The negative mark is balanced by a positive mark on the set key scale while the jaws are closed.
It is common practice to shut the jaws or faces of the system before taking some reading to guarantee a zero reading. If not, please take care of the read. This read is referred to as "zero defect."
There are two forms of zero error:
zero-mistake positive; and
Non-null mistake.
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Hope this helps!
Brainliest would be great!
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With all care,
07x12!
A ball bearing has been selected with the bore size specified in the catalog as 35.000 mm to 35.020 mm. Specify appropriate minimum and maximum shaft diameters to provide a locational interference fit.
Answer:
A ball bearing has been selected with the bore size specified in the catalog as 35.000 mm to 35.020 mm. Specify.... ... has been selected with the bore size specified in the catalog as 35.000 mm to 35.020 mm. Specify appropriate minimum and maximum shaft diameters to provide a locational interference fit.
Explanation:
For the following peak or rms values of some important sine waves, calculate the corresponding other value:
(a) 117 V rms, a household-power voltage in North America
(b) 33.9 V peak, a somewhat common peak voltage in rectifier circuits
(c) 220 V rms, a household-power voltage in parts of Europe
(d) 220 kV rms, a high-voltage transmission-line voltage in North America
Answer:
A) V_peak ≈ 165 V
B) V_rms ≈ 24 V
C) V_peak ≈ 311 V
D) V_peak ≈ 311 KV
Explanation:
Formula for RMS value is given as;
V_rms = V_peak/√2
Formula for peak value is given as;
V_peak = V_rms x √2
A) At RMS value of 117 V, peak value would be;
V_peak = 117 x √2
V_peak = 165.46 V
V_peak ≈ 165 V
B) At peak value of 33.9 V, RMS value would be;
V_rms = 33.9/√2
V_rms = 23.97 V
V_rms ≈ 24 V
C) At RMS value of 220 V, peak value is;
V_peak = 220 × √2
V_peak = 311.13 V
V_peak ≈ 311 V
D) At RMS value of 220 KV, peak value is;
V_peak = 220 × √2
V_peak = 311.13 KV
V_peak ≈ 311 KV
13- Convert the following numbers to the indicated bases. List all intermediate steps.
a- (36459080)10 to octal
b- (20960032010 to hexadecimal
c- (2423233303003040)s to base
25 36459080/8= 4557385 0/8 209600320/16=13100020 + 0/16 (2423233303003040)5 (36459080)10 =( 18 (209600320)10=( 1)16 (2423233303003040)5=( )125
Answer:
Following are the conversion to this question:
Explanation:
In point (a):
[tex]\to \frac{36459080}{8} = 4557385 + \ \ \ \ \ \ \ \ \ \frac{0}{8}\\\\\to \frac{4557385}{8} = 569673 + \ \ \ \ \ \ \ \ \ \frac{1}{8}\\\\\to \frac{569673}{8} = 71209+ \ \ \ \ \ \ \ \ \ \frac{1}{8}\\\\\to \frac{71209}{8}=8901+\ \ \ \ \ \ \ \ \ \ \ \frac{1}{8}\\\\\to \frac{8901}{8}=1112+ \ \ \ \ \ \ \ \ \ \ \ \ \frac{5}{8}\\\\\to \frac{1112}{8}=139+ \ \ \ \ \ \ \ \ \ \ \frac{0}{8}\\\\\to \frac{139}{8}=17+ \ \ \ \ \ \ \ \ \ \ \frac{3}{8}\\\\\to \frac{17}{8}=2+ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{8}\\\\[/tex]
