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1 year ago

combination or combinations of which method allow in principle measure the density of the extrasolar planet?

1 answer:

8 0

We can determine a planet's orbital period and separation from its star using any detection method. The transit method can yield sizes, whereas the astrometric and doppler approaches can provide minimum masses.

We can calculate average density by combining the transit and doppler approaches. Numerous physical properties, including the semi-major axis, stellar mass, star radius, planet radius, eccentricity, and inclination, are calculated from these observable data. The mass of the planet is also calculated using the star's combined radial velocity readings.

List briefly the planetary characteristics that, in theory, can be detected with the present detection techniques. We can determine a planet's orbital period and separation from its star using any detection method. The transit method can yield sizes, whereas the astrometric and doppler approaches can provide minimum masses.

To know more about orbital period

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A man wears convex lens glasses of focal length 30cm in order to correct his eyes defect. Instead of the optimum 25cm, his dista

omeli [17]

Answer:

14 cm

Explanation:

F = (frac{uv}{u – v})

F = +ve

v = -ve

30 = (frac {25 {times} (-v)}{25 – (-v)})

v = (frac {25 {times} (-v)}{25+v})

v = 14cm

(Note that either negative or positive values go to show the positioning and hence, they are not a strong necessity in your final answer.)

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2 years ago

wich of the following are commonly distributed by veterinary assistants in typical veterinary practice?

vredina [299]

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3 years ago

How many total oxygen atoms are in the compound Molybdenum (V) Dichromate?

alekssr [168]

Answer:

$1.2646\times10^{25}\ atoms$

Explanation:

-The chemical formula for Molybdenum (V) Dichromate is $Mo(Cr_2O_7)_3$

-There are 21 moles of oxygen per one mole of Molybdenum (V) Dichromate

-We apply Avogadro's constant to find the number of atoms of oxygen:

$Avogadro's \ Constant=6.022\times 10^{23} \ mol_1\\\\No\ of \ Atoms=Moles\times Avogadro's \ Constant\\\\=21\times 6.022\times 10^{23} \\\\=1.2646\times10^{25}\ atoms$

Hence, there are $1.2646\times10^{25} \ atoms$

3 0

3 years ago

A particle traveling in a circular path of radius 300 m has an instantaneous velocity of 30 m/s and its velocity is increasing a

hodyreva [135]

Answer:

5 m/s2

Explanation:

The total acceleration of the circular motion is made of 2 components: centripetal acceleration and linear acceleration of 4 m/s2. They are perpendicular to each other.

The centripetal acceleration is the ratio of instant velocity squared and the radius of the circle

$a_c = \frac{v^2}{r} = \frac{30^2}{300} = \frac{900}{300} = 3 m/s^2$

So the magnitude of the total acceleration is

$a = \sqrt{a_c^2 + a_l^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 m/s^2$

4 0

3 years ago

To avoid breakdown of the capacitors, the maximum potential difference to which any of them can be individually charged is 125 V

aleksley [76]

Answer:

The maximum energy stored in the combination is 0.0466Joules

Explanation:

The question is incomplete. Here is the complete question.

Three capacitors C1-11.7 μF, C2 21.0 μF, and C3 = 28.8 μF are connected in series. To avoid breakdown of the capacitors, the maximum potential difference to which any of them can be individually charged is 125 V. Determine the maximum energy stored in the series combination.

Energy stored in a capacitor is expressed as E = 1/2CtV² where

Ct is the total effective capacitance

V is the supply voltage

Since the capacitors are connected in series.

1/Ct = 1/C1+1/C2+1/C3

Given C1 = 11.7 μF, C2 = 21.0 μF, and C3 = 28.8 μF

1/Ct = 1/11.7 + 1/21.0 + 1/28.8

1/Ct = 0.0855+0.0476+0.0347

1/Ct = 0.1678

Ct = 1/0.1678

Ct = 5.96μF

Ct = 5.96×10^-6F

Since V = 125V

E = 1/2(5.96×10^-6)(125)²

E = 0.0466Joules

8 0

3 years ago