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amm1812
3 years ago
6

1. The core of a red dwarf will eventually turn into a(n): a) medium mass star b) white dwarf c) planetary nebula d) sun e) blac

k dwarf​
Physics
2 answers:
myrzilka [38]3 years ago
8 0

Answer:

I think planetary nebula but I'm not 100% sure

Explanation:

m_a_m_a [10]3 years ago
3 0

Answer:

B

Explanation:

White dwarf. I love this type of science.

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A force F making an angle ø with the horizontal is acting on an object resting on the table. Which statement is true for the mot
Viktor [21]
The statement that is true for the motion of sliding the object on the table is : its independent of all forces acting on the object

The only thing that it depend on is the forces acting along both the x-axis and the y-axis.

hope this helps
7 0
4 years ago
5 examples of how intertia is used in real life
Mariulka [41]

Answer:

- your body moving forwards when sudden brakes/force is applied e.g. sudden braking in a car

- feeling a backwards force when something moves quickly from rest e.g a bus

- when you stir a cup of tea, it continues to swirl for a short period of time even though you have stopped

- when shaking a tree, it's leaves or fruit fall down

- satellites which move due to the inertia of motion

hope this helped :)

8 0
3 years ago
If earth did not rotate how would air at the equator move?
Dmitry_Shevchenko [17]
Heat rises, and it is warmer at the equator, so I think warm air would rise at the equator and move towards the cooler poles.
3 0
4 years ago
Read 2 more answers
What do a control group and an experiment group have in common ?
Vlad [161]
<span>Whatever the focus of the experiment is, plus any others factors that might influence the outcome of the experiment. If you are testing a new cancer drug, the experimental group and the control group must both be people with the same type of cancer, and both be a representative distribution of the population, all races, genders, ages, etc. You want the only difference in the two groups to be what you are studying, i.e. the effects of the drug.</span>
7 0
3 years ago
Read 2 more answers
What is the orbital period of a spacecraft in a low orbit near the surface of mars? The radius of mars is 3.4×106m.
valkas [14]
<h2>Answer: 56.718 min</h2>

Explanation:

According to the Third Kepler’s Law of Planetary motion<em> </em><em>“The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”. </em>

In other words, this law states a relation between the orbital period T of a body (moon, planet, satellite) orbiting a greater body in space with the size a of its orbit.

This Law is originally expressed as follows:

T^{2}=\frac{4\pi^{2}}{GM}a^{3}   (1)

Where;

G is the Gravitational Constant and its value is 6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}

M=6.39(10)^{23}kg is the mass of Mars

a=3.4(10)^{6}m  is the semimajor axis of the orbit the spacecraft describes around Mars (assuming it is a <u>circular orbit </u>and a <u>low orbit near the surface </u>as well, the semimajor axis is equal to the radius of the orbit)

If we want to find the period, we have to express equation (1) as written below and substitute all the values:

T=\sqrt{\frac{4\pi^{2}}{GM}a^{3}}    (2)

T=\sqrt{\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(6.39(10)^{23}kg)}(3.4(10)^{6}m)^{3}}    (3)

T=\sqrt{11581157.44 s^{2}}    (4)

Finally:

T=3403.1099s=56.718min    This is the orbital period of a spacecraft in a low orbit near the surface of mars

6 0
3 years ago
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