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trasher [3.6K]
3 years ago
5

How can you prevent static sparks while filling your boat's fuel tank?

Physics
2 answers:
kondor19780726 [428]3 years ago
5 0

Answer:

Not to have any contact between fuel pump and tank.

Explanation:

Static spark occurs when there is a imbalance of electric charges at surface of any material. This can even  result into high damage especially while filling boat tank, hence measures should be taken to prevent static sparks.

* Make sure not to have any contact between the brim or surface of fuel pump and the tank's opening.

* Fill the tank carefully without spilling the fuel in water.

* Never fill the tank up to its maximum capacity.

dangina [55]3 years ago
4 0
To prevent static sparks that could occur while you fill fuel is that, a person who is filling the tank should remember that the nozzle of the fuel pump hose should always be in connected or in contact to the tank. This will prevent static sparks from happening to the boat that is going to be used and prevent harm to the person using or refilling the fuel.
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6) Find the speed a spherical raindrop would attain by falling from 4.00 km. Do this:a) In the absence of air dragb) In the pres
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We are asked to determine the velocity of a rain drop if it falls from 4 km.

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Part B. we are asked to determine the velocity if there is air drag. To do that we will use the following formula:

F_d=\frac{1}{2}C\rho_{air}Av^2

Where:

\begin{gathered} F_d=drag\text{ force} \\ C=\text{ constant} \\ \rho_{air}=\text{ density of air} \\ A=\text{ area} \\ v=\text{ velocity} \end{gathered}

We need to determine the drag force. To do that we will use the following free-body diagram:

Since the velocity that the raindrop reaches is the terminal velocity and its a constant velocity this means that the acceleration is zero and therefore the forces are balanced:

F_d=mg

Now, we determine the mass of the raindrop using the following formula:

m=\rho_{water}V

Where:

\begin{gathered} \rho_{water}=\text{ density of water} \\ V=\text{ volume} \end{gathered}

The volume is the volume of a sphere, therefore:

m=\rho_{water}(\frac{4}{3}\pi r^3)

Since the diameter of the raindrop is 3 millimeters, the radius is 1.5 mm or 0.0015 meters. Substituting we get:

m=(0.98\times10^3\frac{kg}{m^3})(\frac{4}{3}\pi(0.0015m)^3)

Solving the operations:

m=1.39\times10^{-5}kg

Now, we substitute the values in the formula for the drag force:

F_d=(1.39\times10^{-5}kg)(9.8\frac{m}{s^2})

Solving the operations:

F_d=1.36\times10^{-4}N

Now, we substitute in the formula:

1.36\times10^{-4}N=\frac{1}{2}C\rho_{air}Av^2

Now, we solve for the velocity:

\frac{1.36\times10^{-4}N}{\frac{1}{2}C\rho_{air}A}=v^2

Now, we substitute the values. We will use the area of a circle:

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Substituting the radius:

\frac{1.36\cdot10^{-4}N}{\frac{1}{2}(0.45)(1.21\frac{kg}{m^{3}})(\pi(0.0015m)^2)}=v^2

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Therefore, the velocity is 8.4 m/s

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