A concrete pump of 120-cy/hr max. output is used to place 250 cy of concrete. It is supplied by 8 cy transit-mix trucks arriving
1 answer:
Answer:
Completing the operation will require 9.765625 hours.
Explanation:
First, let's check which of the stages will be the process limitation:
Regarding the pump, it can perform up to 120 cy/h. The trucks will deliver 8 cy every 15 min, i.e., 32 cy/h, then this is the smaller capacity within the whole process.
Now, let's establish a simple relation:
Placement speed x Efficiency = Placed quantity / time,
solving for time. we have
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Answer:
Explanation:
Given that
Q= 5 L/min
1 L = 10⁻³ m³/s
1 min = 60 s
Q=0.083 x 10⁻³ m³/s
d= 6 μm
v= 1 mm/s
So the discharge flow through one tube
q = A v
A=2.8 x 10⁻¹¹ m²
v= 1 x 10⁻³ m/s
q= 2.8 x 10⁻¹⁴ m³/s
Lets take total number of tube is n
Q= n q
n=Q/q
Surface area A
A= π d L
Solution:
f ( t )= 20 S ( t ) + 55/30 tS ( t )− 55/30 ( t − 30 ) S ( t − 30 )
• Taking the Laplace Transform:
F ( s ) = 20/s + 55/30 ( 1/s^2 ) – 55/30 ( 1/s^2) e^-30s = 20/s + 55/30 ( 1/s^2 ) ( 1 – e^-30s)
f ( t )= 20 S ( t ) + 55/30 tS ( t )− 55/30 ( t − 30 ) S ( t − 30 )
• Taking the Laplace Transform:
F ( s ) = 20/s + 55/30 ( 1/s^2 ) – 55/30 ( 1/s^2) e^-30s = 20/s + 55/30 ( 1/s^2 ) ( 1 – e^-30s)
Answer:
Explanation:
b ) The problem is based on Doppler's effect of sound
f = f₀ x (V - v₀) /( )
f is apparent frequency ,f₀ is real frequency , V is velocity of sound , v₀ is velocity of observer going away , is velocity of source going away
778 = 840 x (340 - 14)/ (340 + )
340 + = 341.18
= 1.18 m /s
it will go away from the observer or the cyclist.
speed of train = 1.18 m /s
a )
For a stationary observer v₀ = 0
f = f₀ x V /( )
= 840 x 340 / (340 + 1.180)
= 837 Hz