Answer:
13.95
Explanation:
Given :
Vector A polar coordinates = ( 7, 70° )
Vector B polar coordinates = ( 4, 130° )
To find A . B we will
A ( r , ∅ ) = ( 7, 70 )
A = rcos∅ + rsin∅
therefore ; A = 2.394i + 6.57j
B ( r , ∅ ) = ( 4, 130° )
B = rcos∅ + rsin∅
therefore ; B = -2.57i + 3.06j
Hence ; A .B
( 2.394 i + 6.57j ) . ( -2.57 + 3.06j ) = 13.95
Answer:
1st part: Section W18X76 is adequate
2nd part: Section W21X62 is adequate
Explanation:
See the attached file for the calculation
Answer:
hello your question is incomplete attached below is the complete question
A) optimum compressor ratio = 9.144
B) specific thrust = 2.155 N.s /kg
C) Thrust specific fuel consumption = 1670.4 kg/N.h
Explanation:
Given data :
Mo = 2.1 , To = 220k , Tt4 = 1700 k, hpr = 42000 kj/kg, Cp = 1.004 kj/ kg.k
γ = 1.4
attached below is the detailed solution
Answer:
The amount of energy transferred to the water is 4.214 J
Explanation:
The given parameters are;
The mass of the object that drops = 5 kg
The height from which it drops = 86 mm (0.086 m)
The potential energy P.E. is given by the following formula
P.E = m·g·h
Where;
m = The mass of the object = 5 kg
g = The acceleration de to gravity = 9.8 m/s²
h = The height from which the object is dropped = 0.086 m
Therefore;
P.E. = 5 kg × 9.8 m/s² × 0.086 m = 4.214 J
Given that the potential energy is converted into heat energy, that raises the 1 g of water by 1°C, we have;
The amount of energy transferred to the water = The potential energy, P.E. = 4.214 J.