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ohaa [14]
2 years ago
5

Orion Balatik Big Dipper Apparent Brightness Absolute Brightness Polaris Star Trail Malihe Gemini Buwaya 1. Local constellation

that signifies the start of rainy season in the Philippines. 2. Local constellation that appears in the month of March. 3. It is seen in the Philippines during the months of April and May. 4. The constellation where we can find Merak and Dubhe. 5. It is the brightes star in the constelation Ursa Minor (Little Dipper). 6. It is a type of photograph that utilizes long-exposure times to capture the apparent motion of stars in the night sky due to the rotation of the Earth. 7. The star's brightness as seen from the Earth. 8. The brightness of the stars if they are at the same standard distance from Earth. 9. This constellation means hunter and is prominent in the night sky all over the world during winter. 10. The local constellation that appears on the month of February and signifies start of planting and setting of traps to protect the crops from animals.​
Physics
1 answer:
Roman55 [17]2 years ago
4 0
I’m nit sure about this but I say it’s 4 the constellation where we can find Merak and Dubhe.
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NASA has asked your team of rocket scientists about the feasibility of a new satellite launcher that will save rocket fuel. NASA
kkurt [141]

Answer:

The answer is "q=0.0945\,C".

Explanation:

Its minimum velocity energy is provided whenever the satellite(charge 4 q) becomes 15 m far below the square center generated by the electrode (charge q).

U_i=\frac{1}{4\pi\epsilon_0} \times \frac{4\times4q^2}{\sqrt{(15)^2+(5/\sqrt2)^2}}

It's ultimate energy capacity whenever the satellite is now in the middle of the electric squares:

U_f=\frac{1}{4\pi\epsilon_0}\ \times \frac{4\times4q^2}{( \frac{5}{\sqrt{2}})}

Potential energy shifts:

= U_f -U_i \\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{\sqrt{(15)^2+( \frac{5}{\sqrt{2})^2)}}\right ) \\\\   =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ 15 +( \frac{5}{2})}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{30+5}{2})}}\right )\\\\

=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{35}{2})}}\right )\\\\=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{17.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 24.74- 5 }{87.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 19.74- 5 }{87.5}}\right )\\\\ =\frac{4q^2}{\pi\epsilon_0}\left ( 0.2256 }\right )\\\\= \frac{0.28 \times q^2}{ \epsilon_0}\\\\=q^2\times31.35 \times10^9\,J

Now that's the energy necessary to lift a satellite of 100 kg to 300 km across the surface of the earth.

=\frac{GMm}{R}-\frac{GMm}{R+h} \\\\=(6.67\times10^{-11}\times6.0\times10^{24}\times100)\left(\frac{1}{6400\times1000}-\frac{1}{6700\times1000} \right ) \\\\ =(6.67\times10^{-11}\times6.0\times10^{26})\left(\frac{1}{64\times10^{5}}-\frac{1}{67\times10^{5}} \right ) \\\\=(6.67\times6.0\times10^{15})\left(\frac{67 \times 10^{5} - 64 \times 10^{5}  }{ 4,228 \times10^{5}} \right ) \\\\

=( 40.02\times10^{15})\left(\frac{3 \times 10^{5}}{ 4,228 \times10^{5}} \right ) \\\\ =40.02 \times10^{15} \times 0.0007 \\\\

\\\\ =0.02799\times10^{10}\,J \\\\= q^2\times31.35\times10^{9} \\\\ =0.02799\times10^{10} \\\\q=0.0945\,C

This satellite is transmitted by it system at a height of 300 km and not in orbit, any other mechanism is required to bring the satellite into space.

6 0
3 years ago
What have psychologists learned about perception from optical illusions ?
ruslelena [56]

Answer:

Explanation:

That an optical illusion somehow interferes with the way we see things. Even simple illusions can completely fool us. If you search out the term, you'll see all kinds of them.

Most critically we see one thing and know another to be true. But knowing the truth doesn't help us. We still see and believe the truth of the illusion.

5 0
3 years ago
Read 2 more answers
What is the amount of charge when 13.5a is flowing for 2 1/2 hours
abruzzese [7]

Answer:

Quick Answer:

a.  8.9Ω

b.  1.2×104 C

Explanation:

4 0
3 years ago
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For sprinters running at 12 m/s around a curved track of radius 26 m, how much greater (as a percentage) is the average total fo
Snezhnost [94]

Answer:

114.86%

Explanation:

In both cases, there is a vertical force equal to the sprinter's weight:

Fy = mg

When running in a circle, there is an additional centripetal force:

Fx = mv²/r

The net force is found with Pythagorean theorem:

F² = Fx² + Fy²

F² = (mv²/r)² + (mg)²

F² = m² ((v²/r)² + g²)

F = m √((v²/r)² + g²)

Compared to just the vertical force:

F / Fy

m √((v²/r)² + g²) / mg

√((v²/r)² + g²) / g

Given v = 12 m/s, r = 26 m, and g = 9.8 m/s²:

√((12²/26)² + 9.8²) / 9.8

1.1486

The force is about 114.86% greater (round as needed).

5 0
3 years ago
An alarm clock draws 0.5 A of a current when connected to a 120 volt circuit. Calculate its resistance
wlad13 [49]

Resistance = (voltage) / (current) =

                       (120 V) / (0.5 A)  =  <em>240 ohms</em>


4 0
3 years ago
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