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Otrada [13]
4 years ago
10

The asteroid belt circles the sun between the orbits of mars and jupiter. one asteroid has a period of 6.2 earth years. what are

the asteroid's orbital radius and speed?
Physics
1 answer:
mash [69]4 years ago
5 0
<span>T² = 4π²rÂł / GM Solved for r : r = [GMT² / 4π²]â…“ Where G is the universal gravitational constant, M is the mass of the sun, T is the asteroid's period in seconds, and r is the radius of the orbit. Covert 5.00 years to seconds : 5.00years = 5.00years(365days/year)(24.0hours/day)(6... = 1.58 x 10^8s The radius of the orbit then is : r = [(6.67 x 10^-11Nâ™m²/kg²)(1.99 x 10^30kg)(1.58 x 10^8s)² / 4π²]â…“ = 4.38 x 10^11m The orbital speed can be found from the circular velocity formula : v = âš[GM / r] = âš[(6.67 x 10^-11Nâ™m²/kg²)(1.99 x 10^30kg) / 4.38 x 10^11m = 1.74 x 10^4m/s</span>
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A hollow metal cylinder has inner radius a, outer radius b, length L, and conductivity σ. The current I is radially outward from
ankoles [38]

Answer

current density is given by

I = \int J.da

where J = σ E

I = \int J.da\\ =\int \sigma E .da\\ = \sigma E \int da \\ =\sigma E \int_0^{2\pi} rL

now,

E=\dfrac{I}{2\pi r L \sigma}

a) Electric field strength at the inner surface of an iron cylinder

a = 0.5 cm = 0.005 m             b = 2.3 = 0.023

L = 10 cm = 0.1 m                     I = 27 A

E=\dfrac{I}{2\pi a L \sigma}

E=\dfrac{27}{2\pi\times 0.005\times 0.1\times 10^7}

      E = 8.59 x 10⁻⁴ V/m

b) Electric field strength at the outer surface of an iron cylinder

a = 0.5 cm = 0.005 m             b = 2.3 = 0.023

L = 10 cm = 0.1 m                     I = 27 A

E=\dfrac{I}{2\pi b L \sigma}

E=\dfrac{27}{2\pi\times 0.023\times 0.1\times 10^7}

      E =1.87 x 10⁻⁴ V/m

7 0
3 years ago
Help me please!!!!!!!!!!
Ksenya-84 [330]

Answer:

B because 17100 is the half of the carbon

7 0
3 years ago
A straightforward method of finding the density of an object is to measure its mass and then measure its volume by submerging it
Thepotemich [5.8K]

Answer:

<em>The density of rock = 3.37 g/cm³</em>

Explanation:

Density: Density can be defined as the ratio of the mass of a body to the volume. The S.I unit of density is kg/m³. It can be expressed mathematically as ,

D = M/V............................................... Equation 1

Where D = density of the body, M = mass of the body, V = volume of the body.

From Archimedes' principle, a body will displace a volume of water equal to  its volume.

Therefore, Volume of the object = volume of water displaced

<em>Given: M = 300 g, V = volume of water displace = 89.0 cm³.</em>

<em>Substituting these values into equation 1</em>

<em>D = 300/89</em>

<em>D = 3.37 g/cm³</em>

<em>The density of rock = 3.37 g/cm³</em>

4 0
4 years ago
Amazon rectangular bar of low carbon steel A-36 is exposed to an axial strees of 150 MPa. What is the original length of the bar
kolbaska11 [484]

Answer:

1.8m

Explanation:

Let the Elastics of the steel ASTM-36 E = 200000 MPa

The strain of the bar when subjected to 150 MPa is

\epsilon = \frac{\sigma}{E} = \frac{150}{200000} = 0.00075

Therefore, if the bar elongates by 1.35 mm, then the original length L would be:

\epsilon = \frac{\Delta L}{L}

L = \frac{\Delta L}{\epsilon} = \frac{1.35}{0.00075} = 1800 mm or 1.8m

5 0
3 years ago
The Event Horizon Telescope needs a 22 micro-arcsecond resolution to view the event horizon regions around black holes. If the a
likoan [24]

Answer:

14869817.395 m

Explanation:

\theta=22 microarcsecond

λ = Wavelength = 1.3 mm

Converting to radians we get

22\times 10^{-6}\frac{\pi}{180\times 3600}\ radians

From Rayleigh Criterion

\theta=1.22\frac{\lambda}{D}\\\Rightarrow D=1.22\frac{\lambda}{\theta}\\\Rightarrow D=1.22\frac{1.3\times 10^{-3}}{22\times 10^{-6}\frac{\pi}{180\times 3600}}\\\Rightarrow D=14869817.395\ m

Diameter of the effective primary objective is 14869817.395 m

It is not possible to build one telescope with a diameter of 14869817.395 m. But, we need this type of telescope. So, astronomers use an array of radio telescopes to achieve a virtual diameter in order to observe objects that are the size of supermassive black hole's event horizon.

7 0
3 years ago
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