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statuscvo [17]
3 years ago
11

Operators must slow to no wake speed when within what distance of a u.s. naval vessel?

Physics
2 answers:
Alex Ar [27]3 years ago
6 0
Recreational boaters have a role in keeping the waterways safe and secure. Violators of the restrictions can expect a quick and severe response. The operator should not approach within yards and slow to minimum speed within 500 yards of any of united states naval vessel. If the vessel need to pass within 100 yards of a united states naval vessel for safe passage, must contact the united states naval vessel or the united states coast guard escort vessel on VHF-FM channel 16. In addition,

• The operator must perceive and avoid all security areas.

• Avoid commercial port operation areas particularly those that include military cruise line or petroleum amenities. 

• The operator must perceive and avoid other controlled areas near dams, power plants, etc. 

• The operator of the vessel must not discontinue or anchor underneath bridges or in the channel. 
Genrish500 [490]3 years ago
6 0

Answer:

500 Yards or 457.2 Meters

Explanation:

The waves created behind a boat because of its movement in the water is called as "Wake". More the speed of the vessel, more severe is the wake.The speed at which the boat will be able to move and navigate but will not be creating wake is known as  "Slow - No Wake"  speed. As per Homeland Security restrictions, any vessel must slow down to "slow-no wake" within 500 yards of any US Naval Vessel.

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One mole of iron (6 x 10^23 atoms) has a mass of 56 grams, and its density is 7.87 grams per cubic centimeter, so the center-to-
choli [55]

Answer:

Explanation:

Given that:

length l = 2.3 m

a = 0.12 cm = 0.12  \times 10^{-2} \ m

x = 1.17 \ cm = 1.17 \times 10^{-2}\ m

m = 149 kg

\delta = 7.87 \ g/cm^3

da = 2.28 \times 10^{-10}\ m

F_{net} = F-mg\\ \\0 = F - mg \\ \\  F = mg \\ \\ k_sx = mg \\ \\

∴

k_s = \dfrac{149(9.8)}{1.17 \times 10^{-2}} \\ \\  k_s = 124803.42  \ N /m

N_{chain} = \dfrac{A_{wire}}{A_{atom}} = \dfrac{A_w}{da^2}

N_{chain} = \dfrac{(a)^2}{(da)^2} = (\dfrac{a}{da})^2

N_{chain} =  (\dfrac{0.12 \times 10^{-2} }{2.28 \times 10^{-10}})^2

N_{chain} = 2.77 \times 10^{13}

N_{bond} = \dfrac{L}{da} \\ \\  = \dfrac{2.3}{2.28 \times 10^{-10}} \\ \\ N_{bond} = 1.009 \times 10^{10}

\text{Finally; the stiffness of a single interatomic spring is:}

k_{si} =\dfrac{N_{bond}}{N_{chain}}\times k_s

k_{si} =\dfrac{(1.009 \times 10^{10})}{2.77*10^{13}}}\times (124803.42)

\mathbf{k_{si} =45.46 \ N/m}

4 0
3 years ago
If a voltmeter has a less than ideal resistance, say 1 MΩ, and is used to measure the voltage across a resistor of a comparable
Naddik [55]

Answer:

As the difference between the resistance of voltmeter and the resistance being measured gets reduced the error in the reading of the voltmeter gets increased.

Explanation:

An ideal voltmeter has infinite parallel resistance and because of this it doesn't draw any current from the circuit of measurement which means it will measure the exact voltage across the elements.

But practically speaking, a real voltmeter doesn't has infinite resistance therefore, all the practical voltmeters face loading effect to some extent.

As the difference between the resistance of voltmeter and the resistance being measured gets reduced the error in the reading of the voltmeter gets increased. This is why we want to have a greater value of voltmeter resistance, ideally infinite so that the corresponding error is minimized.

Lets consider the given scenario,

A voltmeter has 1 MΩ parallel resistance and the resistance of of measuring element is 500 kΩ or 0.5 MΩ

lets suppose the supplied voltage is 1 V.

First lets assume that the voltmeter is ideal and it has infinite resistance, so in this case voltmeter will measure a voltage of 1 V across the 0.5 MΩ resistor.

Now consider the loading effect, when we connect the voltmeter across the 0.5 MΩ resistor they both become parallel so the resistance is

R = (1*0.5)/(1+0.5)

R = 0.33 MΩ

As you can see the voltmeter will see a reduced resistance and the corresponding voltage also reduces because resistance and voltage are directly proportional.

Therefore, it is preferred to have a very high parallel resistance of the voltmeter.

8 0
3 years ago
Two uncharged metal spheres, spaced 25.0 cm apart, have a capacitance of 26.0 pF. How much work would it take to move 12.0 nC of
viktelen [127]

Answer:

W=2.76\times 10^{-6}\ J

Explanation:

Given that,

The distance between two spheres, r = 25 cm = 0.25 m

The capacitance, C = 26 pF = 26×10⁻¹² F

Charge, Q = 12 nC = 12 × 10⁻⁹ C

We need to find the work done in moving the charge. We know that, work done is given by :

U=\dfrac{Q^2}{2C}

Put all the values,

U=\dfrac{(12\times 10^{-9})^2}{2\times 26\times 10^{-12}}\\\\U=2.76\times 10^{-6}\ J

So, the work done is 2.76\times 10^{-6}\ J.

8 0
3 years ago
A football is thrown vertically with an initial velocity of 33m/s. Calculate the velocity of the ball when it’s height increased
tatyana61 [14]

The final velocity becomes 31.48 m/s

<u>Explanation:</u>

Given:

Initial velocity, u = 33 m/s

Height, h = 5m

Final velocity, v = ?

According to Newton's law:

v² - u² = 2gh

where,

g is the acceleration due to gravity and

g = 9.8 m/s²

On substituting the values we get:

v^2 - (33)^2 = 2 X -9.8 X 5\\\\v^2 - 1089 = -98\\\\v^2 = 991\\\\v = 31.48 m/s

Therefore, the final velocity becomes 31.48 m/s

5 0
3 years ago
Turn the ignition switch to start and release the key immediately or you could destroy the______________.
vredina [299]

Starter

Explanation:

Turn the ignition switch to start and release the key immediately or you could destroy the starter.

The car starter is used to cause ignition in the internal combustion engine in order to fire the piston and cause mechanical motion. The starter is used to start the cyclic process of the internal combustion engine.

  • Once the engine starts by igniting the starter, it is best to release it.
  • The starter ensures that the spark plug is engaged and the motor is brought into work.
  • If the ignition is still engaged, the process continues repeatedly and it can damage the starter of the car.

learn more:

Automobile brainly.com/question/2599962

#learnwithBrainly

5 0
3 years ago
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