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.
The well-known equation...
E = m c²... does not address the origin of the mass involved.
Converting 1 kg of any mass entirely into energy generates
(1kg) · (c²) Joules of energy.
E = (1 kg) · (c²) = (1 kg) · (299,792,458 m/s)²
E = 8.9876 x 10¹⁶ Joules
To simplify, this equates to the energy needed to keep a 100-watt light bulb illuminated for about 10,402,259,010 days.
(This is roughly 28.5 million years, based on the current understanding of days and years.)
There's an absence of circuit diagrams.
Initially, this causes worry for a moment, until we remember that we have no understanding of the experiment mentioned in the problem either, rendering such worries unnecessary.
Response:
83%
Clarification:
At the surface, the weight can be expressed as:
W = GMm / R²
where G denotes the gravitational constant, M represents the Earth's mass, m signifies the shuttle's mass, and R is the Earth's radius.
When in orbit, the weight is given by:
w = GMm / (R+h)²
where h indicates the shuttle's altitude above Earth's surface.
The weight ratio is as follows:
w/W = R² / (R+h)²
w/W = (R / (R+h))²
For R = 6.4×10⁶ m and h = 6.3×10⁵ m:
w/W = (6.4×10⁶ / 7.03×10⁶)²
w/W = 0.83
Thus, the shuttle maintains 83% of its weight as it orbits.
Response:
a) 
, b) 
Clarification:
a) The absolute pressure at a depth of 27.5 meters is:
b) The force applied by the water is:
