jmath) \dfrac{\rm m}{\rm s} \implies {v_2}^2 = \|\vec v_2\|^2 = 80.0\dfrac{\mathrm m^2}{\mathrm s^2}

Then the total work is

W = \dfrac{2.40\,\rm kg}2 \left(80.0\dfrac{\rm m^2}{\rm s^2} - 49.32\dfrac{\rm m^2}{\rm s^2}\right)  \approx \boxed{36.8\,\rm J}

5 0
2 years ago
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