Answer:
2
Explanation:To find the solution we must first make the transformations to international units,
In this way:
Inches of length (L) to feet =
So we can identify the area by hour:
Once the area is identified, we now identify the number of passes required,
Passes required = Area x number of passes
We transform speed to international units
So we can identify the covered area
where
m = drum widht
p = efficiency
In this way we obtain the number of rollers given by:
No. of Rollers = Passes required / covered area
No of rollers = 48136/40269
No of rollers = 1.19, that is, 2
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Answer:
C=76 in (answer in inches to the nearest inch)
Explanation:
The statement of question tells us only 2.5% players to be taller than the cut off height C. It means 97.5% are smaller than the cut off height C. We first have to look in standardized normal distribution tables to find the corresponding value of 97.5% which is 1.96.
Now:
z=1.96
mean=μ=70.6
Standard deviation=σ=2.8
Formula:
C=(1.96*2.8)+70.6
C=76.088 in
C=76 in (answer in inches to the nearest inch)
Answer:
An opamp is an operation amplifier. It takes an input signal and amplifies it on the output side.
An ideal opamp should have infinite impedance at its input, infinite gain on the output, and zero impedance on the output
Answer:
So these are the RTL representation:
MAR<----X
MBR<-----AC
M[MAR]<------MBR
Control signal sequence are:
P3T0:MAR<----X
P2P3 P4T1:MBR<-----AC
P0P1 P3T2:M[MAR]<------MBR
Explanation:
STORE X instruction is used for storing the value of AC to the memory address pointed by X. This operation can be done by using the Register Transfers at System Level and this can be represented by using a notation called Register Transfer Language RTL. Let us see what are the register transfer operations happening at the system level.
1. First of all the address X has to be tranfered on to the Memory Address Register MAR.
MAR<----X
2. Next we have to tranfer the contents of AC into the Memory Buffer Register MBR
MBR<-----AC
3. Store the MBR into memory where MAR points to.
M[MAR]<------MBR
So these are the RTL representation:
MAR<----X
MBR<-----AC
M[MAR]<------MBR
Control signal sequence are:
P3T0:MAR<----X
P2P3 P4T1:MBR<-----AC
P0P1 P3T2:M[MAR]<------MBR
Answer:
(a) 151.84 kJ
(b) 2.922 kJ/K
Explanation:
(a) The parameters given are;
Mass of hydrogen gas, H₂ = 0.1 kg = 100 g
Molar mass of H₂ = 2.016 g/mol
Number of moles of H₂ = 100/2.016 = 49.6 moles
V₁ = mRT/P = 0.1×4.12×300/1000 = 0.1236 m³
P₁/P₂ = (V₂/V₁)^k
V₂ = (P₁/P₂)^(1/k)×V₁ =0.1236 × (1000/500)^(1/1.4) = 0.3262 m³
Boundary work done = (V₂ - V₁)(P₂ + P₁)/2 = (0.3262 - 0.1236)*(500 + 1000)/2 = 151.84 kJ
(b) Entropy generated ΔS = Cv · ㏑(T₂/T₁) + R ·㏑(v₂/v₁)
=10.18 × ㏑(270/300) + 4.12 ·㏑(0.3262/0.1236) = 2.922 kJ/K.