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timofeeve [1]
2 years ago
15

3. What Can occur between the sun and Mars.

Physics
2 answers:
seraphim [82]2 years ago
3 0

Answer:

Solar conjunction is the period when Earth and Mars, in their eternal march around the Sun, are obscured from each other by the fiery orb of the Sun itself. Like dancers on either side of a huge bonfire, the two planets are temporarily invisible to each other

ExtremeBDS [4]2 years ago
3 0

i think the answer is B:convection

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Three importance of SI system​
diamong [38]

Answer:

Firstly they are, by design, easy to use in most scientific and engineering calculations; you only ever have to consider multiples of 10. If I’m given a measurement of 3.4 kilometres, I can instantly see that it’s 3′400 metres, or 0.0034 Megametres, or 3′400′000 millimetres. It’s not even necessary to use arithmetic, I just have to remember the definitions of the prefixes (“kilo” is a thousand, “megametre” is a million, “milli” is a thousandth) and shift the decimal point across to the left or the right. This is especially useful when we’re considering areas, speeds, energies, or other things that have multiple units; for instance,

1 metre^2 = (1000millimetre)^2 = 1000000 mm^2.

If we were to do an equivalent conversion in Imperial, we would have

1 mile^2 = (1760 yards)^2

and we immediately have to figure out what the square of 1760 is! However, the fact that SI is based on multiples of 10 has the downside that we can’t consider division by 3, 4, 8, or 12 very easily.

Secondly they are (mostly) defined in terms of things that are (or, that we believe to be) fundamental constants. The second is defined by a certain kind of radiation that comes from a caesium atom. The metre is defined in terms of the second and the speed of light. The kelvin is defined in terms of the triple point of water. The mole is the number of atoms in 12 grams of carbon-12. The candela is defined in terms of the light intensity you get from a very specific light source. The ampere is defined using the Lorentz force between two wires. The only exception is the kilogram, which is still defined by the mass of a very specific lump of metal in a vault in France (we’re still working on a good definition for that one).

Thirdly, most of the Imperial and US customary units are defined in terms of SI. Even if you’re not personally using SI, you are probably using equipment that was designed using SI.

8 0
4 years ago
An object that is negatively charged could contain only electrons with no accompanying protons. O True False​
marta [7]

Answer:

community services high school graduates jevs human service providers continue

6 0
3 years ago
How long does it take to get from a new moon to a first quarter moon
Andru [333]

'First Quarter' is the phase that appears one quarter of the
period of time from one new moon to the next one.

The total period of the moon's phases is 29.53 days (rounded).

One quarter of it is (29.53 / 4) =  <em>7 days  9.2 hours</em>   (rounded)


5 0
4 years ago
Do the planets orbit the sun in a perfect circle
GaryK [48]

Answer:

All the planets in the Solar System orbit the Sun. When planets orbit the Sun, they don’t move around in a perfect circle. They all follow an 'elliptical path', meaning it looks more like an oval.

Explanation:

7 0
3 years ago
Read 2 more answers
A 75-g bullet is fired from a rifle having a barrel 0.540 m long. choose the origin to be at the location where the bullet begin
lyudmila [28]
Part a) The work done by the gas on the bullet is the integral of the force in dx, where x is the distance covered by the bullet inside the barrel with respect to the origin:
W= \int\limits^{0.540m}_{0} {F} \, dx =  \int\limits^{0.540m}_{0} {(16000+10000x-26000x^2)} \, dx =
=16000x+10000  \frac{x^2}{2} - 26000  \frac{x^3}{3}
By substituting the length of the barrel, L=0.540 m, we find the total work done by the gas on the bullet:
W=16000(0.540m)+10000  \frac{(0.540m)^2}{2} - 26000  \frac{(0.540m)^3}{3}  =
=8733 J=8.73 kJ

part b) The resolution of the problem is the same, we just have to use the new length of the barrel (L=0.95 m) inside the final formula, and we find the new value of the work:
W=16000(0.95m)+10000  \frac{(0.95m)^2}{2} - 26000  \frac{(0.95m)^3}{3}  =
=12280 J=12.28 kJ
5 0
3 years ago
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