Answer:
Mining would go under Industry organization.
Answer:
Assumption:
1. The kinetic and potential energy changes are negligible
2. The cylinder is well insulated and thus heat transfer is negligible.
3. The thermal energy stored in the cylinder itself is negligible.
4. The process is stated to be reversible
Analysis:
a. This is reversible adiabatic(i.e isentropic) process and thus 
From the refrigerant table A11-A13
sat vapor
m=
b.) We take the content of the cylinder as the sysytem.
This is a closed system since no mass leaves or enters.
Hence, the energy balance for adiabatic closed system can be expressed as:
ΔE
ΔU
![w_{b, out} =m([tex]u_{1} -u_{2](https://tex.z-dn.net/?f=w_%7Bb%2C%20out%7D%20%3Dm%28%5Btex%5Du_%7B1%7D%20-u_%7B2)
)
workdone during the isentropic process
=5.8491(246.82-219.9)
=5.8491(26.91)
=157.3993
=157.4kJ
Answer:
Molar mass -56.0774 g/mol
There is a way to calculate this
atomic number for calcium is 20
atomic number for oxygen is 8
The molar mass for calcium is 40
The molar mass for oxygen is 16
CALCIUM OXIDE
40. +. 16
=56g/mol
1 molecule = 1×56=56g/mol
2 molecules=2×56 =112g/mol
Answer with Explanation:
The relation between power and energy is
Since the nuclear reactor operates at 1200 MW throughout the year thus the energy produced in 1 year equals
Now from the energy mass equivalence we have
where
'c' is the speed of light in free space
Thus equating both the above values we get
Since it is given that 1 kg of mass is 34% effective thus the mass reuired for the reactor is
Thus 1.235 kg of nuclear fuel is reuired for operation.
Answer:
The solution is written in Python
- binary = ""
- decimal = 13
- quotient = int(decimal / 2)
- remainder = decimal % 2
- binary = str(remainder) + binary
-
- while(quotient >0):
- decimal = int(decimal / 2)
- quotient = int(decimal / 2)
- remainder = decimal % 2
- binary = str(remainder) + binary
-
- print(binary)
Explanation:
Firstly, we declare a variable <em>binary</em> and initialize it with an empty string (Line 1). This <em>binary </em>is to hold the binary string.
Next, we declare variable <em>decimal, quotient </em>and<em> remainder </em>(Line 2-4). We assign a test value 13 to decimal variable and then get the first quotient by dividing decimal with 2 (Line 3). Then we get the remainder by using modulus operator, % (Line 4). The first remainder will be the first digit joined with the binary string (Line 5).
We need to repeat the process from Line 3-5 to get the following binary digits. Therefore create a while loop (Line 7) and set a condition that if quotient is bigger than 0 we keep dividing decimal by 2 and calculate the quotient and remainder and use the remainder as a binary digit and join it with binary string from the front (Line 9-11).
At last, we print the binary to terminal (Line 13).
