Answer:
Runner A will be 0.05 km from the flagpole, and runner B will be 0.07 km from the flagpole
Explanation:
We can find when their paths will cross as follows:
Where:
is the final position
is the initial position
v₀ is the initial speed
t is the time
a is the acceleration = 0 (since they are running with a constant velocity)
When their paths cross we have:
Now we can find the final distance of each runner.
Therefore, runner A will be 0.05 km from the flagpole, and runner B will be 0.07 km from the flagpole.
I hope it helps you!
Answer:
B is the answer
Explanation:
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<u>Answer</u>:
316.67 Hz
<u>Explanation</u>:
Given:
Wave velocity ( v ) = 95 m / sec
wavelength ( λ ) = 0.3 m
We have to calculate Frequency ( f ) :
We know:
v = λ / t [ f = 1 / t ]
v = λ f
= > f = v / λ
Putting values here we get:
= > f = 95 / 0.3 Hz
= > f = 950 / 3 Hz
= > f = 316.67 Hz
Hence, frequency of sound is 316.67 Hz.
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:
