Answer:
Mars2 illuminated side of the planet is very heat, dark side very cool
Venus 2 a warm planet with a constant temperature across the entire surface
Explanation:
For this hypothetical case, when changing the planets they are changed with their current characteristics.
Case of Mars2
In this case, there is a planet with a very thin atmosphere, so the solar radiation reaches the ground without damping it, causing a lot of noise, so the illuminated side of the planet is very heat and when the dark side turns due to the little atmosphere it loses everything the heat for which it is very cold.
This thermal stress between the two sides of the planet continues constantly creating possible fruit trees in its rocky systems.
Case of Venus 2
The planet has a high atmospheric density, but it is very far from the sun, so the amount of radiation that arrives slightly warms the planet, but due to the thin atmosphere the losses for the dark period are very small, so the entire planet it is heated until it reaches an almost uniform temperature over its entire surface.
In this case we have a warm planet with a constant temperature across the entire surface, regardless of which side is lit.
stable equilibrium, if displaced from equilibrium, it experiences a net force or torque in a direction opposite to the direction of the displacement.
unstable equilibrium, if displaced it experiences a net force or torque in the same direction as the displacement from equilibrium. A system in unstable equilibrium accelerates away from its equilibrium position if displaced even slightly.
neutral equilibrium, is when an equilibrium is independent of displacements from its original position.
Have a good day, hope this helps
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Answer:
W = M g = 150 kg * 9.81 m/s^2 = 1470 N
You were only given 3 significant figures in the question.
Answer:
1) P₁ = -2 D, 2) P₂ = 6 D
Explanation:
for this exercise in geometric optics let's use the equation of the constructor
where f is the focal length, p and q are the distance to the object and the image, respectively
1) to see a distant object it must be at infinity (p = ∞)
q = f₁
2) for an object located at p = 25 cm
We can that in the two expressions we have the distance to the image, this is the distance where it can be seen clearly in general for a normal person is q = 50 cm
we substitute in the equations
1) f₁ = -50 cm
2)
= 0.06
f₂ = 16.67 cm
the expression for the power of the lenses is
P = 
where the focal length is in meters
1) P₁ = 1/0.50
P₁ = -2 D
2) P₂ = 1 /0.16667
P₂ = 6 D
