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zloy xaker [14]
3 years ago
8

How to calculate depth of sea whan a ship receives sound produced by its fathometer after 3.5 second

Physics
2 answers:
HACTEHA [7]3 years ago
5 0

Answer:

2600 m

Explanation:

A fathometer produces a sound wave and then detects the echo.  It takes 3.5 seconds for the echo to reach the ship, so that means it takes half the time (1.75 seconds) to reach the ocean floor.

The speed of sound in seawater is approximately 1500 m/s, so the depth of the ocean at that point is:

d = 1500 m/s × 1.75 s

d = 2625 m

Rounding to two significant figures, the depth is approximately 2600 m.

EleoNora [17]3 years ago
3 0

Answer: the depth is around d = 1.75s*1450m/s = 2537.5 m

Explanation:  A fathometer creates a sound pulse and receives the echo of that sound, and in this way, the instrument can obtain a difference of time (the time between the wave is released and received)

Now, knowing that the velocity of the sound in water is around 1450 m/s

now, the lecture is 3.5 seconds, but in this time the wave goes to the bottom and bounces, so the time needed to reach the bottom is half that:

3.5s/2 = 1.75s

then the distance traveled by the wave in that lapse of time is:

d = 1.75s*1450m/s = 2537.5 m

this is a simplification of the math that the device does.

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The international space station travels at a distance of about 250 miles above Earth’s surface and at a speed of 17,500 miles pe
Arisa [49]

Answer:

In this case we are dealing with the pythagorean theorm involving right angled triangles. This theorm states that a^2 + b^2 = c^2 which means the square of the hypotenuse (side c, opposite the right angle) is equal to the square of the remaining two sides.

In this case we will say that a = 3963 miles which is the radius of the earth. c is equal to the radius of the earth plus the additional altitude of the space station which is 250 miles; therefore, c = 4213 miles. We must now solve for the value b which is equal to how far an astronaut can see to the horizon.

(3963)^2 + b^2 = (4213)^2

b^2 = 2,044,000

b = 1430 miles.

The astronaut can see 1430 miles to the horizon.

Explanation:

:D hopes this Helps

3 0
3 years ago
What is the magnification when an object is placed at 2f from the pole of the convex mirror? 
Gekata [30.6K]

Answer:

Linear magnification = 1/3

Explanation:

Given:

Convex mirror

Object's distance from pole = 2f

Find:

Linear magnification

Computation:

Object distance, u = −2f

So,

1/v + 1/u = 1/f

1/v + 1/(-2f) = 1/f

1/v = 1/f + 1/2f

BY taking LCM

1/v = 3 / 2f

v = 2f / 3

Magnification, M = -v / u

So,

Magnification, M = (2f / 3) / 2f

Magnification, M = 2f / 6f

Magnification, M = 2 / 6

Linear magnification = 1/3

 

5 0
3 years ago
Spurs"" are probably the result of ____.
egoroff_w [7]

Spurs are probably the result of <u>self-sustaining</u> <u>star formation.</u>

<h3>What is the formation of gaseous spurs in spiral galaxies?</h3>

The gigantic form of the magnificent doppelganger spiral patterns that spiral outward from the galactic cores gave spiral galaxies their name. These light arms of spiral galaxies are frequently seen in optical pictures to be speckled with bright star-forming areas at regular intervals.

Smaller structures spread forth and rearward into the interarm area from each major spiral arm. Spiral-arm also known as spurs are the name given to these substructures. Sometimes the spurs are also filled with star-forming clusters. As a consequence, we may draw the conclusion that spurs most likely emerge from self-sustaining star formation.

Learn more about the spiral galaxies here:

brainly.com/question/13956361

#SPJ11

5 0
2 years ago
How do I go about this?
Anna71 [15]

Hi there!

(a)

Recall that:
W = F \cdot d = Fdcos\theta

W = Work (J)
F = Force (N)
d = Displacement (m)

Since this is a dot product, we only use the component of force that is IN the direction of the displacement. We can use the horizontal component of the given force to solve for the work.

W =248(56)cos(30) = 12027.36 J

To the nearest multiple of ten:
W_A = \boxed{12030 J}

(b)
The object is not being displaced vertically. Since the displacement (horizontal) is perpendicular to the force of gravity (vertical), cos(90°) = 0, and there is NO work done by gravity.

Thus:
\boxed{W_g = 0 J}

(c)
Similarly, the normal force is perpendicular to the displacement, so:
\boxed{W_N = 0 J}

(d)

Recall that the force of kinetic friction is given by:
F_{f} =\mu_k mg

Since the force of friction resists the applied force (assigned the positive direction), the work due to friction is NEGATIVE because energy is being LOST. Thus:
W_f = -\mu_k mgd\\W_f = - (0.1)(56)(9.8)(56) = -3073.28 J

In multiples of ten:
\boxed{W_f = -3070 J}

(e)
Simply add up the above values of work to find the net work.

W_{net} = W_A + W_f \\\\W_{net} = 12027.36 + (-3073.28) = 8954.08 J

Nearest multiple of ten:
\boxed{W_{net} = 8950 J}}

(f)
Similarly, we can use a summation of forces in the HORIZONTAL direction. (cosine of the applied force)
F_{net} = F_{Ax} - F_f

W = F_{net} \cdot d = (F_{Ax} - F_f)

W = (F_Acos(30) - \mu_k mg)d\\W = (248cos(30) - 0.1(56)(9.8)) * 56 \\\\W = 8954.08 J

Nearest multiple of ten:
\boxed{W_{net} = 8950 J}

5 0
2 years ago
Red clothes will---the red light.A)Reflect B)refract C)Absorb D)transmission E)Dispersion
posledela

Red clothes look red  because they REFLECT the red light, and absorb light of other colors.

7 0
3 years ago
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