Answer:
p2 = 9.8×10^4 Pa
Explanation:
Total pressure is constant and PT = P = 1/2×ρ×v^2
So p1 + 1/2×ρ×(v1)^2 = p2 + 1/2×ρ×(v2)^2
from continuity we have ρ×A1×v1 = ρ×A2×v2
v2 = v1×A1/A2
and
r2 = 2×r1
then:
A2 = 4×A1
so,
v2 = (v1)/4
then:
p2 = p1 + 1/2×ρ×(v1)^2 - 1/2×ρ×(v2)^2 = p1 + 1/2×ρ×(v1)^2 - 1/2×ρ×(v1/4)^2
p2 = 3.0×10^4 Pa + 1/2×(1000 kg/m^3)×(12m/s)^2 - 1/2×(1000kg/m^3)×(12^2/16)
= 9.75×10^4 Pa
= 9.8×10^4 Pa
Therefore, the pressure in the wider section is 9.8×10^4 Pa
Answer:
A simple machine consisting of an axle to which a wheel is fastened so that torque applied to the wheel winds a rope or chain onto the axle, yielding a mechanical advantage equal to the ratio of the diameter of the wheel to that of the axle.
She say give tow points and write the equation of the line with the two give points (1,5) and (-3,-5 )
Given :
Mass of block , M = 20 kg .
Force applied , F = 80 N .
Acceleration of block , 
.
To Find :
The coefficient is Kinetic force friction between the block and the table .
Solution :
We know , Force equation on block is given by :
Therefore , coefficient is Kinetic force friction between the block and the table is 0.15 .
Hence , this is the required solution .
The correct answer for this question is this one:
<span>A snowstorm was predicted in Chicago. The possible upper air temperature, surface temperature, and air pressure of Chicago on that day. Normal atmospheric pressure is 29.9 inches of mercury. </span><em>I'm pretty sure the answer is 40 for upper air, 29 for surface temp, and 30 for air pressure. </em>Hope this helps answer your question and have a nice day ahead.
