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4 years ago

14

Explain how the mass of a planet affects the motion of the planet around the sun?

1 answer:

vodomira [7]4 years ago

6 0

Answer:

The gravitational attraction of the Sun is what holds the planets in their elliptical orbits. So to explain this the mass effects the motion of the planets because the strength of gravitational force depends of the mass.

Explanation:

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Two train whistles have identical frequencies of 1.85 102 Hz. When one train is at rest in the station and the other is moving n

Makovka662 [10]

Answer:

Explanation:

frequency of whistle = 1.85 x 10² = 185 Hz

frequency of beat heard = 8 beat /s . No of beat produced is equal to difference of frequencies of two sound source . Here difference is created due to Doppler effect . One of the train is moving so it will have apparent frequency which is different one from its original frequency .

When the moving train is approaching the observer , its frequency will be higher . As beat is heard at the rate of 8 beats / s , apparent frequency of approaching train will be 185 + 8 = 193 Hz .

Applying Doppler's formula of apparent frequency ,

193 = 185 x V / ( V - v ) , where V is velocity of sound and v is velocity of train .

193 V - 193 v = 185 V

193 v = 8 V

v = 8 x V / 193

= 8 x 343 / 193

= 14.21 m /s

Second possibility is that apparent velocity is less ie 185 - 8 = 177 Hz

In that case moving train will be moving away from observer . If its velocity be v

177 = 185 x V / ( V + v )

177 V + 177 v = 185 V

v = 8 x 343 / 177

= 15.50 m /s .

6 0

4 years ago

Consider the elementary gas-phase reversible reaction A 3C Pure A enters at a temperature of 400 K and a pressure of 10 atm. At

xxMikexx [17]

Answer:

39%

Explanation:

The equilibrium equation is:

$A\rightleftharpoons 3C$

The initial concentration of A can be calculated from the ideal gas equation:

$pV=nRT\\\\n/v=p/(RT)\\\\C_A=\dfrac{10atm}{0.08206(atm-dm^3/K-mol)/times 400K}\\\\C_A=0.304mol/liter$

Determine the conversion of substance A to substance C using an ICE table and the Kc constant:

A ⇄ 3C

I 0.304 0

C - x + 3x

E 0.304 - x 3x

$K_c=0.25=\dfrac{(3x)^3}{(0.304-x)}$

Solve for x:

You need to use a graphing calculator:

108x³ = 0.304 - x

108x³ + x - 0.304 = 0

x ≈ 0.1195 mol/liter

Then:

$C_A=0.304mol/liter-0.1195mol/liter=0.1845mol/liter\\\\C_C=3\times 0.304mol/liter=0.912mol/liter$

The equilbrium conversion is:

$\% = [0.304mol/liter-0.1845mol/liter]/(0.304mol/liter)\times 100$

$\% \approx 39\%$

3 0

3 years ago

If the weight of an object that is submerged in a fluid is 10 N and the buoyant force on it is

Nady [450]

Answer: the object should be overcome by buoyancy and rise in the fluid.

Explanation:

5 0

3 years ago

An ethernet cable is 3.80 m long and has a mass of 0.210 kg. A transverse pulse is produced by plucking one end of the taut cabl

Tasya [4]

Answer:

$T=94.54N$

Explanation:

The tension in a cable is given by:

$T=\mu v^2(1)$

Where $\mu$ is the mass density of the cable and v is the speed of the cable's pulse. These values are defined as:

$\mu=\frac{m}{L}(2)\\v=\frac{d}{t}$

The pulse makes four trips down and back along the cable, so $d=4(2L)$

$v=\frac{8L}{t}(3)$

Replacing (2) and (3) in (1), we calculate the tension in the cable:

$T=\frac{m}{L}(\frac{8L}{t})^2\\T=\frac{64mL}{t^2}\\T=\frac{64(0.21kg(3.80m))}{(0.735s)^2}\\T=94.54N$

3 0

3 years ago

If you and a friend are standing at opposites ends of a gymnasium and one of you claps, will the other hear the clap at the same

34kurt

Answer:

No. Your friend will not hear the clap when he/she sees it.

Explanation:

It takes time for sound waves to go through a large area. It takes longer if people are in the area rather than it is empty.

3 0

3 years ago