Ideally, the resistance of an ammeter should be:

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

Alex Ar [27]

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.

6 0

3 years ago

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A 6.0-kilogram block, sliding to the east across a horizontal, frictionless surface with a momentum of 30.0 kilogram · meters pe

Lina20 [59]

The final speed of the block after the collision with the obstacle is $\boxed{3.33\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}}$ .

Further Explanation:

Given:

The mass of the block is $6.0\,{\text{kg}}$ .

The initial momentum of the block is $30\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/ {\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}$ .

The impulse imparted by the obstacle is $10\,{\text{N}} \cdot {\text{s}}$ .

Concept:

The block is sliding towards east and the impulse imparted by the obstacle is towards the obstacle is towards west on the block. It means that the impulse exerted by the obstacle will reduce the momentum of the block.

According to the impulse momentum theorem, the rate of change of momentum of the body is equal to the impulse imparted to the body.

The expression for the impulse momentum theorem is.

${p_f} - p{ & _i} = I$ …… (1)

Substitute $30\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/{\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}$ for ${p_i}$ and $- 10\,{\text{N}} \cdot {\text{s}}$ for $I$ in equation (1).

$\begin{aligned}{p_f} &= - 10\,{\text{N}} \cdot {\text{s}} + 30\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/{\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}} \\&= 20\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/{\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}\\\end{aligned}$

The final momentum of the block can be expressed as:

${p_f} = m{v_f}$ …… (2)

Substitute $20\text{kg}\;\text{m/s}$ for ${p_f}$ and $6.0\,{\text{kg}}$ for $m$ in equation (2).

$\begin{aligned}20 &= 6 \times {v_f} \\ {v_f}&= \frac{{20}}{6}\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}\\&= 3.33\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}} \\ \end{aligned}$

Thus, the final speed of the block after the collision with the obstacle is $\boxed{3.33\;\text{m/s}}$ .

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Answer Details:

Grade: High School

Chapter: Impulse-momentum theorem

Subject: Physics

Keywords: Impulse, imparted, obstacle, speed, momentum, the obstacle, impulse-momentum theorem, frictionless surface, speed of block after collision.

5 0

3 years ago

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Consider three force vectors Fi with magni- tude 53 N and direction 116º, F2 with mag- nitude 57 N and direction 217°, and F3 wi

Flauer [41]

Answer:

a. Fnet =37.67N

b. The direction = 133.4 from the x axis counter clockwise.

c. Option 2

Explanation:

Given that F1 is 53N at 116°, then it will be at a direction of 116-90=26° in the second quadrant.

Given that F2 is 57N at 116°, then it will be at a direction of 217-180=37° in the third quadrant..

Given that F1 is 71N at 20°, then it is in the first quadrant.

a. Fnet= F1+F2+F3

Fnet= -F1sin26i+F2cos26j-F2cos37i-F2sin37j+F3cos20i+F3sin20j

Fnet= 53sin26i+53cos26j-57cos37i-57sin37j+71cos20i+71sin20j

Resolving the vectors into x and y components.

Fnet= -2.04i+37.62j

Magnitude of the vector

Fnet= √((-2.04)^2+(37.62)^2)

Fnet= 37.67N

Fnet is approximately 38N.

b. Direction of the Fnet.

Angle=arctan(y/x)

Angle=arctan(-37.61/2.04)

Angle= -43.37°

The angle is in the negative x axis and positive y axis.

Then the direction becomes 180-43.37

Therefore, the direction of the net force is 133.37°.

c. The instantaneous velocity of a body is always in the direction of the net force at that instant. Option 2 is correct.

Fnet=ma

Fnet= mv/t

So the velocity is in the direction of the Fnet.

3 0

3 years ago

Calculate the charge that flows through the cell in 1 minute. Each filament lamp has a power of 3 W and a resistance of 12 Ω

lapo4ka [179]

Answer:

24 Coulumbs

Explanation:

Given data

time= 1 minute= 6 seconds

P=2 W

R= 12 ohm

We know that

P= I^2R

P/R= I^2

2/12= I^2

I^2= 0.166

I= √0.166

I= 0.4 amps

We know also that

Q= It

substitute

Q= 0.4*60

Q= 24 Columbs

Hence the charge is 24 Coulumbs

5 0

3 years ago

By the term universe astronomers mean

OverLord2011 [107]

'Universe' means 'Everything'. That is, all matter, all space, all time.

6 0

4 years ago