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3 years ago

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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According to the law of refraction, light passing from air into a piece of glass at an angle of 30 degrees will cause the light

pantera1 [17]

Answer:

bend toward the normal line

Explanation:

When light passes from a less dense to a more dense substance, (for example passing from air into water), the light is refracted (or bent) towards the normal. In your question the light is moving from rarer to denser medium

4 0

3 years ago

(20 points) You are at the center of a boat and have been rowing the boat for a long time. You weigh only 80 kg and your 120 kg

valina [46]

Answer:

Explanation:

From the given information:

Let the first weight be $m_ 1$ = 80 kg

The weight of the buddy be $m_2$ = 120 kg

The weight of Bubba be $m_3$ = 60 kg

Also, since you and Budda are a distance of 4m to each other, then the length to which both meet buddy will be:

$x_1 = x_3 = \dfrac{4}{2} \\ \\ = 2$

The length of the boat be $x_2$ = 4 m

∴

We can find the center of mass of the system by using the formula:

$X_{CM} = \dfrac{m_1x_1+m_2x_2+m_3x_3}{m_1+m_2+m_3} \\ \\ X_{CM} = \dfrac{(80 \times 2)+(120\times4)+(60\times2)}{80+120+60} \\ \\ X_{CM} = \dfrac{160+480+120}{260} \\ \\ \mathbf{X_{CM} = 2.923}$

4 0

2 years ago

A radioactive material has a count rate of 400 per minute. It has a half life of 40 years. How long will it take to decay to a r

cestrela7 [59]

Answer:

160 years.

Explanation:

From the question given above, the following data were obtained:

Initial count rate (Cᵢ) = 400 count/min

Half-life (t½) = 40 years

Final count rate (Cբ) = 25 count/min

Time (t) =?

Next, we shall determine the number of half-lives that has elapse. This can be obtained as follow:

Initial count rate (Cᵢ) = 400 count/min

Final count rate (Cբ) = 25 count/min

Number of half-lives (n) =?

Cբ = 1/2ⁿ × Cᵢ

25 = 1/2ⁿ × 400

Cross multiply

25 × 2ⁿ = 400

Divide both side by 25

2ⁿ = 400/25

2ⁿ = 16

Express 16 in index form with 2 as the base

2ⁿ = 2⁴

n = 4

Thus, 4 half-lives has elapsed.

Finally, we shall determine the time taken for the radioactive material to decay to the rate of 25 counts per minute. This can be obtained as follow:

Half-life (t½) = 40 years

Number of half-lives (n) = 4

Time (t) =?

n = t / t½

4 = t / 40

Cross multiply

t = 4 × 40

t = 160 years.

Thus, it will take 160 years for the radioactive material to decay to the rate of 25 counts per minute.

7 0

2 years ago

What is the disadvantage of the parallax method, especially for studying distant parts of the Galaxy?

Katena32 [7]

Answer and Explanation:

Parallax method is used for finding the distance of objects in space there are two types of parallax method that is stellar parallax and trigonometric parallax.The disadvantage of using parallax method is that it can can not reach so far in the Galaxy due to this reason parallax method is generally not used for measuring distance in galaxy.

6 0

3 years ago

A thermometer is removed from a room where the temperature is 70° F and is taken outside, where the air temperature is 10° F. Af

vekshin1

Answer:

$T=51.64^\circ F$

$t=180.10s$

Explanation:

The Newton's law in this case is:

$T(t)=T_m+Ce^{kt}$

Here, $T_m$ is the air temperture, C and k are constants.

We have

$70^\circ F$ in $t=0$

So:

$T(0)=70^\circ F\\T(0)=10^\circ F+Ce^{k(0)}\\70^\circ F=10^\circ F+C\\C=70^\circ F-10^\circ F=60^\circ F$

And we have $60^\circ F$ in $t=30 s$ , So:

$T(30)=60^\circ F\\T(30)=10^\circ F+(60^\circ F)e^{k(30)}\\60^\circ F=10^\circ F+(60^\circ F)e^{k(30)}\\50^\circ F=(60^\circ F)e^{k(30)}\\e^{k(30)}=\frac{50^\circ F}{60^\circ F}\\(30)k=ln(\frac{50}{60})\\k=\frac{ln(\frac{50}{60})}{30}=-0.0061$

Now, we have:

$T=10^\circ F+(60^\circ F)e^{-0.0061t}(1)$

Applying (1) for $t=1 min=60s$ :

$T=10^\circ F+(60^\circ F)e^{-0.0061*60}\\T=10^\circ F+(60^\circ F)0.694\\T=10^\circ F+41.64^\circ F\\T=51.64^\circ F$

Applying (1) for $T=30^\circ F$ :

$30^\circ F=10^\circ F+(60^\circ F)e^{-0.0061t}\\30^\circ F-10^\circ F=(60^\circ F)e^{-0.0061t}\\-0.0061t=ln(\frac{20}{60})\\t=\frac{ln(\frac{20}{60})}{-0.0061}=180.10s$

8 0

3 years ago