Answer:
Conic Sections
a conic section is a curve which is obtained when a surface performs an intersection with a plane. The types of conic sections include hyperbola, parabola and ellipse. A circle can also be considered as a conic section.
Conic Solids on the other hand are the set of points on any segment between a region and a point which is not present in the plane of the base. They are solids with one base.
Build a 32-bit accumulator circuit. The circuit features a control signal inc and enable input en. If en is 1 and inc is 1, the circuit increments the stored value by an amount specified by an input A[31:0] on the next clock cycle. If en is 1 and inc is 0 the circuit decrements the stored value by the amount specified in the input A on the next clock cycle. If en is 0, the circuit simply stores its current value without modification. The circuit has the following interface:______.
Input clock governs the state transitions in the circuit upon each rising edge.
Input clear is used as a synchronous reset for the stored value.
Input inc controls whether the value stored is to be incremented or decremented.
Input en is a control signal that activates the values increment/decrement
Input A determines how much to increment or decrement by
Output value is a 32-bit signal that can be used to read the stored value at any time.
* Note: Use any combination of combinational or sequential logic. It may be helpful to look into D Flip Flops and Registers.
Sorry need.points I'm new
You have available three blocks of different material, at various temperatures. They are, respectively, a 2 kg block of iron at 600 K, a 3 kg block of copper at 800 K and a 10 kg block of granite at 300 K. The heat capacities for the three materials are 0.460 (iron), 0.385 (copper), and 0.790 (granite), in kj/(kg*K), all independent of temperature. For solids, the heat capacities at constant pressure and constant volume can be assumed to be equal, Cp=Cv. what is the minimum temperature that could be obtained in any one of the block? what is the maximum temperature that could be obtained? no heat or work interactions with the enviroment are allowed.
Answer:
max temp = 711.32 k
mini temp = 331.29 k
Explanation:
Given data:
2kg block of Iron : temperature = 600k , C = 0.460 kJ/kgk
3 kg block of copper : temp = 800k , C = 0.385 KJ /kgk
10 kg block of granite : temp = 300k , C = 0.790 KJ/kgk
Cp = Cv at constant pressure and constant volume
Determine the minimum temperature that is obtained in any one of the block
considering the heat transfer equation
Q = mC ( T2 - T1 )
attached below is a detailed solution of the problem
A frequenter of a pub had observed that the new barman poured in average 0.47 liters of beer into the glass with a standard deviation equal to 0.09 liters instead of a half a liter with the same standard deviation. The frequenter had used a random sample of 47 glasses of beer in his experiment. Consider the one-sided hypothesis test for volume of beer in a glass: H0: u=0.5 against H1: u<0.5. Determine the P-value of this test.
Round your answer to four decimal places (e.g. 98.7654).
Answer:
P-value = 0.0011
Explanation:
Formula for the test statistic is;
z = (x¯ - μ)/(σ/√n)
We have;
Sample mean;x¯ = 0.47
Population mean; μ = 0.5
Standard deviation; σ = 0.09
Sample size; n = 47
Thus;
z = (0.46 - 0.5)/(0.09/√47)
z = -3.05
From z-distribution table attached, the p-value corresponding to z = -3.05 is;
P = 0.00114
To four decimal places gives;
P-value = 0.0011
The steam requirements of a manufacturing facility are being met by a boiler whose rated heat input is 5.5 x 3^106 Btu/h. The combustion efficiency of the boiler is measured to be 0.7 by a hand-held flue gas analyzer. After tuning up the boiler, the combustion efficiency rises to 0.8. The boiler operates 4200 hours a year intermittently. Taking the unit cost of energy to be $4.35/10^6 Btu, determine the annual energy and cost savings as a result of tuning up the boiler.
Answer:
Energy Saved = 6.93 x 10⁹ Btu
Cost Saved = $ 30145.5
Explanation:
The energy generated by each boiler can be given by the following formula:
[tex]Annual\ Energy = (Heat\ In)(Combustion\ Efficiency)(Operating\ Hours)[/tex]
Now, the energy saved by the increase of efficiency through tuning will be the difference between the energy produced before and after tuning:
[tex]Energy\ Saved = (Heat\ In)(Efficiency\ After\ Tune - Efficiency\ Before\ Tune)(Hours)[/tex][tex]Energy\ Saved = (5.5\ x\ 3\ x\ 10^{6}\ Btu/h)(0.8-0.7)(4200\ h)[/tex]
Energy Saved = 6.93 x 10⁹ Btu
Now, for the saved cost:
[tex]Cost\ Saved = (Energy\ Saved)(Unit\ Cost)\\Cost\ Saved = (6.93\ x\ 10^{9}\ Btu)(\$4.35/10^{6}Btu)\\[/tex]
Cost Saved = $ 30145.5