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2 years ago

Sb-28 a collision could occur when the distance decreases and bearing between two vessels does what?

Physics

1 answer:

USPshnik [31]2 years ago

4 0

A collision happens when the distance between two boats decreases and the bearings aren't changing.

To avoid collisions, the operators should master the Rules of the Road. The operators should be able to determine if his boat is a Stand-On Vessel (maintains course and speed) or the give-way vessel (the "burdened" vessel which gives way to the stand-on vessels),and maneuver accordingly.

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Describe how radio waves are different from sound waves.

Lilit [14]

Answer:

sound and radio waves are completely different phenomena.

Explanation:

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1 year ago

Read 2 more answers

Four identical balls are thrown from the top of a cliff, each with the same speed. The

Jlenok [28]

Answer:

the speed and the kinetic energy of the first and fourth ball are equal, while the speed and kinetic energy of the second and third balls are equal

Explanation:

The kinetic energy, K.E. = (1/2) × m × v²

The velocity of the ball, v = u × sin(θ)

Where;

u = The initial velocity of the ball

θ = The reference angle

1) For the ball thrown straight up, we have;

θ = 90°

∴ v = u

The final velocity of the ball as it strikes the ground is v₂ = u² + 2gh

Where;

h = The height of the cliff

∴ K.E. = (1/2) × m × (u² + 2gh)²

2) For the second ball thrown 30° to the horizontal, we have;

K.E. = (1/2) × m × ((u×sin30)² + 2·g·h)² = K.E. = (1/2) × m × ((0.5·u)² + 2·g·h)²

3) For the third ball thrown at 30° below the horizontal, we have;

K.E. = (1/2) × m × ((u×sin30)² + 2·g·h)² = K.E. = (1/2) × m × ((0.5·u)² + 2·g·h)²

4) For the fourth ball thrown straight down, we have;

K.E. = (1/2) × m × (u² + 2gh)²

Therefore, as the ball strike the ground, the speed and the kinetic energy of the first and fourth ball are equal, while the speed and kinetic energy of the second and third balls are equal

Learn more about object kinetic energy of object if free fall here;

brainly.com/question/14872097

6 0

3 years ago

A 2300 kg sailboat is moving west at 5.5 m/s when an eastward wind

pogonyaev

The boat is initially at equilibrium since it seems to start off at a constant speed of 5.5 m/s. If the wind applies a force of 950 N, then it is applying an acceleration a of

950 N = (2300 kg) a

a = (950 N) / (2300 kg)

a ≈ 0.413 m/s²

Take east to be positive and west to be negative, so that the boat has an initial velocity of -5.5 m/s. Then after 11.5 s, the boat will attain a velocity of

v = -5.5 m/s + a (11.5 s)

v = -0.75 m/s

which means the wind slows the boat down to a velocity of 0.75 m/s westward.

5 0

3 years ago

I want to answer these questions about vectors

stellarik [79]

Answer:

the first one answer is no

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3 years ago

A well-thrown ball is caught in a well-padded mitt. If the deceleration of the ball is 2.10×10^{4} 4 m/s^{2} 2 , and 1.85 ms (1

Paha777 [63]

Answer:

u = - 38.85 m/s^-1

Explanation:

given data:

acceleration = 2.10*10^4 m/s^2

time = 1.85*10^{-3} s

final velocity = 0 m/s

from equation of motion we have following relation

v = u +at

0 = u + 2.10*10^4 *1.85*10^{-3}

0 = u + (21 *1.85)

0 = u + 38.85

u = - 38.85 m/s^-1

negative sign indicate that the ball bounce in opposite directon

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2 years ago