Assume that a specific hard disk drive has an average access time of 16ms (i.e. the seek and rotational delay sums to 16ms) and
a throughput or transfer rate of 185MBytes/s, where a megabyte is measured as 1024.
Required:
What is the average access time for a hard disk spinning at 360 revolutions per second with a seek time of 10 milliseconds?
Required:
What is the average access time for a hard disk spinning at 360 revolutions per second with a seek time of 10 milliseconds?
1 answer:
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Answer:
Q= 4.6 × 10⁻³ m³/s
actual velocity will be equal to 8.39 m/s
Explanation:
density of fluid = 900 kg/m³
d₁ = 0.025 m
d₂ = 0.05 m
Δ P = -40 k N/m²
C v = 0.89
using energy equation
under ideal condition v₁² = 0
v₂² = 88.88
v₂ = 9.43 m/s
hence discharge at downstream will be
Q = Av
Q =
Q =
Q= 4.6 × 10⁻³ m³/s
we know that
hence , actual velocity will be equal to 8.39 m/s