<span><span>centic<span>10-2</span></span><span>millim<span>10-3</span></span><span>microu [footnote 2]<span>10-6</span></span><span>nanon<span>10-<span>9
</span></span></span></span>
Answer:
i answered this question not that long ago....it's D
Explanation:
<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 
of a body (moon, planet, satellite) orbiting a greater body in space with the size 
of its orbit.
This Law is originally expressed as follows:
(1)
Where;
is the Gravitational Constant and its value is 
is the mass of Mars
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:
(2)
(3)
(4)
Finally:
This is the orbital period of a spacecraft in a low orbit near the surface of mars
Answer:
2 moles
Explanation:
2 moles becz
1 mole of oxygen = 16
.: 2 moles of oxygen = 36
:)
Answer:
Vy = V sin theta = 30 * ,574 = 17.2 m/s
t1 = 17.2 / 9.8 = 1.76 sec to reach max height
Max height = 17.2 * 1.76 - 1/2 * 4.9 * 1.76^2 = 15.1 m
H = V t - 1/2 g t^2 = 1.2 * 9.8 * 1.76^2 = 15.1 m
Time to fall from zero speed to ground = rise time = 1.76 sec
Vx = V cos 35 = 24.6 m / sec horizontal speed
Time in air = 1.76 * 2 = 3.52 sec before returning to ground
S = 24.6 * 3.52 = 86.6 m
