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ikadub [295]
3 years ago
5

What are some of the challenges that you might face if you wanted to observe and study stars from Earth?

Physics
2 answers:
schepotkina [342]3 years ago
7 0
<span>The earth moves, which is unstoppable and if the earth moves the telescope won't be able to see it clearly because the telescope needs to be able to move at the same pace as earth to keep up to the objects</span>
AlexFokin [52]3 years ago
4 0
Another main issue with astronomical sighting is light pollution. The lights from our cities and such drown out the light of the stars. You could also say clouds, humidity, fog, or anything of the kind.
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A car travles a distance of 540km in 6 hours. What speed did it travel at?
Tju [1.3M]
Simply,
To figure out the speed the car traveled to get 540km from point A in 6 hours, You’d have to solve 540 divided by 6.
540 divided by 6 = 90

So the answer would be 90km per hour,
Assuming you want the answer in km.
4 0
3 years ago
An airplane is flying at a velocity of 122.4 m/s. It is getting ready to land so it slows down by accelerating at a rate of -2.8
PIT_PIT [208]

Answer:

115.12

Explanation:

-2.8=v-122.4/2.6

v=115.12

8 0
3 years ago
Assume that a satellite orbits mars 150km above its surface. Given that the mass of mars is 6.485 X 10^23kg, and the radius of m
Kisachek [45]
<span>3598 seconds The orbital period of a satellite is u=GM p = sqrt((4*pi/u)*a^3) Where p = period u = standard gravitational parameter which is GM (gravitational constant multiplied by planet mass). This is a much better figure to use than GM because we know u to a higher level of precision than we know either G or M. After all, we can calculate it from observations of satellites. To illustrate the difference, we know GM for Mars to within 7 significant figures. However, we only know G to within 4 digits. a = semi-major axis of orbit. Since we haven't been given u, but instead have been given the much more inferior value of M, let's calculate u from the gravitational constant and M. So u = 6.674x10^-11 m^3/(kg s^2) * 6.485x10^23 kg = 4.3281x10^13 m^3/s^2 The semi-major axis of the orbit is the altitude of the satellite plus the radius of the planet. So 150000 m + 3.396x10^6 m = 3.546x10^6 m Substitute the known values into the equation for the period. So p = sqrt((4 * pi / u) * a^3) p = sqrt((4 * 3.14159 / 4.3281x10^13 m^3/s^2) * (3.546x10^6 m)^3) p = sqrt((12.56636 / 4.3281x10^13 m^3/s^2) * 4.458782x10^19 m^3) p = sqrt(2.9034357x10^-13 s^2/m^3 * 4.458782x10^19 m^3) p = sqrt(1.2945785x10^7 s^2) p = 3598.025212 s Rounding to 4 significant figures, gives us 3598 seconds.</span>
8 0
3 years ago
N4M.6 A board has one end wedged under a rock having a mass of 380 kg and is supported by another rock that touches the bottom s
Hitman42 [59]

Answer:

it is safe to stand at the end of the table

Explanation:

For this exercise we use the rotational equilibrium condition

         Στ = 0

         W x₁ - w x₂ - w_table x₃ = 0

         M x₁ - m x₂ - m_table x₃ = 0

where the mass of the large rock is M = 380 kg and its distance to the pivot point x₁ = 850 cm = 0.85m

the mass of the man is 62 kg and the distance

            x₂ = 4.5 - 0.85

            x₂ = 3.65 m

the mass of the table (m_table = 22 kg) is at its geometric center

            x_{cm} = L/2 = 2.25 m

            x₃ = 2.25 -0.85

            x₃ = 1.4 m

let's look for the maximum mass of man

            m_{maximum} = \frac{ M x_1 -m_{table} x_3}{ x_2}

let's calculate

             m_{maximum} = \frac{ 380 \ 0.85 - 22 \ 1.4}{3.65}(380 0.85 - 22 1.4) / 3.65

             m_{maximum} = 80 kg

we can see that the maximum mass that the board supports without turning is greater than the mass of man

             m_{maximum}> m

consequently it is safe to stand at the end of the table

5 0
3 years ago
In an electron cloud, an electron farther east away from the nucleus has?
vladimir2022 [97]

An electron that is far away from the nucleus have higher energy than an electron near the nucleus. Nucleus are positively charged and those electrons near it get attracted; those electrons gain kinetic energy hence reducing their internal energy. The electrons far from nucleus have low kinetic energy hence more internal energy.

8 0
3 years ago
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