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
The force on an object use Socratic glad to help ..
The answer is B.) I took the test.
Answer:
a. -369.36J
b. -123.9J
c. 9.52m
Explanation:
From the expression for kinetic energy
K. E=1/2mv^2
Since the mass is constant, but the velocity changes. Hence the change in kinetic energy is
K.E=1/2*19(3.6²-7.2²)
K.E= -369.36J
b. to determine the workdone by the force,we determine the distance moved.
But the acceleration is from
F=ma ,
a=f/m
a=-13/19
0.68m/s²
the distance moved is
s=v²/2a
s=3.6²/2*0.68
s=9.52m
Hence the work done is
W=force * distance
W=-13*9.52
W=-123.9J
d. the distance moved is
s=v²/2a
s=3.6²/2*0.68
s=9.52m
