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Marat540 [252]
3 years ago
5

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?
Physics
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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The half-wave rectifier circuit of vs(t) = 170 sin(377t) V and a load resistance R = 15Ω. Determine: a. The average load current
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).

6 0
3 years ago
Read 2 more answers
The gauge pressure inside an alveolus with a 200 µm radius is 25 mmHg, while the blood pressure outside is only 10 mmHg. Assumin
alekssr [168]

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.

3 0
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>

Therefore, the magnitude of the tangential velocity

=0.872<em>m/secs</em>

5 0
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

T=\frac{d}{1.50}

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}

For more information on this visit

brainly.com/question/22271063

4 0
2 years ago
Oscilloscope amplitude and frequency problem. Study the above graph. The volts/div dial is set to 2 volts/div and the time/div d
denis-greek [22]

Answer:

Amplitude = 8 Volts

Frequency = 0.067 kHz

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)

Finding Frequency:

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,

Frequency = \frac{1}{Time Period}\\Frequency = \frac{1}{15m}\\Frequency = 0.067 kHz

8 0
3 years ago
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