All the boats operating at night requires Navigation light.
Navigation light helps prevent collisions between boats and see if visibility is poor. The types of boat are : Rowboats, Tug boats, Vessels, Sailboat etc.
Various boats have different lightning color to show its use and side of the boat.
To solve the problem it is necessary to apply the concepts related to the conservation of energy through the heat transferred and the work done, as well as through the calculation of entropy due to heat and temperatra.
By definition we know that the change in entropy is given by

Where,
Q = Heat transfer
T = Temperature
On the other hand we know that by conserving energy the work done in a system is equal to the change in heat transferred, that is

According to the data given we have to,




PART A) The total change in entropy, would be given by the changes that exist in the source and sink, that is



On the other hand,



The total change of entropy would be,



Since
the heat engine is not reversible.
PART B)
Work done by heat engine is given by



Therefore the work in the system is 100000Btu
Explanation:
frequency =speed/wavelength
=5/0.5=10Hz
Yes, the above-given statement is true
<u>Explanation:</u>
- The product of the mass x the velocity will be the same for both. Momentum is the action of a body with a particular mass through space and there is the conservation of momentum.
- Momentum is described as the mass of the object multiplied by its velocity.
- <u>Momentum (p) = Mass (M) * Velocity (v)</u>
- Therefore for two objects with many masses to have a similar momentum, then the lighter one has to be moving quicker than the heavier object.
Answer:
26.9 Pa
Explanation:
We can answer this question by using the continuity equation, which states that the volume flow rate of a fluid in a pipe must be constant; mathematically:
(1)
where
is the cross-sectional area of the 1st section of the pipe
is the cross-sectional area of the 2nd section of the pipe
is the velocity of the 1st section of the pipe
is the velocity of the 2nd section of the pipe
In this problem we have:
is the velocity of blood in the 1st section
The diameter of the 2nd section is 74% of that of the 1st section, so

The cross-sectional area is proportional to the square of the diameter, so:

And solving eq.(1) for v2, we find the final velocity:

Now we can use Bernoulli's equation to find the pressure drop:

where
is the blood density
are the initial and final pressure
So the pressure drop is:
