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lord [1]
3 years ago
11

Our Sun is a G2V star with absolute magnitude 4.8. Suppose that a star of spectral type G2V is observed to have apparent magnitu

de -0.2. How far away is it? A 1 parsec. B 5 parsecs. C 10 parsecs. D 100 parsecs. E 1000 parsecs.
Physics
1 answer:
Fofino [41]3 years ago
8 0

Answer:

Option D

The star is at a distance of 100 parsecs.

Explanation:

The distance can be determined by means of the distance modulus:

M - m = 5log(d) - 5  (1)

Where M is the absolute magnitude, m is the apparent magnitude and d is the distance in units of parsec.

Therefore, d can be isolated from equation 1

log(d) = (M - m + 5)/5

Then, Applying logarithmic properties it is gotten:

d = 10^{(M - m + 5)/5}  (2)

The absolute magnitude is the intrinsic brightness of a star, while the apparent magnitude is the apparent brightness that a star will appear to have as is seen from the Earth.

Since both have the same spectral type is absolute magnitude will be the same.

Finally, equation 2 can be used:

d = 10^{(4.8 - (-0.2)+ 5)/5}  

d = 100 pc    

Hence, the star is at a distance of 100 parsecs.

<em>Key term:</em>

Parsec: Parallax of arc seconds

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You add 500 mL of water at 10°C to 100 mL of water at 70°C. What is the
Sphinxa [80]

Answer:

Option (c) : 20°C

Explanation:

t(final) =  \frac{w1 \times t1 + w2 \times t2}{w1 + w2}

T(final) = 500* 10 + 100*70/600 = 20°C

4 0
3 years ago
An airplane dropped a flare from a height of 2860 feet above a lake. How many seconds did it take for the flare to reach the wat
KATRIN_1 [288]

Answer: 13.2 seconds.

Explanation: using equation of motion; S= ut +1/2at² where u = initial velocity=0

S= distance travelled

a = acceleration due gravity

t= time.

1 foot = 0.305m so,

S= 2860 feet =872.3m

S= ut+1/2 at²

872.3 = 0×t + 1/2×10 × t²

872.3 =0 + 5t²

T²= 872.3/5

T²= 174.46

Take the square root of T we then have;

t = 13.2 seconds to one decimal place.

8 0
3 years ago
Read 2 more answers
A 50.0 Watt stereo emits sound waves isotropically at a wavelength of 0.700 meters. This stereo is stationary, but a person in a
photoshop1234 [79]

Answer:

a) f' = 432 Hz

b) I = 8.12*10^-4 W/m^2

Explanation:

a) To calculate the frequency of sound waves that car receives, you take into account the Doppler effect. In this case (observer moves away of the source) you have the following formula:

f'=f(\frac{v-v_o}{v+v_s})    (1)

where

f: frequency of the source = ?

v: speed of sound = 343 m/s

vo: speed of the observer = 40.0 m/s

vs: speed of the source = 0 m/s (stationary)

You replace the values of all parameters in the equation (1):

To calculate f' you first calculate the frequency of the sound wave, by using the following formula:

v=\lambda f\\\\

v: speed of sound

λ: wavelength = 0.700 m

f=\frac{v}{\lambda}=\frac{343m/s}{0.700m}=480Hz

Next, you replace the values of all parameters in the equation (1):

f'=(490Hz)(\frac{343m/s-40.0m/s}{343m/s})=432Hz

hence, the frequency perceived by the car is 432 Hz

b) To calculate the power of the sound wave, when the car is 70.0 maway from the speaker, you use the following formula:

I=\frac{P}{4\pi r^2}

P: power of the source = 50.0 W

r: distance to the source = 70.0 m

I=\frac{50.0 W}{4\pi(70.0m)^2}=8.12*10^{-4}\frac{W}{m^2}

hence, the intensity is 8.12*10^⁻4 W/m^2

3 0
3 years ago
How many atoms are in each product silver sulfide ​
AlekseyPX

there are 3 atoms in each silver sulfide

5 0
3 years ago
Consider an oblique shock wave with a wave angle of 35 o . Upstream of the wave, the static pressure and temperature are 2,000 l
ziro4ka [17]

Answer:

The pressure is 6570  lbf/ft²

The temperature is 766 ⁰R

The velocity is 2746.7 ft/s

deflection angle behind the wave is 17.56⁰

Explanation:

Speed of air at initial condition:

a_1 = \sqrt{\gamma RT } =  \sqrt{1.4* 1716*520 } = 1117.70 \ ft/s

γ is the ratio of specific heat, R is the universal gas constant, and T is the initial temperature.

initial mach number

M_1 = \frac{v_1}{a_1} = \frac{3355}{1117.7}  = 3

then, M_n = M_1sin \beta = 3sin(35) = 1.721

based on the values obtained, read off the following from table;

P₂/P₁ = 3.285

T₂/T₁ = 1.473

Mₙ₂ = 0.6355

Thus;

P₂ = 3.285P₁ = 3.285(2000) = 6570  lbf/ft²

T₂ = 1.473T₁ = 1.473(520⁰R) = 766 ⁰R

Again; to determine the velocity and deflection angle, first we calculate the mach number.

M_t_1 = M_1cos \beta = 3 cos(35) = 2.458

w_2 = a_1M_t_1 = 2.458(1117.70) = 2746.7 \ ft/s

a_2 = \sqrt{\gamma RT_2} = \sqrt{1.4*1718*766} = 1357.34 \ ft/s

v_2 = a_2M_n_2 = 1357.34(0.6355) = 862.59 \ ft/s

Tan(\beta -\theta) = \frac{v_2}{w_2} = \frac{862.59}{2746.7}  \\\\Tan(\beta -\theta) = 0.314\\\\\beta -\theta= 17.44\\\\\theta = \beta - 17.44 = 17.56^o

6 0
3 years ago
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