The largest positive number that can be represented in 13 bits for the two cases are
A) Unsigned number
<h3>The largest unsigned no can be represented with 13bits as</h3>
In Binary -
In decimal -
In hexadecimal -
B) Signed number
<h3>The largest signed no can be represented with 13bit as</h3>
The range is from
therefore, the maximum value is
In binary -
In hexadecimal -
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Answer
a) 62 percent
b) 40 percent
Explanation:
Original diameter ( d ) = 10.33 mm
Original Gauge length ( L ) = 52.8 mm
diameter at point of fracture ( d ) = 6.38 mm
New gauge length ( L ) = 73.9 mm
<u>Calculate ductility in terms of</u>
a) percent reduction in area
percentage reduction = [ (A - A ) / A ] * 100
A ( initial area ) = π /4 di^2
= π /4 * ( 10.33 )^2 = 83.81 mm^2
A ( final area ) = π /4 df^2
= π /4 ( 6.38)^2 = 31.97 mm^2
hence : %reduction = ( 83.81 - 31.97 ) / 83.81
= 0.62 = 62 percent
b ) percent elongation
percentage elongation = ( L - L ) / L
= ( 73.9 - 52.8 ) / 52.8 = 0.40 = 40 percent
Answer:
so heat loss = 4312 W
cost of heat loss daily is $8.28 per day
Explanation:
given data
slab length L = 11 m
slab wide W = 8 m
thickness t = 0.20 m
temperature top T1 = 17°C
temperature bottom T2 = 10°C
thermal conductivity k = 1.4 W/m-K
efficiency ηf = 0.90
priced Cg = $0.02 / MJ
to find out
rate of heat loss and daily cost of the heat loss
solution
we calculate here heat loss by heat transfer
so apply here formula that is
q = (thermal conductivity × area × temperature difference) / thickness
put here all these value we get heat loss
q =
q =
q = 4312 W
so heat loss = 4312 W
and
cost of the heat loss is express as
cost of heat loss =
put here all these value
cost of heat loss is = × 24 hr/day × 3600 s/hr
cost of heat loss = 8.279
so cost of heat loss daily is $8.28 per day