Answer:
Yes
Explanation:
Given Data
Temprature of source=750°c=1023k
Temprature of sink =0°c=273k
Work produced=3.3KW
Heat Rejected=4.4KW
Efficiency of heat engine(η)=
and
Heat Supplied
η=
η=42.85%
Also the maximum efficiency of a heat engine operating between two different Tempratures i.e. Source & Sink
η=1-
η=1-
η=73.31%
Therefore our Engine Efficiency is less than the maximum efficiency hence the given claim is valid.
Answer:
save space = 1.9626 Gibibits
Explanation:
given data
recording song time = 3 minute = 180 seconds
song per sample = 24 bits per sample
frequency = 96 kHz
surround sound system = 5.1
solution
first we get here required space for 24 bits at 96 kHz that is
required space = 24 × 96 × 10³ × 6 × 180
required space = 2.48832 × bits
as 1 Gib bit = bits
so required space = 2.48832 × bits ÷
bits
required space = 2.3174 Gibibits ...............1
and
space required to save record 16 bit at 44.1 kHz
space required = 16 × 44.1 × 10³ × 3 × 180
space required = 0.381024 × bits
space required = 0.381024 × bits ÷
bits
space required = 0.3548 Gibibits ...........2
so
we get here that save space in 16 bit at 44.1 kHz
save space = 2.3174 Gibibit - 0.3548 Gibibits
save space = 1.9626 Gibibits