[tex]\to \frac{2}{8}=0+ \ \ \ \ \ \ \ \ \ \frac{2}{8}\\\\ \bold{(36459080)_{10}=(213051110)_8}[/tex]
In point (b):
[tex]\to \frac{20960032010}{16} = 13100020+ \ \ \ \ \ \ \ \ \ \frac{0}{16}\\\\\to \frac{13100020}{16} = 818751+ \ \ \ \ \ \ \ \ \ \frac{4}{16}\\\\\to \frac{818751}{16} = 51171+ \ \ \ \ \ \ \ \ \ \frac{15}{16}\\\\\to \frac{51171}{16}=3198+\ \ \ \ \ \ \ \ \ \ \ \frac{3}{16}\\\\\to \frac{3198}{16}=199+ \ \ \ \ \ \ \ \ \ \ \ \ \frac{14}{1}\\\\\to \frac{199}{16}=12+ \ \ \ \ \ \ \ \ \ \ \frac{7}{16}\\\\\to \frac{12}{16}=0+ \ \ \ \ \ \ \ \ \ \ \frac{12}{16}\\\\ \bold{(20960032010)_{10}=(C7E3F40)_{16}}[/tex]
In point (c):
[tex]\to (2423233303003040)_s=(88757078520)_{10}\\\\\to \frac{88757078520}{25}= 3550283140+ \ \ \ \ \ \ \ \ \ \frac{20}{25}\\\\ \to \frac{3550283140}{25}= 142011325+ \ \ \ \ \ \ \ \ \ \frac{15}{25}\\\\\to \frac{142011325}{25}= 5680453+ \ \ \ \ \ \ \ \ \ \frac{0}{25}\\\\\to \frac{5680453}{25}= 227218+ \ \ \ \ \ \ \ \ \ \frac{3}{25}\\\\\to \frac{227218}{25}= 9088+ \ \ \ \ \ \ \ \ \ \frac{18}{25}\\\\\to \frac{9088}{25}= 363+ \ \ \ \ \ \ \ \ \ \frac{13}{25}\\\\[/tex]
[tex]\to \frac{363}{25}= 14+ \ \ \ \ \ \ \ \ \ \frac{13}{25}\\\\\to \frac{14}{25}= 0+ \ \ \ \ \ \ \ \ \ \frac{14}{25}\\\\\bold{(2423233303003040)_s=(EDDI30FK)_{25}}[/tex]
Symbols of Base 25 are as follows:
[tex]0, 1, 2, 3,4,5,6,7,8,9,A,B,C,D,E,F,G,H,I,J,K,L,M,N, \ and \ O[/tex]
At steady state, a refrigerator whose coefficient of performance is 3 removes energy by heat transfer from a freezer compartment at 0 degrees C at the rate of 6000 kJ/hr and discharges energy by heat transfer to the surroundings, which are at 20 degrees C. a) Determine the power input to the refrigerator and compare with the power input required by a reversible refrigeration cycle operating between reservoirs at these two temperatures. b) If electricity costs 8 cents per kW-hr, determine the actual and minimum theoretical operating costs, each in $/day
Answer:
(A)0.122 kW (B) Actual cost = 1.056 $/day, Theoretical cost = 0.234 $/day
Explanation:
Solution
Given that:
The coefficient of performance is =3
Heat transfer = 6000kJ/hr
Temperature = 20°C
Cost of electricity = 8 cents per kW-hr
Now
The next step is to find the power input to the refrigerator and compare with the power input considered by a reversed refrigeration cycle operating between reservoirs at the two temperatures.
Thus
(A)The coefficient of performance is given below:
COP = Heat transfer from freezer/Power input
3 =6000/P
P =6000/3
P= 2000
P = 2000 kJ/hr = 2000/(60*60) kW
= 2000 (3600)kW
= 0.55 kW
Thus
The ideal coefficient of performance = T_low/(T_high - T_low)
= (0+273)/(20-0)
= 13.65
So,
P ideal = 6000/13.65 = 439.6 kJ/hr
= 439.6/(60*60) kW
= 0.122 kW
(B)For the actual cost we have the following:
Actual cost = 0.55 kW* 0.08 $/kW-hr = $ 0.044 per hour
= 0.044*24 $/day
= 1.056 $/day
For the theoretical cost we have the following:
Theoretical cost = 0.122 kW* 0.08 $/kW-hr = $ 0.00976 per hour
= 0.00976*24 $/day
= 0.234 $/day