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3 years ago

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g In a certain binary-star system, each star has the same mass which is 8.2 times of that of the Sun, and they revolve about the

ir center of mass. The distance between them is the 6.2 times the distance between Earth and the Sun. What is their period of revolution in years?

1 answer:

Mademuasel [1]3 years ago

7 0

To solve this problem it is necessary to apply the concepts related to the Third Law of Kepler.

Kepler's third law tells us that the period is defined as

$T^2 = \frac{4\pi^2 d^3}{2GM}$

The given data are given with respect to known constants, for example the mass of the sun is

$m_s = 1.989*10^{30}$

The radius between the earth and the sun is given by

$r = 149.6*10^9m$

From the mentioned star it is known that this is 8.2 time mass of sun and it is 6.2 times the distance between earth and the sun

Therefore:

$m = 8.2*1.989*10^{30}$

$d = 6.2*149.6*10^6$

Substituting in Kepler's third law:

$T^2 = \frac{4\pi^2 d^3}{2}$

$T^2 = \frac{4\pi^2(6.2*149.6*10^9)^3}{2(6.674*10^{-11} )(8.2*1.989*10^30 )}$

$T=\sqrt{\frac{4\pi^2(6.2*149.6*10^9)^3}{2(6.674*10^{-11} )(8.2*1.989*10^30)}}$

$T = 120290789.7s$

$T = 120290789.7s(\frac{1year}{31536000s})$

$T \approx 3.8143 years$

Therefore the period of this star is 3.8years

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irina1246 [14]

Answer:

a average load current = 11.33 A

b rms load current = 8.02A

c true power =962.64 W

d apparent power =962.64 W

e. power factor cosθ =1

Explanation:

Vs (t) = 170 sin(377t) v

Vm =170v

Vrms = 170/√2 =120.23 v

Im = Vm/R = 170/15 = 11.33 A

Irms = Im/ √2 = 11.33/√2 =8.02A

Resistors are electronic components that consume energy

the power in a resistor is given by P =IVcosθ ; in a resistor cosθ =1

P =IV

The electrical power consumed by a resistance, (R) is called the true or real power

and is obtained by multiplying the rms voltage with the rms current.

P= Vrms × Irms

120.03×8.02

P= 962.64 W ; true power

apparent power = Vrms × Irms

=120.03×8.02

= 962.64W ; apparent power

power factor cosθ = true power/ apparent power

cosθ = 962.64/962.64

cosθ = 1

For the purely resistive circuit, the power factor is 1 , because the reactive power is equal to zero (0).

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3 years ago

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The gauge pressure inside an alveolus with a 200 µm radius is 25 mmHg, while the blood pressure outside is only 10 mmHg. Assumin

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Answer:

The surface tension is 0.0318 N/m and is sufficiently less than the surface tension of the water.

Solution:

As per the question:

Radius of an alveolus, R = $200\mu m = 200\times 10^{- 6}\ m$

Gauge Pressure inside, $P_{in} = 25\ mmHg$

Blood Pressure outside, $P_{o} = 10\ mmHg$

Now,

Change in pressure, $\Delta P = 25 - 10 = 15\ mmHg = 1.99\times 10^{3}\ Pa$

Since the alveolus is considered to be a spherical shell

The surface tension can be calculated as:

$\Delta P = \frac{4\pi T}{R}$

$T = \frac{1.99\times 10^{3}\times 200\times 10^{- 6}}{4\pi} = 0.0318\ N/m = 0.318\ mN/m$

And we know that the surface tension of water is 72.8 mN/m

Thus the surface tension of the alveolus is much lesser as compared to the surface tension of water.

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3 years ago

An athlete completes 1 laps around a track with a radius of 25 meters in 180 seconds. What is the magnitude of the athlete's tan

DanielleElmas [232]

Answer:

0.872<em>m/s</em>

Explanation:

Tangential velocity is given by the formula,

$v= 2\pi r/ t$

In the question given,

radius= 25meters

time= 180secs

pie= 3.14

number of laps= 1

The magnitude of tangential velocity equals;

$\frac{1lap* 2 *3.14 *25m}{180secs}$

<em>v </em>= 157<em>m</em>/180<em>secs</em>

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2 years ago

If the average velocity during the athlete's walk back

goblinko [34]

Hence ,From the Guide there are other parameters which with this equation will give the exact time the athlete's walk back

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From the question we are told

If the average velocity during the athlete's walk back to the starting line in Guided Example 2.5 is – 1.50 m/s,

Generally the equation Time spent is mathematically given as

T=\frac{d}{v}

Therefore

$T=\frac{d}{1.50}$

Hence ,From the Guide there are other parameters which with this equation will give the exact time the athlete's walk back

$T=\frac{d}{1.50}$

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Oscilloscope amplitude and frequency problem. Study the above graph. The volts/div dial is set to 2 volts/div and the time/div d

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Answer:

Amplitude = 8 Volts

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Explanation:

Note: The missing picture in question is attached for your review.

Given:

Volts/Div = 2 V/div

Time/Div = 5 msec/div

Finding Amplitude:

Now, as you can see in the attached picture, there are 4 division between two peaks of the waveform, so,

$Amplitude = 4 div/volts * 2 volts/ div )\\Amplitude = 8 Volts$

(Multiplying by 2 V/div because oscilloscope dial is set at 2 V/div)

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As can be seen in attached picture, 3 division are there for one complete cycle of waveform,so,

$Time Period = 3 div * 5msec /div\\Time Perod = 15 msec$

Since,

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3 years ago