Water flows at 10 m/s through a pipe with radius 0.025 m. The pipe goes up to the second floor of the building, 2.5 m higher, an
d the pressure remains unchanged. Which of the following statements is true concerning the velocity of water in the pipe and the radius of the pipe on the second floor?
O A. Since this is a closed system, the velocity of water and area of the pipe are the same on both floors.
OB. On the second floor, the velocity of the water increased and the area of the pipe is larger.
Oc. On the second floor, the velocity of the water increased and the area of the pipe smaller.
OD. On the second floor, the velocity of the water decreased and the area of the pipe is larger.
OE. On the second floor, the velocity of the water decreased and the area of the pipe is smaller.
OF. There is not enough information to determine what happens to the water and pipe size.
O A. Since this is a closed system, the velocity of water and area of the pipe are the same on both floors.
OB. On the second floor, the velocity of the water increased and the area of the pipe is larger.
Oc. On the second floor, the velocity of the water increased and the area of the pipe smaller.
OD. On the second floor, the velocity of the water decreased and the area of the pipe is larger.
OE. On the second floor, the velocity of the water decreased and the area of the pipe is smaller.
OF. There is not enough information to determine what happens to the water and pipe size.
1 answer:
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Answer:
The depth of the well, s = 54.66 m
Given:
time, t = 3.5 s
speed of sound in air, v = 343 m/s
Solution:
By using second equation of motion for the distance traveled by the stone when dropped into a well:
Since, the stone is dropped, its initial velocity, u = 0 m/s
and acceleration is due to gravity only, the above eqn can be written as:
(1)
Now, when the sound inside the well travels back, the distance covered,s is given by:
(2)
Now, total time taken by the sound to travel:
t = t' + t''
t'' = 3.5 - t' (3)
Using eqn (2) and (3):
s = 343(3.5 - t') (4)
from eqn (1) and (4):
Solving the above quadratic eqn:
t' = 3.34 s
Now, substituting t' = 3.34 s in eqn (2)
s = 54.66 m