Well it would help if there was a picture but wild guess i’m going with B or the second choice.
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:
a) P = 86720 N
b) L = 131.2983 mm
Explanation:
σ = 271 MPa = 271*10⁶ Pa
E = 119 GPa = 119*10⁹ Pa
A = 320 mm² = (320 mm²)(1 m² / 10⁶ mm²) = 3.2*10⁻⁴ m²
a) P = ?
We can apply the equation
σ = P / A ⇒ P = σ*A = (271*10⁶ Pa)(3.2*10⁻⁴ m²) = 86720 N
b) L₀ = 131 mm = 0.131 m
We can get ΔL applying the following formula (Hooke's Law):
ΔL = (P*L₀) / (A*E) ⇒ ΔL = (86720 N*0.131 m) / (3.2*10⁻⁴ m²*119*10⁹ Pa)
⇒ ΔL = 2.9832*10⁻⁴ m = 0.2983 mm
Finally we obtain
L = L₀ + ΔL = 131 mm + 0.2983 mm = 131.2983 mm
(False) bearings reduce friction by using smooth balls lubricated with oil or grease that freely roll between a smooth inner and outer surface. The main concept of the ball bearing is that objects that roll past each other produce less friction than if the objects were sliding against each other.
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.