Elijah wants to make some changes to a game that he is creating. Elijah wants to move the bleachers to the right, the coaches to
2 answers:
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Answer:
Explanation:
(a) we take the free surface of the lake to point 1 and the free surfaces of the storage tank to point 2. We also take the lake surface as the reference level (z1= 0), and thus the potential energy at points 1and 2 are pe1= 0 and pe2= gz2. The flow energy at both points is zero since both 1 and 2 are open to the atmosphere (P1= P2= Paytm). Further, the kinetic energy both points is zero (ke1= ke2= 0) since the water at both locations is essentially stationary. The mass flow rate of water and its potential energy at point 2 kg/s70/s)m070.0)(kg/m1000(33===V&&rmkJ/kg196.0/sm1000kJ/kg1m)20)(m/s(9.8122222===gzpe
Then the rate of increase of the mechanical energy of water becomes kW13.7kJ/kg)6kg/s)(0.1970()0()(22inmech,outreach,fluid mech,===-=-=DpempemeemE&&&&The overall efficiency of the combined pump-motor unit is determined from its definition,67.2%or 0.672
NB:
Please check the attached file for clearer work.
Answer:
The force acting on the bullet
The value of impulse on the bullet in the given time interval P = 92.4
Explanation:
Mass of the bullet ( m ) = 200 gr = 0.028 lb
Initial velocity ( U ) = 0
Final Velocity ( V ) = 3300
Force acting on the bullet
⇒
⇒
This is the force acting on the bullet.
Magnitude of the impulse imparted on the bullet -------- (1)
Put the value of F & dt in above equation we get,
P = 84000 × 0.0011
P = 92.4
This is the value of impulse on the bullet in the given time interval.