A 7.1 cm diameter horizontal pipe gradually narrows to 5.4 cm . When water flows through this pipe at a certain rate, the gauge
pressure in these two sections is 32.5 kPa and 20.6 kPa , respectively. What is the volume rate of flow?
1 answer:
Answer:
Q=22.6L/s
Explanation:
First you must consider the continuity equation at points 1 and 2, which indicates that both flows are of equal value, in this way you get an equation between the two flow rates.
Then you raise the Bernoulli equation taking into account that the height is the same, which makes the term h1-h2 zero.
Using the equations above to calculate one of the speeds.
Finally you find the flow by multiplying the speed by the area.
I attached procedure
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Answer:
a. b.
c.
d. The angular acceleration when sitting in the middle is larger.
Explanation:
a. The magnitude of the torque is given by , being r the radius, F the force aplied and
the angle between the vector force and the vector radius. Since
and so
.
b. Since the relation hols, being I the moment of inertia, the angular acceleration can be calculated by
. Since we have already calculated the torque, all left is calculate the moment of inertia. The moment of inertia of a solid disk rotating about an axis that passes through its center is
, being M the mass of the disk. If we assume that a person has a punctual mass, the moment of inertia of a person would be given by
, being
the mass of the person and
the distance from the person to the center. Given all of this, we have
.
c. Similar equation to b, but changing , so
.
d. The angular acceleration when sitting in the middle is larger because the moment of inertia of the person is smaller, meaning that the person has less inertia to rotate.
Answer:
a = 17.68 m/s²
Explanation:
given,
length of the string, L = 0.8 m
angle made with vertical, θ = 61°
time to complete 1 rev, t = 1.25 s
radial acceleration = ?
first we have to calculate the radius of the circle
R = L sin θ
R = 0.8 x sin 61°
R = 0.7 m
now, calculating at the angular velocity
ω = 5.026 rad/s
now, radial acceleration
a = r ω²
a = 0.7 x 5.026²
a = 17.68 m/s²
hence, the radial acceleration of the ball is equal to 17.68 rad/s²