Answer:
See explaination
Explanation:
Please kindly check attachment for the step by step solution of the given problem.
Answer:
Explanation:
The temperature can be defined as the measurement of the intensity of the heat present in the object. Fahrenheit, kelvin and centigrade are the common scale used for measuring Temperature.
Given:
T1=170C
To convert to Kelvin
= 17+273 =290K
T1 = 290K
Pressure (P)= 95KPa
Specific heat ratio = CP/CV= K
WhereK=1.005/0.718
K = 1.4
The final temperature can be calculated using the formula below.
T2 = CP/CV × T1
=. K × T1
T2 = 1.4 × 290
Answer:
3021.7 N/m^2 or 3.022 kN/m^2
Explanation:
The area of the interior column is equivalent to 6*6 = 36 m^2. The length of the structure is 4790 N/m^2. The live load element factor () is 4. The reduced live load will be:
L = =
Therefore, the value of the reduced live load that will be supported by the column is 3021.7 N/m^2 or 3.022 kN/m^2.
This is less than 0.4* = 0.4*4790 = 1916 N/m^2
Given Information:
Capacitor 1 = C₁ = 100 μF
Capacitor 2 = C₂ = 300 μF
Voltage rating of capacitor 1 = 25 V
Voltage rating of capacitor 2 = 35 V
Supply voltage = Vs = 100 V
Required Information:
Which capacitor will be damaged = ?
Answer:
Capacitor 1 will get damaged
Explanation:
The given two capacitors are connected in series, the equivalent capacitance will be
The voltage across the capacitor 1 is
Since the voltage rating of capacitor 1 is 25 volts, it will get damaged because 75 volts are much greater as compared to its voltage rating.
The voltage across the capacitor 2 is
Since the voltage rating of capacitor 2 is 35 volts, it will not get damaged because 25 volts are less as compared to its voltage rating.