A & B
Observe the path of the light ray as it passes through the lenses as shown in the attached images. Concave lenses diverge light rays while the convex lens converges the light rays.
Explanation:
Real images are formed where the rays converge, a property of images by convex lenses. Convex lenses can be used to magnify objects. If the image occurs before the focal point of the lens then the image will be upright but smaller. The images inverts and gets bigger past the focal point.
Virtual images are property of concave lenses. These images appear closer but smaller than the real object.
Learn More:
For more on images formed by lenses check out;
brainly.com/question/6722295
brainly.com/question/12191285
brainly.com/question/12529812
brainly.com/question/11788630
#LearnWithBrainly
Answer:
dababy
Explanation:
dababy is the best and that's the answer
Answer:
Rmax = 3.4 10⁶ m
Explanation:
For this exercise we will use the concept of energy
Initial. On the surface of the luma
Em₀ = K + U
Em₀ = ½ m v² - G m M / R_moon
Final. At the furthest point
Emf = U
Emf = - g m M / R_max
Em₀ = Emf
½ m v² - G m M / R_moon = - G m M / R_max
½ v² + G M (-1 / R_moon + 1 / R_max) = 0 (1)
Let's use the fact that tells us that the speed of the rocket is equal to the speed of a satellite that rotates around the moon near the surface, let's use Newton's second law
F = m a
Acceleration is centripetal
a = v² / r
r= R_moon
G m M / R_moon² = m v² / R_mon
G M / R_moon = v²
We substitute in 1
½ G M / R_moon + G M (1 / R_max - 1 / R_moon) = 0
1 / R_max = 1 / R_moon (1- ½)
R_max = R_moon 2
Rmax = 2 1700 103
Rmax = 3.4 10⁶ m
Lucas works in the biopharming field. (Hope it’s right!)
Answer:
The height of the cliff is approximately 127.31 meters
Explanation:
The given parameters are;
The upward speed of the ball = 30 m/s
The time after the ball is thrown before it lands = 9 s
We have;
The time to maximum height;
v = u - g·t
Where;
v = The velocity at maximum height = 0 m/s
u = Initial velocity = 30 m/s
g = The acceleration due to gravity = 9.81 m/s²
∴ 0 = 30 - 9.81 × t
t = 30/9.81 = 3.06 s
The maximum height can be obtained from;
v² = u² - 2·g·s
v = 0 m/s
u² = 2·g·s
30² = 2 × 9.81 × s
s = 30²/(2×9.81) = 45.87 m
The time from maximum height to landing = 9 - 3.06 = 5.942 s
The height to landing
s = u·t + 1/2·g·t².
Here, u = 0 m/s
s = 0×5.942 + 1/2×9.81×5.942² = 173.1766 m
s = 173.1766 m.
The height of the cliff = 173.1766 m. - 45.87 m. ≈ 127.31 m.