Ask question

Ask question

All categories

3 years ago

11

A boat sends an ultrasound signal of frequency 30 kHz, and receives the signal after 6 s. If the speed of sound in water is 1500

m/s, then find the depth of the ocean.
PLEASE HELP

1 answer:

natka813 [3]3 years ago

3 0

Answer:

4500

Explanation:

hey there!

first we convert the khz to hz=30000hz

we use the formula 2d=distance×time

2d=1500×6

d=9000/2=4500

i think this is the right answer....

hope this helps u :)

You might be interested in

What are the uses of X-rays.

Arturiano [62]

Explanation:

id : 237 514 2470

pass : frmCV5

Join in zoom!❤

I am boy :)

3 0

3 years ago

An experiment is performed to test the effect of three different types of water on the growth of a plant. The test is done by us

natima [27]

I think the correct answer would be D. The tap water in the experiment is one the three test conditions of the independent variable, the type of water. The independent variable in a experiment is the one being manipulated or the one being changed. In this case, it is the type of water.

6 0

3 years ago

Read 2 more answers

Match the technology that uses radio waves to the field in which it is used

Elden [556K]

Answer:

Naturally occurring radio waves are made by lightning or by astronomical objects. Artificially generated radio waves are used for fixed and mobile radio communication, broadcasting, radar and other navigation systems, communications satellites, computer networks and innumerable other applications.

Explanation:

5 0

3 years ago

Topic Gravitational force amd firld strength.. help me please

I am Lyosha [343]

The gravitational force between <em>m₁</em> and <em>m₂</em> has magnitude

$F_{1,2} = \dfrac{Gm_1m_2}{x^2}$

while the gravitational force between <em>m₁</em> and <em>m₃</em> has magnitude

$F_{1,3} = \dfrac{Gm_1m_3}{(15-x)^2}$

where <em>x</em> is measured in m.

The mass <em>m₁</em> is attracted to <em>m₂</em> in one direction, and attracted to <em>m₃</em> in the opposite direction such that <em>m₁</em> in equilibrium. So by Newton's second law, we have

$F_{1,2} - F_{1,3} = 0$

Solve for <em>x</em> :

$\dfrac{Gm_1m_2}{x^2} = \dfrac{Gm_1m_3}{(15-x)^2} \\\\ \dfrac{m_2}{x^2} = \dfrac{m_3}{(15-x)^2} \\\\ \dfrac{(15-x)^2}{x^2} = \dfrac{m_3}{m_2} = \dfrac{60\,\rm kg}{40\,\rm kg} = \dfrac32 \\\\ \left(\dfrac{15-x}x\right)^2 = \dfrac32 \\\\ \left(\dfrac{15}x-1\right)^2 = \dfrac32 \\\\ \dfrac{15}x - 1 = \pm \sqrt{\dfrac32} \\\\ \dfrac{15}x = 1 \pm \sqrt{\dfrac32} \\\\ x = \dfrac{15}{1\pm\sqrt{\dfrac32}}$

The solution with the negative square root is negative, so we throw it out. The other is the one we want,

$x \approx 6.74\,\mathrm m$

5 0

3 years ago

As the train in the image moves to the right which person hears the train horn at a lower pitch?

Svetach [21]

B. C would be your answer

5 0

3 years ago

Read 2 more answers