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composition of the gas: 131 (Inert), 135 (Active)
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What are the prime numbers from 1 to 200?
2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199
Side wall cooling calculation for glass tank furnace?
Typically 1000 cubic feet per minute (per linear running foot of the perimeter at the glass line) of air at a static pressure of 4 inches of water column if the requirement is only for the sidewalls. so for a 6 meter by 14 meter furnace the perimeter would be slightly less than 40 meters (131 feet) so the installed cooling would be 131000 CFM (2 blowers of 65000 CFM) If the crown too is to be cooled a safer calculation would be 1950 cubic feet per minute. It is a trade off. More cooling will give a better life but a lower energy efficiency.
How do you convert the binary value 11000001111111100000000000000000 from IEEE 754-1985 to its decimal equivalent?
To convert 11000001111111100000000000000000 from IEEE 754-1985 to its decimal equivalent, proceed as follows: The most significant bit, bit 31, represents the sign, where 0 denotes positive and 1 denotes negative. So this number is clearly negative. The next 8 bits, from bit 23 to bit 30, represent the exponent. Here we have the value 10000011 which is 131 decimal. However, exponents are biased by 127, thus 131 really means 131-127=3. Thus the exponent is 3. In scientific notation, the exponent tells us how many decimal places to move the decimal point and in what direction. This is no different in binary except we use a binary point instead of a decimal point. Thus an exponent of 3 tells us to move the binary point 3 positions to the right (if it were -3, we'd move three position to the left instead). The remaining 23 bits, from bit 0 to bit 22, represent the mantissa, the fractional component. Here we have 11111100000000000000000. However, a leading 1 bit is always implied so the mantissa is really 24 bits long, with a value of 111111100000000000000000. The binary point is placed to the right of the (implied) leading digit, thus this mantissa has a binary value of 1.11111100000000000000000. Note that the leading 1 bit can always be implied because the digit to the left of the binary point must always be non-zero and the only non-zero digit in binary is 1. The exponent, 3, tells us to move the binary point 3 positions to the right, so we now have 1111.11100000000000000000. We can ignore the trailing zeroes so we now have a binary fraction of 1111.111. Converting floating point binary values to decimal values is no different to converting integer values where every bit position represents some power of 2. Those to the left of the binary point represent increasing powers of 2 starting with 2^0, just as they do in integer conversions. Those to the right represent decreasing powers of 2, starting with 2^-1 (which is 0.5 in decimal). Thus, working from left to right we have: 1 x 2^3 = 8 1 x 2^2 = 4 1 x 2^1 = 2 1 x 2^0 = 1 binary point 1 x 2^-1 = 0.5 1 x 2^-2 = 0.25 1 x 2^-3 = 0.125 Adding these up we get the decimal value 15.857. However, the sign bit indicated the value was negative, so the value is -15.857. Thus 11000001111111100000000000000000 in IEEE 754-185 represents -15.857 in decimal.
The binary number 11 would have a decimel equivelent of?
It depends. If you are using unsigned numbers, then the following assumption is made: 0b11 = 0b00000011, in which case the answer is; 2^1 + 2^0 = 2 + 1 = 3 If you are using signed numbers, than a binary number in the form 0b11 would be interpreted as negative because the leading bit is equal to 1. For signed numbers, the '1' in the leading bit is extended, thus: 0b11 = 0b11111111 In order to interpret this number, negate the number by flipping the bits and adding 1: 0b11111111 0b00000000 (bits flipped) 0b00000001 (added one) The positive representation of 0b11111111 is equal to 0b00000001, which is equal to 1, thus 0b11 = 0b11111111 = -1
If a light bulb is wired in series with a 133-ohm resistor and they are connected across a 131-volt source and the power delivered to the bulb is 22.9 watts then what are the two possible resistances?
