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.
For motion in a circle.
Centripetal acceleration is calculated as mv²/r = mω²r
where v represents linear velocity, r equals radius which is diameter/2 equating to 1/2 or 0.5m
. Here, m is the mass of the object, which is 175g or 0.175kg.
The angular speed, ω, is derived from Angle covered / time
= 2 revolutions per 1 second
= 2 * 2π radians for each second
= 4π radians per second
Thus, Centripetal Acceleration = mω²r = 0.175*(4π)² * 0.5. Utilize a calculator
≈13.817 m/s²
. The acceleration's magnitude is approximately 13.817 m/s² and it is oriented towards the center of the circular path.
The tension in the string equates to m*a
= 0.175*13.817
= 2.418 N
The formula to apply is expressed as:
ρ = nA/VcNₐ
where
ρ signifies density
n represents the number of atoms within a unit cell (for FCC, n=4)
A indicates the atomic weight
Vc stands for the cubic cell volume equal to a³, with a being the side length (for FCC, a = 4r/√2, where r is the radius)\
Nₐ denotes Avogadro's number, which is 6.022×10²³ atoms/mol
Calculating the radius: r = 0.1387 nm*(10⁻⁹ m/nm)*(100 cm/1m) = 1.387×10⁻⁸ cm
Then find a: a = 4(1.387×10⁻⁸ cm)/√2 = 3.923×10⁻⁸ cm
Then calculate V: V = a³ = (3.923×10⁻⁸ cm)³ = 6.0376×10⁻²³ cm³
Now compute density:
ρ = [(4 atoms)(195.08 g/mol)]/[(6.0376×10⁻²³ cm³)(6.022×10²³ atoms/mol)]Finally, we get ρ = 21.46 g/cm³
