The post Fahrenheit to Celsius conversion solved examples – Physics numerical appeared first on Study Material Online.

]]>## Fahrenheit to Celsius Solved Numericals

We use the basic conversion formula for Celsius to Fahrenheit and vice versa.The solved examples listed below will clear your basic doubts and practicing these problems will give you a good command on temperature conversion based solved numerical . When you are confident enough to solve temperature numerical , it would be good practice to try some unsolved numerical online on ONLINE CONVERSION QUIZ . The numerical listed here deal with two main temperature scales: One is **Fahrenheit Scale** ( **°F)** and is used in the US , and the other is the **Celsius Scale ( °C )** which is a part of the Metric System and used in most other countries .

- Water boils at
**100° in Celsius, but 212° in Fahrenheit** - Water freezes at
**0° in Celsius, but 32° in Fahrenheit**

Conversion Rules

**Celsius to Fahrenheit Conversion **

First multiply by 180/100 ( to make it simple , multiply by 9/5 ) and then add 32

**Fahrenheit to Celsius Conversion **

First subtract 32 from the given value and then multiply by 100/180 ( to make it simple , multiply by 9/5 )

Celsius to Fahrenheit Conversion Formula : (°C × ^{9}/_{5}) + 32 = °F

Fahrenheit to Celsius Conversion Formula : (°F − 32) x^{5}/_{9} = °C

Fahrenheit to Celsius Conversion Formula : (°F − 32) x

## Temperature conversion solved numerical # 1

Question : Convert 26° Celsius ( usually room temperature ) to Fahrenheit

Solution : Given = 26°C

Use the formula (°C × ^{9}/_{5}) + 32 = °F

Value of 26° × 9/5 = 234/5 = 46.8

Then add 32 , it would be 46.8 + 32 = **78.8° F**

Fahrenheit to Celsius Conversion Problem # 2

Question : Convert 98.6° Fahrenheit *(often given as convert normal body temperature ) *to Celsius

Solution : Given = 98.6° F

Use the formula = (°F − 32) x ^{5}/_{9} = °C

Value of (°F − 32) = 98.6° − 32 = 66.6

Then 66.6 × 5/9 = 333/9 =** 37° C**

Hence value of normal body temperature in degree Celsius is **37° C**

**Note : Sometimes we use 1.8 instead of 9/5 in the formula as value of 9/5 is 1.8 **

Hence the formulae can be

Celsius to Fahrenheit: °C × 1.8 + 32 = °F

Fahrenheit to Celsius: (°F − 32) / 1.8 = °C

Fahrenheit to Celsius: (°F − 32) / 1.8 = °C

## Fahrenheit Celsius solved numerical # 3

Question : What is the value of 20° Celsius into Fahrenheit

Solution : Given = 20° C

Use the formula = °C × 1.8 + 32 = °F

Value of first part is = 20° C × 1.8 = 36

Then = 36 + 32 = 68 °F

Value of 20° Celsius into Fahrenheit is 68 °F

Celsius to Fahrenheit Conversion # 4

Question : What is the value of 10° Celsius to Fahrenheit .

Solution : Given = 10° C

Use the formula = °C × 1.8 + 32 = °F

°F = 10° C x 1.8 + 32 = 18 +32 = **50° F**

Value in degree Fahrenheit is **50° F**

## Temperature Conversion Solved Examples # 5

Question : The average body temperature of a cat is 38.6 °C in Celsius. Convert this value in Fahrenheit .

Solution : Given = 38.6 °C

Use the formula = °C × 1.8 + 32 = °F => 38.6 °C × 1.8 + 32 = 101.48 °F

Therefore , value in degree Fahrenheit is 101.48 °F

So these are some of the Fahrenheit to Celsius conversion and vice versa solved examples to assist you in your mathematics and physics assignments . If you want us to solve some specific conversion numerical , feel free to comment in the comment box and we will surely post the solution for your questions . One can even share these with their friends who are having hard time to solve these temperature conversion problems. For numerical based on other topics, stay tuned and get the updates .

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]]>The post Solved Examples on Celsius Kelvin Temperature Conversions- Physics Assignments appeared first on Study Material Online.

]]>Most of the calculations in numerical require the temperature to be in the Kelvin temperature scale and Often the temperature in data is given in Celsius. Here is one simple formula which is used for the Celsius Kelvin conversion .

Formula:

K = ºC + 273.15

where

K is the temperature in Kelvin

°C is the temperature in Celsius

Note : Fahrenheit and Celsius temperature scale use the degree (°) symbol but Kelvin temperature scale does not use degree (°).

Celsius Kelvin Problem

The boiling temperature of liquid oxygen at normal pressure at -182.96 °C. What is the value in Kelvin Scale ?

**Solution # **

K = ºC + 273.15

ºC= -182.96

Put the value of ºC in the formula : K = -182.96 + 273.15

Value in Kelvin Scale : K = 90.19 K

The temperature at which liquid oxygen boils under normal pressure is 90.19 K .

Temperature Conversion Problem

While we are solving ideal gas problems , in order to make the calculations easier we take its value as 300 K . What is the value of this temperature in Celsius?

**Solution #**

Use the Formula K = ºC + 273.15

Put the value of K as 300 K in the above formula .

