Answer:
A
Explanation:
The change in the system’s gravitational potential energy would be zero
Answer:
1) q₁ = 12.987 cm
, b) L = 17.987 cm
, c) m = 179.87
Explanation:
We can solve the geometric optics exercises with the equation of the constructor
1 / f = 1 / p + 1 / q
where f is the focal length, p and q are the distance to the object and the image respectively.
Let's apply this equation to our case
1) f = 5mm = 0.5 cm
p₁ = 5.2 mm = 0.52 cm
h = 0.1 mm = 0.01 cm
1 / q₁ = 1 / f- 1 / p
1 / q₁ = 1 / 0.5 - 1 / 0.52 = 2 - 1.923
1 / q₁ = 0.077
q₁ = 12.987 cm
2) in this part they tell us that the eyepiece creates an image at infinity, therefore the object that comes from being at the focal length of the eyepiece
p₂ = 5 cm
The absolute thing that goes through the two lenses is
L = q₁ + p₂
L = 12.987 +5
L = 17.987 cm
3) This lens configuration forms the so-called microscope, whose expression for the magnifications
m = -L / f_target 25 cm / f_ocular
m = - 17.987 / 0.5 25 / 5.0
m = 179.87
Maintaining an orbit has nothing to do with the satellite's speed.
ANY (tangential) speed is enough to stay in orbit, if the satellite
just stays away from Earth's atmosphere.
If the satellite completely stops moving 'sideways' at all, and just
hangs there, then the forces of gravity between the satellite and
the Earth will pull them together ... the satellite will fall into the
atmosphere and then to the ground.
Answer:
Increasing the launch height increases the downward distance, giving the horizontal component of the velocity greater time to act upon the projectile and hence increasing the range.
Explanation:
