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Flura [38]
3 years ago
5

Which type of waves is important to have for a surfer who wants to do aerial maneuvers? A. Plunging waves B. Surging waves C. Or

bital waves D. Spilling waves
Physics
1 answer:
Sladkaya [172]3 years ago
8 0

Aerial maneuvers in surfing demand taking the surfboard into the air. The best type of waves for such maneuvers are:

- the section of a wave coming towards you should be  smaller than the section you are on

- you should try to find closeout sections of waves (  formations that do not allow a tubular ride )

- the wave should be either wedging or bowling in towards you, it should not be a straight wave.

A.PLUNGING WAVES

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Digiron [165]

Answer:

i) acceleration from B to D is 0, because the velocity is constant (stays the same)

ii) whatever units of distathat might be, we can calculate the number:

for 4 time-steps (2 to 6) the velocity is 6 per time step, that makes 24 distance units in these 4 time steps. it's the same the area underneath the graph.

there is also the vertical line from 0 to 2. we can calculate that distance like the area of a triangle with 2*6 / 2 = 6

the total distance from 0 to D is therefore 30

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The difference in electronegativity values between H and O is 1.4. What type of bond will these elements form?
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C I completed the same quiz
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3 years ago
A.) If its booster rockets accelerate the space shuttle at 15m/s2, how high will it be one minute after launch?
poizon [28]

Answer:

27,000 m

450 m/s

Explanation:

Assuming the initial velocity is 0 m/s:

v₀ = 0 m/s

a = 15 m/s²

t = 60 s

A) Find: Δy

Δy = v₀ t + ½ at²

Δy = (0 m/s) (60 s) + ½ (15 m/s²) (60 s)²

Δy = 27,000 m

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5 0
3 years ago
One normal afternoon in undisclosed City X, the chocolate factory workers overload the Easter egg machine. A group of angsty tee
musickatia [10]

Answer:

The correct option is;

C. 1,715 m

Explanation:

We are given the information from the group of teen at the City edge

Time of arrival of explosion sound = 5 s after sighting

Time of sighting explosion = 5 s before hearing the boom

Speed of sound in air ≈ 343 m/s

Speed of light = 299,792 km/s

Therefore, distance covered by sound in 5 seconds is given by the following equation;

Speed = \frac{Distance}{Time}

\therefore 343 \ m/s= \frac{Distance}{5 \, s}

Hence Distance = 343 m/s × 5 s = 1715 m

To check, we compare the time it would take for the light to cover 1715 m

That is Time = \frac{Distance}{Speed} =  \frac{1715}{299,792,000} = 0.00000572 \, s which is instantaneous hence the distance can be approximated by the time duration for the speed of sound.

Therefore, the distance of the students from the factory is approximately 1,715 m

8 0
4 years ago
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