First, we assume you are asking for the resistance of the bulb. Second, there can be only one possible resistance. The bulb has one and only one resistance. The equation for solving this problem may be a quadratic or even a cubic equation, which may have multiple roots, but only ONE can be a real-world answer. Third, let's list everything we know about the circuit. Let's label the power source Vcc. We'll label the known resistor R, and the resistance of the bulb Rb. We'll label the power dissipated by the bulb as simply P. We'll also label the current flowing through the bulb as I. What do we know about power? The power dissipated by a resistor is given by three well-known formulas: P = IV, P = I2R, or P = V2/R. (The second and third are derived from the first.) The second one will come in handy for us right now. We can write: [Equ. 1] P = I2Rb What else do we know? Since it's a series circuit, the current flowing through the bulb, I, is the same as the current flowing through the resistor. And we know from Ohm's Law that I = V/RT, where RT is the sum of the two resistances in the circuit: R + Rb. So, we can write: [Equ. 2] I = Vcc/(R + Rb) So, substituting for I in Equ. 1 with the I we found in Equ. 2 we have: P = [Vcc/(R + Rb)]2Rb = [Vcc2/(R + Rb)2]Rb = Vcc2Rb/(R2 + 2RRb + R2b) Phew! The algebra's getting a bit hairy. We need to do a few more manipulations. We can swap the P on the left side of the equal sign with the denominator of the fraction on the right side of the equal sign, so we write: R2 + 2RRb + R2b = Vcc2Rb/P We now multiply both sides of the equation by P and we write: PR2 + 2PRRb + PR2b = Vcc2Rb Stay with me now; we're almost there. If we subtract Vcc2Rb from both sides and gather similar terms, we can write: PR2b+ (2PR - Vcc2)Rb + PR2 = 0 We can now substitute all the known values into the equation above. We know Vcc = 131, R = 133, and P = 22.9. So, we can write: 22.9R2b - 11070Rb + 405078 = 0 So, we have quadratic equation, which will have two roots for Rb. Using the quadratic formula, we determine that there are two real roots: Rb = 443.5 and 39.9. The second result is thrown out since it is not consistent with Equ. 1 and Equ. 2.
What is the difference between 79 and 131?
52
What is the difference between 131 and 29?
94
What is the difference between a 131 paragraph and a 131 essay?
A 131 paragraph is typically a paragraph with 131 words, while a 131 essay is an essay that consists of 131 words in total. The main difference lies in the scope and structure of the content - a paragraph focuses on a specific topic within an essay, which is a formal piece of writing that presents an argument or explores a theme in more detail.
How far is Galway from Mullingar?
About 81 kilometres or 131 miles is the distance between Galway city and Mullingar.About 81 kilometres or 131 miles is the distance between Galway city and Mullingar.About 81 kilometres or 131 miles is the distance between Galway city and Mullingar.About 81 kilometres or 131 miles is the distance between Galway city and Mullingar.About 81 kilometres or 131 miles is the distance between Galway city and Mullingar.About 81 kilometres or 131 miles is the distance between Galway city and Mullingar.About 81 kilometres or 131 miles is the distance between Galway city and Mullingar.About 81 kilometres or 131 miles is the distance between Galway city and Mullingar.About 81 kilometres or 131 miles is the distance between Galway city and Mullingar.About 81 kilometres or 131 miles is the distance between Galway city and Mullingar.About 81 kilometres or 131 miles is the distance between Galway city and Mullingar.
How many odd number between 120 and 131?
5. "Between" implies that 131 is not included.
What is the percentage increase from 115 to 131?
Start: 115, End: 131 Difference = +16 Percentage = Difference ÷ Start x 100% = +16/115 x 100% = +13.9%
What is the sum of all prime number between 40 and 50?
The prime numbers between 40 and 50 are 41, 43, 47. So their sum is 131.
How many multiple of 131 are in between 1 and 100000?
763 of them.
What is the only prime number between 113 and 131?
127.
What number is between 129 and 326?
130, 131 onwards to 325
What are 2 composite numbers between 100 and 200?
101 and 131
What are prime numbers between 130 and 140?
131, 137, 139
