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USPshnik [31]
3 years ago
8

A sonar echo returns to a submarine 3.20 s after being emitted. What is the distance to the object creating the echo? (Assume th

at the submarine is in the ocean, not in fresh water.)
Physics
2 answers:
EastWind [94]3 years ago
8 0

Well that's a problem.

The answer depends on the speed of the sonar signal through the water, but the speed of sound in seawater is not a constant value. It varies by a few percent from place to place, from season to season, from morning to evening, and also with the depth of the water.

I'll use the round figure of 1,500 m/s just to show how to handle the problem.

Traveling at an average speed of 1,500 m/s for 3.2 seconds, the ping of the sonar covers

(1,500 m/s) x (3.2 sec) = 4,800 meters

But that's <u>not</u> the distance to the object that reflects the ping back.

Don't forget that the sound had to cover the distance <u>twice</u> . . . once from the sub to the target, and then RETURN to the sub.  So the actual distance from the sub to the target is half of that.

<em>Distance</em> = (4,800 meters / 2) = <em>2,400 meters</em> .

Feliz [49]3 years ago
6 0

Answer: 2,464m

Explanation:

Echo made by a source is dependent on the velocity of the sound in air/water, the distance between the source and the reflector and the time taken

Using the echo formula

2x = vt

Where x is the distance between the source and the reflector

v is the velocity of the sound in sea water is approximately= 1540m/s (This varies from place to place and depending on the season and nature of water)

t is the time taken

t = 3.20s

Substituting this datas into the equation to get 'x'

2x = 1540 × 3.20

x = 1540×3.20/2

x = 2,464m

The distance between the object creating the echo is 2,464m

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5 0
4 years ago
A rectangular coil of dimensions 5.40cm x 8.50cm consists of25 turns of wire. The coil carries a current of 15.0 mA.
Kazeer [188]

Answer:

(a) Magnetic moment will be 17.212\times 10^{-4}A-m^2

(b) Torque will be 6.024\times 10^{-4}N-m

Explanation:

We have given dimension of the rectangular 5.4 cm × 8.5 cm

So area of the rectangular coil A=5.4\times 8.5=45.9cm^2=45.9\times 10^{-4}m^2

Current is given as i=15mA=15\times 10^{-3}A

Number of turns N = 25

(A) We know that magnetic moment is given by magnetic\ moment=NiA=25\times 45.9\times 10^{-4}\times 15\times 10^{-3}=17.212\times 10^{-4}A-m^2

(b) Magnetic field is given as B = 0.350 T

We know that torque is given by \tau =BINA=0.350\times 15\times 10^{-3}\times 25\times 45.9\times 10^{-4}=6.024\times 10^{-4}N-m

4 0
3 years ago
A child is sliding down a slide at the playgound. is mechanicalenergy conserved
Flauer [41]

No. Mechanical energy is not conserved.  There's quite a bit of friction on the slide.  So some of the potential energy is lost to heat on the way down, and the child arrives at the bottom with hot pants and less kinetic energy than you might expect.

5 0
3 years ago
Biologists use optical tweezers to manipulate micron-sized objects using a beam of light. In this technique, a laser beam is foc
vekshin1

Answer:

Explanation:

Part A) Using

light intensity I= P/A

A= Area= π (Radius)^2= π((0.67*10^-6m)/(2))^2= 1.12*10^-13 m^2

Radius= Diameter/2

P= power= 10*10^-3=0.01 W

light intensity I= 0.01/(1.12*10^-13)= 9*10^10 W/m^2

Part B)  Using

I=c*ε*E^2/2

rearrange to solve for E= \sqrt{((I*2)/(c*ε))

c is the speed of light which is 3*10^8 m/s^2

ε=permittivity of free space or dielectric constant= 8.85* 10^-12 F⋅m−1

I= the already solved light intensity= 8.85*10^10 W/m^2

amplitude of the electric field E= \sqrt{(9*10^10 W/m^2)*(2) / (3*10^8 m/s^2)*(8.85* 10^-12 F⋅m−1)

---> E= \sqrt{(1.8*10^11) / (2.66*10^-3) = \sqrt{(6.8*10^13) = 8.25*10^6 V/m    

 

8 0
3 years ago
Radio signals travel at a rate of 3 × 108 meters per second. how many seconds would it take for a radio signal to travel from a
nlexa [21]
<span>3.2x10^-2 seconds (0.032 seconds)
   This is a simple matter of division. I also suspect it's an exercise in scientific notation, so here is how you divide in scientific notation:

   9.6 x 10^6 m / 3x10^8 m/s

   First, divide the significands like you would normally.
 9.6 / 3 = 3.2

   And subtract the exponent. So
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   So the answer is 3.2 x 10^-2
 And since the significand is less than 10 and at least 1, we don't need to normalize it.

   So it takes 3.2x10^-2 seconds for the radio signal to reach the satellite.</span>
6 0
4 years ago
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