Answer:
a) 
b)
c) 
d) Treat the humans as though they were points or uniform-density spheres.
Explanation:
Given:
- mass of Mars,
- radius of the Mars,
- mass of human,
a)
Gravitation force exerted by the Mars on the human body:
where:
= gravitational constant
b)
The magnitude of the gravitational force exerted by the human on Mars is equal to the force by the Mars on human.
c)
When a similar person of the same mass is standing at a distance of 4 meters:
d)
The gravitational constant is a universal value and it remains constant in the Universe and does not depends on the size of the mass.
- Yes, we have to treat Mars as spherically symmetric so that its center of mass is at its geometric center.
- Yes, we also have to ignore the effect of sun, but as asked in the question we have to calculate the gravitational force only due to one body on another specific body which does not brings sun into picture of the consideration.
If the mass of the sun is 1x, at least one planet will fall into the habitable zone. if I place a planet in orbits 1, 3, 5 , 6 and all planets will orbit the sun successfully.
<h3>
What are planets?</h3>
Planets are the large spherical shaped objects that rotate about the Sun in the elliptical orbits.
Planets are shaped from Planetary cloud. The dust storm and gases gathers under its own weight. The dense matter beginnings pivoting at high paces and accumulates more mass. The center structures, the star and rest of it ultimately levels into a curved plate from which planet is formed.
Thus, if I place a planet in orbits 1, 3, 5 , 6 and all planets will orbit the sun successfully.
Learn more about planets.
brainly.com/question/14581221
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Answer:
The behavior of droplets trapped in geometric structures is essential to droplet manipulation applications such as for droplet transport. Here we show that directional droplet movement can be realized by a V-shaped groove with the movement direction controlled by adjusting the surface wettability of the groove inner wall and the cross sectional angle of the groove. Experiments and analyses show that a droplet in a superhydrophobic groove translates from the immersed state to the suspended state as the cross sectional angle of the groove decreases and the suspended droplet departs from the groove bottom as the droplet volume increases. We also demonstrate that this simple grooved structure can be used to separate a water-oil mixture and generate droplets with the desired sizes. The structural effect actuated droplet movements provide a controllable droplet transport method which can be used in a wide range of droplet manipulation applications.
Explanation:
BOOM NOW I WINNNNNNNNNNNn
Answer:
A) ω = 6v/19L
B) K2/K1 = 3/19
Explanation:
Mr = Mass of rod
Mb = Mass of bullet = Mr/4
Ir = (1/3)(Mr)L²
Ib = MbRb²
Radius of rotation of bullet Rb = L/2
A) From conservation of angular momentum,
L1 = L2
(Mb)v(L/2) = (Ir+ Ib)ω2
Where Ir is moment of inertia of rod while Ib is moment of inertia of bullet.
(Mr/4)(vL/2) = [(1/3)(Mr)L² + (Mr/4)(L/2)²]ω2
(MrvL/8) = [((Mr)L²/3) + (MrL²/16)]ω2
Divide each term by Mr;
vL/8 = (L²/3 + L²/16)ω2
vL/8 = (19L²/48)ω2
Divide both sides by L to obtain;
v/8 = (19L/48)ω2
Thus;
ω2 = 48v/(19x8L) = 6v/19L
B) K1 = K1b + K1r
K1 = (1/2)(Mb)v² + Ir(w1²)
= (1/2)(Mr/4)v² + (1/3)(Mr)L²(0²)
= (1/8)(Mr)v²
K2 = (1/2)(Isys)(ω2²)
I(sys) is (Ir+ Ib). This gives us;
Isys = (19L²Mr/48)
K2 =(1/2)(19L²Mr/48)(6v/19L)²
= (1/2)(36v²Mr/(48x19)) = 3v²Mr/152
Thus, the ratio, K2/K1 =
[3v²Mr/152] / (1/8)(Mr)v² = 24/152 = 3/19
The spiral structure emerges when galactic clusters (open), H II regions and O & B type stars (young stars) are used as tracers. We know this to be true as other pinwheel galaxies exhibit the same patterns across these tracers as in the milky way.
