Final answer:
The linear speed of an ultracentrifuge is calculated to be about 0.524 km/s, while Earth's linear speed in its orbit is calculated to be approximately 29.7 km/s, verifying the initial information presented. The correct option is b).
Step-by-step explanation:
The question asks us to verify the linear speed of an ultracentrifuge and Earth in its orbit. The linear speed of a point on the circumference of a rotating object can be calculated using the formula v = r⋅ω, where v is the linear speed, r is the radius, and ω is the angular velocity. In case (a), to find the linear speed of a point on an ultracentrifuge 0.100 m from its center rotating at 50,000 revolutions per minute (rpm), we first convert the rpm to radians per second (rad/s), given that 1 revolution is 2π radians and 1 minute is 60 seconds.
For the ultracentrifuge:
ω = 50,000 rev/min × (2π rad/rev) × (1 min/60 s) ≈ 5,236 rad/s
The linear speed v = 0.100 m × 5,236 rad/s ≈ 524 m/s, which is equivalent to 0.524 km/s.
In case (b), the average linear speed of Earth in orbit can be calculated by using the orbital circumference divided by the orbital period (one year). Given the average distance from Earth to the Sun is 1.5 x 10¹¹ meters, and Earth's orbit is approximately circular, the circumference of Earth's orbit is 2π×1.5 x 10¹¹ meters. The Earth takes about 365.25 days to complete one orbit, which equals 365.25×24×60×60 seconds. We calculate the speed v as the orbit's circumference divided by the time in seconds, which results in roughly 29.7 km/s. The correct options are (a) 0.524 km/s for the ultracentrifuge and (b) 29.7 km/s for Earth's orbit. Thus, we see that the ultracentrifuge has a linear speed of about 0.50 km/s as given, and Earth's linear speed in its orbit is indeed close to 30 km/s.