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GaryK [48]
2 years ago
13

Which of these properties is the best measure of a star's brightness? (1 point)

Physics
1 answer:
storchak [24]2 years ago
6 0

Answer:

ABSOLUTE MAGNITUDE, C

Explanation:

Absolute magnitude is how bright the star is from a distance of 10 parsecs. Apparent magnitude is how bright the star appears from Earth, which can vary, because some stars may be farther or closer away. Age and size are not very accurate either. For example, at the end of their lifetimes, very big start may explode in a violent and bright supernova, and outshine the stars near them. As for size, as many stars several times smaller than the sun shine much brighter than it, making it inaccurate.

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In an emergency situation, firemen need to respond as quickly as possible. If a fireman is responding from the second floor, how
rjkz [21]

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4m/s

Explanation:

May be different considering how long the pole is and how heavy the firefighter is.

8 0
3 years ago
Two small insulating spheres with radius 3.50×10−2 m are separated by a large center-to-center distance of 0.555 m. One sphere i
xz_007 [3.2K]

Answer:

E = 7.83 \times 10^5 N/C

Explanation:

Since we know that two sphere is oppositely charged so net electric field at the mid point of two balls will be sum of the electric field due to each ball at the mid point

So we know that

E = \frac{kq_1}{r^2} + \frac{kq_2}{r^2}

here we know that

q_1 = 4.20 \mu C

q_2 = 2.50 \mu C

r = \frac{0.555}{2}

so we have

E = \frac{(9\times 10^9)(4.20 + 2.50) \times 10^{-6}}{0.2775^2}

E = 7.83 \times 10^5 N/C

4 0
3 years ago
A five sentence summary about compounds <br><br><br> please it is do In 5min HELP ME PLZZZZ
borishaifa [10]

Explanation:

Compounds are substances composed of two or more kind of atoms or elements joined together in a definite grouping. The properties of compounds are distinct from those of the individual elements that are combined in its make up.

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7 0
3 years ago
A plane cruising at 233 m/s accelerates at 17 m/s 2 for 4.8 s. What is its final velocity? Answer in units of m/s. 013 (part 2 o
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Answer:

Final velocity will be 314.6 m/sec

Distance traveled = 1314.24 m

Explanation:

We have given initial velocity u = 233 m/sec

Acceleration a=17m/sec^2

Time t = 4.8 sec

From first equation of motion v=u+at, here v is final velocity, u is initial velocity and t is time

So v=233+17\times 4.8=314.6m/sec

Now we have to find distance traveled

From second equation of motion

S=ut+\frac{1}{2}at^2=233\times 4.8+\frac{1}{2}\times 17\times 4.8^2=1314.24m

So distance traveled in given time will be 1314.24 m

4 0
3 years ago
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