Answer:
hahaha dude but thnx for points
Mierda me dices solo necesito puntos
Answer:
the pressure reading when connected a pressure gauge is 543.44 kPa
Explanation:
Given data
tank volume (V) = 400 L i.e 0.4 m³
temperature (T) = 25°C i.e. 25°C + 273 = 298 K
air mass (m) = 3 kg
atmospheric pressure = 98 kPa
To find out
pressure reading
Solution
we have find out pressure reading by gauge pressure
i.e. gauge pressure = absolute pressure - atmospheric pressure
first we find absolute pressure (p) by the ideal gas condition
i.e pV = mRT
p = mRT / V
p = ( 3 × 0.287 × 298 ) / 0.4
p = 641.44 kPa
so
gauge pressure = absolute pressure - atmospheric pressure
gauge pressure = 641.44 - 98
gauge pressure = 543.44 kPa
Answer:
Vector C = 1.334i + 8.671j + 2k or 1.334x + 8.671y + 2z
Explanation:
The concept applied to solve the question is cross product of vector, AXB since vector C is perpendicular to vector A and B and this is solved by applying the 3X3 determinant method.
A detailed step by step explanation is attached below.
Answer:
-2/√3 atan ((2t + 1)/√3) + C
Explanation:
∫ (t − 1) / (1 − t³) dt
Factor the difference of cubes:
∫ (t − 1) / ((1 − t)(1 + t + t²)) dt
Divide:
∫ -1 / (1 + t + t²) dt
-∫ 1 / (t² + t + 1) dt
Complete the square:
-∫ 1 / (t² + t + ¼ + ¾) dt
-∫ 4 / (4t² + 4t + 1 + 3) dt
-∫ 4 / ((2t + 1)² + 3) dt
If u = 2t + 1, du = 2 dt:
-∫ 2 / (u² + 3) du
Use an integral table, or use trigonometric substitution:
-2 (1/√3) atan (u/√3) + C
-2/√3 atan (u/√3) + C
Substitute back:
-2/√3 atan ((2t + 1)/√3) + C