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3 years ago

Waves begin to "feel bottom" when the depth of water is

2 answers:

8 0

I think the correct answer would be one half the wavelength. Waves would "feel bottom" when the water is at the depth of 0.5 of the wavelength. "Feel bottom" is a term used to describe that the depth of water affects the wave properties. Hope this answers the question.

Likurg_2 [28]3 years ago

6 0

Equal to one-half the wavelength is the right answer.

Water is regulated by many approaches such as depth gauges (based on force) a sounding line and by sonar range. Prior to the modern era, sailors would cast metal weighted materials into the ocean. Submarines made a new demand for depth estimation, to need to know the depth of boats in the water. However, waves start to "feel bottom" when the depth of water is equivalent to one-half the wavelength. As the wave approaches a lightweight bank, its track is adjusted. This involution made the tides to become inactive. As the flow becomes steeper, the width of the wave diminishes.

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An overhead door is guided by wheels at a and b that roll in horizontal and vertical tracks. when θ = 40°, the velocity of wheel

Karo-lina-s [1.5K]

I think the situation is modeled by the scenario in the attached image. Some specific values seem to be missing (like the height of door $d$ )...

The door forms a right triangles that satisfies

$\tan\theta=\dfrac ab\implies\sec^2\theta\dfrac{\mathrm d\theta}{\mathrm dt}=\dfrac{b\frac{\mathrm da}{\mathrm dt}-a\frac{\mathrm db}{\mathrm dt}}{b^2}$

We also have

$\tan\theta=\dfrac ab\implies\cos\theta=\dfrac bd$

so if you happen to know the height of the door, you can solve for $b$ and $a$ .

$d$ is fixed, so

$a^2+b^2=d^2\implies2a\dfrac{\mathrm da}{\mathrm dt}+2b\dfrac{\mathrm db}{\mathrm dt}=0\implies\dfrac{\mathrm da}{\mathrm dt}=-\dfrac ba\dfrac{\mathrm db}{\mathrm dt}$

We can solve for the angular velocity $\dfrac{\mathrm d\theta}{\mathrm dt}$ :

$\dfrac{\mathrm d\theta}{\mathrm dt}=\cos^2\theta\dfrac{b\left(-\frac ba\frac{\mathrm db}{\mathrm dt}\right)-a\frac{\mathrm db}{\mathrm dt}}{b^2}=-\dfrac1a\dfrac{\mathrm db}{\mathrm dt}$

At the point when $\theta=40^\circ$ and $\dfrac{\mathrm db}{\mathrm dt}=1.8$ ft/s, we get

$\dfrac{\mathrm d\theta}{\mathrm dt}=-\dfrac{1.8}a\dfrac{\rm deg}{\rm s}=-\dfrac{1.8}{d\sin40^\circ}\dfrac{\rm deg}{\rm s}$

6 0

3 years ago

9. A car driver brakes gently. Her car slows down front --

sleet_krkn [62]

Answer:

9) This is a case of deceleration

10)-0.8 ms-2

b) acceleration is the change in velocity with time

11)

a) 100 ms-1

b) 100 seconds

12) 10ms-1

13) more information is needed to answer the question

14) - 0.4 ms^-2

15) 0.8 ms^-2

Explanation:

The deceleration is;

v-u/t

v= final velocity

u= initial velocity

t= time taken

20-60/50 =- 40/50= -0.8 ms-2

11)

Since it starts from rest, u=0 hence

v= u + at

v= 10 ×10

v= 100 ms-1

v= u + at but u=0

1000 = 10 t

t= 1000/10

t= 100 seconds

12) since the sprinter must have started from rest, u= 0

v= u + at

v= 5 × 2

v= 10ms-1

14)

v- u/t

10 - 20/ 25

10/25

=- 0.4 ms^-2

15)

a=v-u/t

From rest, u=0

8 - 0/10

a= 8/10

a= 0.8 ms^-2

7 0

3 years ago

What is trapezord mean

igor_vitrenko [27]

A quadrilateral with only one pair of parallel sides.
<span>a small carpal bone in the base of the hand, articulating with the metacarpal of the index finger.
</span>

5 0

3 years ago

If a cars velocity is slowing down is it considered a positive or negative acceleration?

Natali [406]

Answer:

Negative

Explanation:

Observe that the object below moves in the positive direction with a changing velocity. An object which moves in the positive direction has a positive velocity. If the object is slowing down then its acceleration vector is directed in the opposite direction as its motion (in this case, a negative acceleration).

4 0

3 years ago

The noise level coming from a pig pen with

sweet [91]

Answer:

The decibel of the remaining pigs is 51.5 dB.

Explanation:

Decibel (dB) is a unit of measure of the intensity of a given sound.

Number of pigs = 199, noise level = 74.3 dB.

Given that the intensity (I) of the sound from the pen is proportional to the number of pigs (N), thus:

I $\alpha$ N

I = kN

where k is the constant of proportionality.

⇒ k = $\frac{I}{N}$

= $\frac{74.3}{199}$

k = 0.3734

When 61 numbers of pigs were removed, the number of remaining pigs (N) squealing at their original level is 138.

Thus, the becibel level (I) of the remaining pigs can be determined by:

I = kN

= 0.3734 × 138

= 51.53 dB

The becibel level (I) of the remaining pigs is 51.53 dB.

6 0

3 years ago