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4 years ago

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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A man wears convex lens glasses of focal length 30cm in order to correct his eyes defect. Instead of the optimum 25cm, his dista

omeli [17]

Answer:

14 cm

Explanation:

F = (frac{uv}{u – v})

F = +ve

v = -ve

30 = (frac {25 {times} (-v)}{25 – (-v)})

v = (frac {25 {times} (-v)}{25+v})

v = 14cm

(Note that either negative or positive values go to show the positioning and hence, they are not a strong necessity in your final answer.)

So happy that i could help you!

Now this question could turn out to be easy for you!!

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2 years ago

What is the strength of the electric field ep 0.90 mm from a proton?

dem82 [27]

The electric field strength is inversely related to the square of the distance.so the strength of the electric field is given by

$E=\frac{1}{4\pi \epsilon _{0} } \frac{q}{r^{2} } = \frac{k q}{r^{2} }$

Here, $\frac{1}{4\pi \epsilon _{0} } = k$ is constant depend upon medium and its value is $9.0 \times10^{9} \ N m^2/C^2$ and q is charge and r is the distance.

Given $r = 0.90 mm = 9.0 \times 10^{-4} m$ and we know the charge of proton, $q = 1.6\times 10^{-19} \ C$ .

Therefore,

$E=\frac{9.0 \times10^{9} \ N m^2/C^2 \times1.6\times 10^{-19} \ C }{(9.0 \times 10^{-4} m)^2} = 0.177 \times 10^{-2} \ N/C \\\\ E= 1.77 \times 10^{-3} N/C$

4 0

3 years ago

A car is moving with an initial relocity of

MA_775_DIABLO [31]

Answer:

The final acceleration of the car, v = 70 m/s

Explanation:

Given,

The initial velocity of the car, u = 20 m/s

The acceleration of the car, a = 10 m/s²

The time period of travel, t = 5 s

Using the I equations of motion

v = u + at

= 20 + 10(5)

= 20 + 50

= 70 m/s

Hence, the final acceleration of the car, v = 70 m/s

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3 years ago

Which tool is used to track which organisms are carriers of a specific trait through several generations?

maw [93]

Answer:

answer is a pedigree chart :)

Explanation:

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4 years ago

Read 2 more answers

Describe all the ways that newtons laws can apply in a car crash

Harman [31]

Newtons first law - Objects in the car at rest (The human) will remain at rest unless affected by an unbalanced force. Well the unbalanced force would be the crash and this would set the human in motion and they would ether fly out the car if not wearing a seat belt or if wearing one they would get bad whip lash

Newtons second law - With more mass requires more force, so since the human is pretty light or even if heavy in a big crash there will be so much more from it that this will send the human flying.

Newtons 3rd law - Objects A puts force onto objects b and object b excretes the same amount of force back onto object a, so in a crash the human would hit the car hard and the car would excrete the same amount of force back on the human which would really damage him/her

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3 years ago