Answer:
31.4 mm²
Explanation:
The ability of a telescope or eye to gather light can be expressed by the formula,

where d signifies the diameter of the pupil.
In bright daylight, the usual size of the pupil is 3 mm.

Conversely, in darkness, the diameter typically enlarges to 7 mm.

This indicates an increase in light-gathering capacity.

Thus, the amount of light the eye can capture is 31.4 mm².
<span>We will apply the momentum-impulse theorem here. The total momentum along the x-direction is defined as p_(f) = p_(1) + p_(2) + p_(3) = 0.
Therefore, p_(1x) = m1v1 = 0.2 * 2 = 0.4. Additionally, p_(2x) = m2v2 = 0 and p_(3x) = m3v3 = 0.1 *v3, where v3 represents the unknown speed and m3 signifies the mass of the third object, which has an unspecified velocity.
In the same way, for the particle of 235g, the y-component of the total momentum is described with p_(fy) = p_(1y) + p_(2y) + p_(3y) = 0.
Thus, p_(1y) = 0, p_(2y) = m2v2 = 0.235 * 1.5 = 0.3525 and p_(3y) = m3v3 = 0.1 * v3, where m3 is the mass of the third piece.
Consequently, p_(fx) = p_(1x) + p_(2x) + p_(3x) = 0.4 + 0.1v3; yielding v3 = 0.4/-0.1 = - 4.
Similarly, p_(fy) = 0.3525 + 0.1v3; thus v3 = - 0.3525/0.1 = -3.525.
Therefore, the x-component of the speed of the third piece is v_3x = -4 and the y-component is v_3y = 3.525.
The overall speed is calculated as follows: resultant = âš (-4)^2 + (-3.525)^2 = 5.335</span>
As the parachutist is descending at a steady rate
we can conclude that

Acceleration indicates the change in velocity
given the constant velocity in this scenario

Thus, in this situation, we find the acceleration to be zero
It’s understood from Newton's second law

where a is equal to 0


Here, the force due to gravity
equals the force due to buoyancy
Hence, we can deduce

therefore

as such the upward force is counteracted by the downward force.
<span>A centripetal force maintains an object's circular motion. When the ball is at the highest point, we can assume that the ball's speed v is such that the weight of the ball matches the required centripetal force to keep it moving in a circle. Hence, the string will not become slack.
centripetal force = weight of the ball
m v^2 / r = m g
v^2 / r = g
v^2 = g r
v = sqrt { g r }
v = sqrt { (9.80~m/s^2) (0.7 m) }
v = 2.62 m/s
Thus, the minimum speed for the ball at the top position is 2.62 m/s.</span>