Response: a. The mirrors and eyepiece of a large telescope are designed with spring-loaded components to quickly return to a predetermined position.
Justification:
Adaptive optics refers to a technique employed by various astronomical observatories to compensate in real-time for the atmospheric turbulence that impacts astronomical imaging.
This is executed by integrating advanced deformable mirrors into the telescope's optical pathway, operated by a set of computer-controlled actuators. This allows for obtaining clearer images despite the atmospheric fluctuations that create distortions.
It is crucial to note that this process requires a moderately bright reference star located closely to the object being studied.
However, locating such stars is not always feasible, prompting the use of a strong laser beam directed at the upper atmosphere to create artificial stars.
Answer:
The snowball's speed after the impact is 3 m/s
Explanation:
Given the following:
mass of each ball
m₁ = 0.4 Kg
m₂ = 0.6 Kg
initial speed of both balls = v₁ = 15 m/s
Speed of 1 Kg mass post-collision =?
Applying conservation of momentum
m₁ v₁ - m₂ v₁ = (m₁+m₂) V
A negative velocity indicates that the second ball moves in the opposite direction.
0.4 x 15 - 0.6 x 15 = (1) V
Therefore,
V = - 3 m/s
Consequently,
The snowball's speed following the collision is 3 m/s
Answer:
(a) the coefficient of friction is 0.451
This was derived using the energy conservation principle (the total energy in a closed system remains constant).
(b) No, the object stops 5.35 m away from point B. This is due to the spring's expansion only performing 43 J of work on the block, which isn't sufficient compared to the 398 J required to overcome friction.
Explanation:
For more details on how this issue was resolved, refer to the attached material. The solution for part (a) separates the body’s movement into two segments: from point A to B, and from B to C. The total system energy originates from the initial gravitational potential energy, which transforms into work against friction and into work compressing the spring. A work of 398 J is needed to counteract friction over the distance of 6.00 m. The energy used for this is lost since friction is not a conservative force, leaving only 43 J for spring compression. When the spring expands, it exerts a work of 43 J back on the block, which is only sufficient to move it through a distance of 0.65 m, stopping 5.35 m short of point B.
Thank you for your attention; I trust this is beneficial to you.
The volumes are 200cm3 and 0.0002m3
Answer:
Height (h) = 17 m
Velocity (v) = 18.6 m/s
Explanation: This problem can be solved using kinematic motion equations.
Given Data
Initial velocity (u) = 0
Acceleration (a) = g
Time (t) = 1.9 seconds
First, we calculate the height.

Then, we find the final velocity

The acceleration graph is a linear representation described by y=9.8, as it remains constant:
The velocity graph can be represented by y=9.8x (where y signifies velocity and x indicates time):
The displacement graph can be described as y=4.9x^2 (with x as time and y as displacement):
These graphs apply exclusively from x=0 to x=1.9, so disregard other sections of the graphs.