Answer:
Its approx location is (5.18,1.9)
Explanation:
Using F( 5,2) = ( xy-1, y²-11)
 = ( 5*2-¹, 2²-11)
 = (9,-5)
 = so at point t=1.02
 (5,2)+(1.02-1)*(9,-5)
 (5,2)+( 0.02)*(9,-5)
 (5+0.18, 2-0.1)
 = ( 5.18, 1.9)
 
        
             
        
        
        
Answer:
A) ≥ 325Kpa
B) ( 265 < Pe < 325 ) Kpa 
C) (94 < Pe < 265 )Kpa
D)  Pe < 94 Kpa
Explanation:
Given data :
A large Tank : Pressures are at 400kPa and 450 K 
Throat area = 4cm^2 ,  exit area = 5cm^2
<u>a) Determine the range of back pressures that the flow will be entirely subsonic</u>
The range of flow of back pressures that will make the flow entirely subsonic 
will be ≥ 325Kpa
attached below is the detailed solution 
<u>B) Have a shock wave</u> 
The range of back pressures for there to be shock wave inside the nozzle 
= ( 265 < Pe < 325 ) Kpa 
attached below is a detailed solution
C) Have oblique shocks outside the exit 
= (94 < Pe < 265 )Kpa
D) Have supersonic expansion waves outside the exit 
= Pe < 94 Kpa
 
        
             
        
        
        
Answer:
Explanation:
The amount of force needed needs to be greater than all the forces acting in the opposite direction that the bowling ball was thrown. This includes air resistance, floor friction, gravity, and any other force involved. As long as the force acting on the bowling ball that is causing it to go in the direction of the pins is slightly greater than the opposite acting forces then it will continue in that direction. Since no values are provided we cannot calculate the actual precise value of force needed.
 
        
             
        
        
        
Answer:
-48 N
Explanation:
mass of door (m) = 4 kg
acceleration of the door = 12 m/s^{2}
force exerted by the person = 48 N
From Newton's third law of motion, action and reaction are equal but opposite. Therefore the force exerted on the door by the person which is 48 N will be the same as the force exerted on the person by the door but opposite in its direction, and this would be - 48 N