Answer:
In general a cache memory is useful because the speed of the processor is higher than the speed of the ram . so reducing the number of memory is desirable to increase performance .
Explanation:
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Answer:
gauge pressure is 133 kPa
Explanation:
given data
initial temperature T1 = 27°C = 300 K
gauge pressure = 300 kPa = 300 × 10³ Pa
atmospheric pressure = 1 atm
final temperature T2 = 77°C = 350 K
to find out
final pressure
solution
we know that gauge pressure is = absolute pressure - atmospheric pressure so
P (gauge ) = 300 × 10³ Pa - 1 ×
Pa
P (gauge ) = 2 ×
Pa
so from idea gas equation
................1
so
P2 = 2.33 ×
Pa
so gauge pressure = absolute pressure - atmospheric pressure
gauge pressure = 2.33 ×
- 1.0 ×
gauge pressure = 1.33 ×
Pa
so gauge pressure is 133 kPa
Answer:
a)6.8 KPa
b)0.264 gallon
c)47.84 Pa.s
Explanation:
We know that
1 lbf= 4.48 N
1 ft =0.30 m
a)
Given that
P= 1 psi
psi is called pound force per square inch.
We know that 1 psi = 6.8 KPa.
b)
Given that
Volume = 1 liter
We know that 1000 liter = 1 cubic meter.
1 liter =0.264 gallon.
c)

Answer:
Heat losses by convection, Qconv = 90W
Heat losses by radiation, Qrad = 5.814W
Explanation:
Heat transfer is defined as the transfer of heat from the heat surface to the object that needs to be heated. There are three types which are:
1. Radiation
2. Conduction
3. Convection
Convection is defined as the transfer of heat through the actual movement of the molecules.
Qconv = hA(Temp.final - Temp.surr)
Where h = 6.4KW/m2K
A, area of a square = L2
= (0.25)2
= 0.0625m2
Temp.final = 250°C
Temp.surr = 25°C
Q = 64 * 0.0625 * (250 - 25)
= 90W
Radiation is a heat transfer method that does not rely upon the contact between the initial heat source and the object to be heated, it can be called thermal radiation.
Qrad = E*S*(Temp.final4 - Temp.surr4)
Where E = emissivity of the surface
S = boltzmann constant
= 5.6703 x 10-8 W/m2K4
Qrad = 5.6703 x 10-8 * 0.42 * 0.0625 * ((250)4 - (25)4)
= 5.814 W