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3 years ago

How might engineers use tidal data when building a bridge or a dock?

Physics

1 answer:

kifflom [539]3 years ago

7 0

Answer:

Explanation:

1) Make sure the service portion of the structure stays above the highest probable water level.

2) Make sure the support structure in contact with the water can withstand the tidal flows

3) Consider whether the added restriction of the proposed structure will altar the height of, or water velocity contained in, the tidal exchanges.

4) Consider whether a dock needs to have mechanisms to compensate for large sea level changes.

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What is the gauge pressure of the water right at the point p, where the needle meets the wider chamber of the syringe? neglect t

Helen [10]

Missing details: figure of the problem is attached.

We can solve the exercise by using Poiseuille's law. It says that, for a fluid in laminar flow inside a closed pipe,

$\Delta P = \frac{8 \mu L Q}{\pi r^4}$

where:

$\Delta P$ is the pressure difference between the two ends

$\mu$ is viscosity of the fluid

L is the length of the pipe

$Q=Av$ is the volumetric flow rate, with $A=\pi r^2$ being the section of the tube and $v$ the velocity of the fluid

r is the radius of the pipe.

We can apply this law to the needle, and then calculating the pressure difference between point P and the end of the needle. For our problem, we have:

$\mu=0.001 Pa/s$ is the dynamic water viscosity at $20^{\circ}$

L=4.0 cm=0.04 m

$Q=Av=\pi r^2 v= \pi (1 \cdot 10^{-3}m)^2 \cdot 10 m/s =3.14 \cdot 10^{-5} m^3/s$

and r=1 mm=0.001 m

Using these data in the formula, we get:

$\Delta P = 3200 Pa$

However, this is the pressure difference between point P and the end of the needle. But the end of the needle is at atmosphere pressure, and therefore the gauge pressure (which has zero-reference against atmosphere pressure) at point P is exactly 3200 Pa.

8 0

3 years ago

Determine the mass of a sphere with a radius of 8 cm having a density of 7500 kg / m3

Aliun [14]

Answer:

Explanation:

V=2140CM3

M=VOLUME*DENSITY

7500kg/m3=7.5g/cm3

2140cm3*7.5g/cm3=16050

16050g

7 0

2 years ago

How do you find the net force acting on an object?

weqwewe [10]

Answer:

C. Add all the force vectors

Explanation:

The net force acting on an object is the vector sum of all the forces on the object.

Remember, Newton's first law tells us a body at rest will remain at rest or that in uniform motion will continue in motion unless acted by unbalanced forces.These unbalanced forces act in all direction towards the body thus to get the net force you require a summation of all these force with respect to their magnitudes and directions.

For example a force of 3N towards the East direction acting on a body and another force of 2N towards the West direction on the same body will generate a net force of 1N towards the East direction.

4 0

4 years ago

All of following are types of kinetic energy EXCEPT for: *

Valentin [98]

Light energy is not kinetic energy

7 0

3 years ago

If you drop a rock off a cliff and it hits the ground 3 seconds later how tall is that cliff

skelet666 [1.2K]

Answer:

Basic kinematics, negating drag and assuming ideal conditions, we use the equation:

d=vi*t+1/2*a*t^2

Since vi is 0 (we know this because you’re dropping it, not throwing it)…

…and the only acceleration acting on it is gravity, a=9.8 m/s^2…

…we get

d=1/2(9.8)(5)^2

Explanation:

Some quick mental math tells us that this is about 125 m.

Plugging it in, we find it to be 122.5 m.

6 0

3 years ago