Answer:
The fluids speed at a) 
and b) 
are 
and 
respectively
c) Th volume of water the pipe discharges is: 
Explanation:
To solve a) and b) we should use flow continuity for ideal fluids:
(1)
With Q the flux of water, but Q is 
using this on (1) we have:
(2)
With A the cross sectional areas and v the velocities of the fluid.
a) Here, we use that point 2 has a cross-sectional area equal to 
, so now we can solve (2) for 
:
b) Here we use point 2 as 
:
c) Here we need to know that in this case the flow is the volume of water that passes a cross-sectional area per unit time, this is 
, so we can write:
, solving for V:
Answer:
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Explanation:
Q=l^2 times R times t
Where Q - heat, I -current, R - resistance and t is time
If you increase I twice (and it's squared), than Q gonna went up 4 times (2 squared).
Choose last
Answer:
The distance covered by puck A before collision is 
Explanation:
From the question we are told that
The label on the two hockey pucks is A and B
The distance between the two hockey pucks is D 18.0 m
The speed of puck A is 
The speed of puck B is 
The distance covered by puck A is mathematically represented as
=> 
The distance covered by puck B is mathematically represented as
=> 
Since the time take before collision is the same
substituting values
=> 
=>
The energy transfer in terms of work has the equation:
W = mΔ(PV)
To be consistent with units, let's convert them first as follows:
P₁ = 80 lbf/in² * (1 ft/12 in)² = 5/9 lbf/ft²
P₂ = 20 lbf/in² * (1 ft/12 in)² = 5/36 lbf/ft²
V₁ = 4 ft³/lbm
V₂ = 11 ft³/lbm
W = m(P₂V₂ - P₁V₁)
W = (14.5 lbm)[(5/36 lbf/ft²)(4 ft³/lbm) - (5/9 lbf/ft²)(11 lbm/ft³)]
W = -80.556 ft·lbf
In 1 Btu, there is 779 ft·lbf. Thus, work in Btu is:
W = -80.556 ft·lbf(1 Btu/779 ft·lbf)
<em>W = -0.1034 BTU</em>
