Which part of the ear receives the signal from the cochlea and transmits it to the brain?

As part of an interview for a summer job with the Coast Guard, you are asked to help determine the search area for two sunken sh

marshall27 [118]

Answer:

The resultant velocity is $v_t=10 knots$

Explanation:

Apply the law of conservation of momentum

$M_L *v_L + M_f * V_f = (M_L + M_f) v_t$

Where $M_L$ is the mass of the Luxury Liner = 40,000 ton

$v_L$ is the velocity of Luxury Liner = 20 knots due west

$M_f$ mass of freighter = 60,000

$v_f$ is the velocity of freighter = 10 knots due north

Apply the law of conservation of momentum toward the the west direction

$v_f = 0 \ knots$

So the equation would be

$M_L *v_L = (M_L + M_f) v_t$

Substituting values

$40000*20 = (40000+ 60000)v_t_w$

Where $v_t_w$ the final velocity due west

Making $v_t_w$ the subject

$v_t_w = \frac{40,000* 20}{(40000 + 60000)}$

$= 8 \ knots$

Apply the law of conservation of momentum toward the the north direction

$v_L = 0 \ knots$

So the equation would be

$M_f *v_f = (M_L + M_f) v_t_n$

Where $v_t_n$ the final velocity due north

Making $v_t_n$ the subject

$v_t_n = \frac{60,000* 10}{(40000 + 60000)}$

$= 6 \ knots$

The resultant velocity is

$v_t = \sqrt{v_t_w^2 + v_t_n^2}$

$= \sqrt{8^2 +6^2}$

$v_t=10 knots$

8 0

3 years ago

Light is incident on an air-water interface at an angle of 30 degree to the normal. What angle does the refracted ray make with

Vikentia [17]

Answer:

22 degree

Explanation:

Angle of incidence, i = 30 degree

the refractive index of water with respect to air is 4/3.

As the ray of light travels from rarer medium to denser medium, that mean air to water, the refraction takes place.

According to Snell's law,

Refractive index of water with respect to air = Sin i / Sin r

Where, r be the angle of refraction

4 / 3 = Sin 30 / Sin r

0.75 = 2 Sin r

Sin r = 0.375

r = 22 degree

Thus, the angle of refraction is 22 degree.

6 0

3 years ago

Need help really fast ! can give brainliest if correct :> thanks <3

Vlada [557]

Answer:

2. Move faster

Explanation:

Because you lighten the weight and pushed at the same speed it is easier to push the 400-grams than the 800-grams.

Have a wonderful day!

8 0

3 years ago

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What if is a forecasting game, below are actions on your document and all you have to do is predict what will happen after the a

Talja [164]

1. Triple click on the paragraph.

By extension, the triple click can allow you to select a line and the quadruple click a section (paragraph).

2. Pressing Ctrl + A.

Ctrl+A | Ctrl+E: Select all. Depending on the programs and the translations, it will be one or the other.

3. Double click within the word.

In most text editors, a double-click on a word selects it in its entirety.

4. Pressing key combinations Shift + arrow key.

If you press one of the arrow keys with the Shift key pressed, you can select a section of text.

5. Pressing key combinations Ctrl + End.

Ctrl + End: Go to the end of the document or window.

Ctrl + Home: Go to the beginning of the document or window.

Ctrl + Insert: Copy the selected text.

Please mark me as brainliest.

<h3>➥ I hope I have helped you, greetings! </h3><h3>Atte: ღTheGirlSadღ </h3>

7 0

2 years ago

Read 2 more answers

A tugboat tows a ship at a constant velocity. The tow harness consists of a single tow cable attached to the tugboat at point A

Y_Kistochka [10]

Answer:

The tensions in $T_{BC}$ is approximately 4,934.2 lb and the tension in $T_{BD}$ is approximately 6,035.7 lb

Explanation:

The given information are;

The angle formed by the two rope segments are;

The angle, Φ, formed by rope segment BC with the line AB extended to the center (midpoint) of the ship = 26.0°

The angle, θ, formed by rope segment BD with the line AB extended to the center (midpoint) of the ship = 21.0°

Therefore, we have;

The tension in rope segment BC = $T_{BC}$

The tension in rope segment BD = $T_{BD}$

The tension in rope segment AB = $T_{AB}$ = Pulling force of tugboat = 1200 lb

By resolution of forces acting along the line A_F gives;

$T_{BC}$ × cos(26.0°) + $T_{BD}$ × cos(21.0°) = $T_{AB}$ = 1200 lb

$T_{BC}$ × cos(26.0°) + $T_{BD}$ × cos(21.0°) = 1200 lb............(1)

Similarly, we have for equilibrium, the sum of the forces acting perpendicular to tow cable = 0, therefore, we have;

$T_{BC}$ × sin(26.0°) + $T_{BD}$ × sin(21.0°) = 0...........................(2)

Which gives;

$T_{BC}$ × sin(26.0°) = - $T_{BD}$ × sin(21.0°)

$T_{BC}$ = - $T_{BD}$ × sin(21.0°)/(sin(26.0°)) ≈ - $T_{BD}$ × 0.8175

Substituting the value of, $T_{BC}$ , in equation (1), gives;

- $T_{BD}$ × 0.8175 × cos(26.0°) + $T_{BD}$ × cos(21.0°) = 1200 lb

- $T_{BD}$ × 0.7348 + $T_{BD}$ ×0.9336 = 1200 lb

$T_{BD}$ ×0.1988 = 1200 lb

$T_{BD}$ ≈ 1200 lb/0.1988 = 6,035.6938 lb

$T_{BD}$ ≈ 6,035.6938 lb

$T_{BC}$ ≈ - $T_{BD}$ × 0.8175 = 6,035.6938 × 0.8175 = -4934.1733 lb

$T_{BC}$ ≈ -4934.1733 lb

From which we have;

The tensions in $T_{BC}$ ≈ -4934.2 lb and $T_{BD}$ ≈ 6,035.7 lb.

8 0

4 years ago