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bonufazy [111]
4 years ago
14

Using Figure 25-2, determine how Giant stars differ from main sequence stars.

Physics
2 answers:
erastova [34]4 years ago
8 0
Main sequence stars lay on a diagonal going from upper left angle to lower right angle. Part of main sequence is our Sun.

When we observe this diagram we need to compare giant stars and main sequence stars. GIant stars are positioned to upper right when compared to main sequence. From labels on coordinate axis we can see that giant stars generaly have lower temperature than main sequence stars. They also have higher brightness and lower magnitude (meaning that they are more bright in night sky).
lora16 [44]4 years ago
6 0

Answer:

Giant stars differ from main sequence stars in having greater absolute magnitudes for the same temperatures.

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What is drag? In what situations might it act on the force of gravity?
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3 years ago
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Which element has an atomic number of 9 and an atomic mass of 19?
yan [13]

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If the car passes point A with a speed of 20 m/s and begins to increase its speed at a constant rate of at = 0.5 m/s2 , determin
IceJOKER [234]

Answer:

1.68 \frac{m}{s^2}

Explanation:

Please find the image for the question as attached file.

Solution -

Given -

First of all we will calculate the velocity at point C,

As per newton's third law of motion-

V_C^2 = V_A^2 + 2 a_t (S_C - S_A)\\

Substituting the given values in above equation, we get -

V_C^2 = 20^2 + 2*0.5*(100-0)\\V_C = 22.361 \frac{m}{s}

Now we will determine the radius of curvature for the curve shown in the attached image

Y = 16 - \frac{1}{625} X^2\\

Differentiating on both the sides, we get -

\frac{dy}{dx} = -3.2 (10^-3) X\\\frac{d^2y}{d^2x} =  -3.2 (10^-3)\\Curve = \frac{[1+(\frac{dy}{dx})^2]^{\frac{3}{2}}  }{\frac{d^2y}{d^2x}} \\Curve = 312.5meter

Acceleration on curved path

a = \frac{V_C^2}{Curve} \\a = \frac{22.361^2}{312.5} \\a= 1.60 \frac{m}{s^2}

Final acceleration

a_f = \sqrt{0.5^2 + 1.6^2} \\a_f = 1.68\frac{m}{s^2}

5 0
3 years ago
A sample of an ideal gas has a volume of 2.37 L at 2.80×102 K and 1.15 atm. Calculate the pressure when the volume is 1.68 L and
iogann1982 [59]

Answer:

p_2 = 1.76 atm

Explanation:

given data:

v_1 = 2.37 L

v_2 = 1.68 L

p_1 =1.15 atm

p_2 = ?

t_1 = 280 K

t_2 = 304 K

from Gas Law Equation

, WE HAVE

\frac{p_1 v_1}{t_1} =\frac{p_2 v_2}{t_2}

Putting the values

\frac{1.15*2.37}{280}  =\frac{p_2 *1.68}{304}

9.733*10^{-3} = \frac{p_2 *1.68}{304}

9.733*10^{-3}*304 = p_2*1.68

\frac{9.733*10^{-3}*304}{1.68} =p_2

p_2= 1.76 atm

7 0
4 years ago
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