300 = °C + 273.15

°C = 300 – 273.15

Value of °C = 26.85 °C

Value of this temperature in °C is 26.85 ° .

Celsius to Kelvin Conversion

Convert value 25.0 °C to Kelvin temperature scale.

**Solution #**

Use the Formula K = ºC + 273.15

Put the value of ºC as 25.0

25.0 + 273.15 = 298.15

Value of 25ºC in Kelvin Scale is 298.15

Note : Sometimes to make the calculations easy 273.15 is treated as 273.0

Temperature Conversion Numerical

State the value of 40.0 °C in Kelvin scale

**Solution #**

Use the Formula K = ºC + 273.15

Put the value of ºC as 40.0 °C

K = 40.0 + 273.15 = 277.15

Celsius Kelvin Problem

Convert -20.0 °C into Kelvin scale temperature .

**Solution #**

Use the Formula K = ºC + 273.15

Put the value of ºC as -20.0 in the above equation .

K = -20.0 + 273.15 = 217.15

Value of -20.0 °C in Kelvin is 217.15

Kelvin to Celsius Temperature

Convert 298 K into degrees Celsius .

**Solution #**

Use the Formula K = ºC + 273.15

Value of K is 298

298 = ºC + 273.15 => ºC = 298- 273.15 = 24.85

Value of K in degree Celsius is 24.85 ºC

Kelvin Celsius Solved Problems

Convert -145.0 °C to K

**Solution #**

Use the Formula K = ºC + 273.15

Put value of ºC as -145.0 °C

K= -145.0 °C + 273.15

Solution is 128.15

What is the value of 252 K in degree Celsuis .

**Solution #**

Use the Formula K = ºC + 273.15

252 = °C + 273.15

°C = 252 – 273.15 = – 21.15

Value in degree Celsius is – 21.15

These are some of the solved numerical related temperature scale measurements.It covers all the basic questions that can be asked in the exam based on temperature conversions . Moreover it will be very helpful in physics assignments related to Celsius kelvin temperature conversions . If you have any doubt or questions related to these conversion problems , do comment and we will revert as soon as possible . For more problems and their solutions , stay connected .

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]]>Today, we make use of three temperature scales for measurements .These are : Fahrenheit, Celsius and Kelvin. Fahrenheit temperature scale was discovered by German physicist Daniel Gabriel Fahrenheit and in our next post we will deal with some of the basic facts and conversions between Fahrenheit and other temperature scales .In this post we will deal with other two scales i.e Kelvin and Celsius .

Image Source : EquipmentExplained.com

Celsius temperature / centigrade temperature scale takes 0 degree as the freezing point of water and 100 degree celsius as the boiling point of water for the reference . It was invented by Anders Celsius and this scale is used in scientific work everywhere. Then comes SI unit of temperature : Kelvin temperature which is defined as 1/ 273.16 of the triple point ( triple point is basically an equilibrium among the solid, liquid, and gaseous phases) of pure water. It has symbol K and does not have any degree sign unlike other two .But all these temperature scales are related to each other and can be easily converted from one temperature scale to another by use of simple formulas . For pictorial representation on how they are related to each other , one can visit **abyss.uoregon.edu**

While dealing with the problem if you are given more the two types of temperature scale then you must change all the scales into S.I unit that is Kelvin Scale . One thing more ,sometimes in order to confuse the teenage kids , your teacher might put a temperature in the problem which you really need not to use . You should be confident enough that you do not need this temperature value.

Here is the formula for temperature conversion between these two temperature scales :

**Kelvin = Celsius + 273.15. **

**K= Degree (C ) + 273.15. **

For ease , number of times 273 is used instead of 273.15. Also always says for example 270 Kelvins instead of 270 degrees Kelvin.

**Celsius Kelvin Solved Example #1:**

Convert 25.0 °C to Kelvin.

**Solution** :

*Kelvin = Celsius + 273*

K = 25.0 °C + 273 = 298.0 K

**Celsius Kelvin Tricky Questions # 2 **

Temperatures on one of the planet is measured between the range of 85-90 Kelvin . Define the temperature limits in , degrees Celsius?

**Solution :**

*Kelvin = Celsius + 273.15*

Case 1 : When lower limit is 85 K

K = °C + 273.15

By putting the values in the formula

85= °C + 273.15 => °C = 85 – 273.15

Value of lower limit in °C is -188.15

Case 2 : When upper limit is 90 K

90 = °C + 273.15

°C= 90 – 273.15 => °C = -183.15

The temperature on one of the planet has temperature range of (-183.15 to -188.15) °C

**Celsius to Kelvin Problem #3**

Do the conversion of 27° C into Kelvin.

**Solution**

K = °C + 273

Put C= 27 and you will get 27 + 273 = 300

Therefore answer is 300 K

**Celsius and Kelvin Temperature Numerical # 4**

Convert 375 K to degrees Celsius.

**Solution **

*Kelvin = Celsius + 273.15.*

375 K = °C *+ 273 *

Therefore

Celsius : 375 − 273 = 102 °C

**Kelvin- Celsius Example #5**

Convert value of −50 °C to Kelvin. (this is a negative value )

**Solution**

According to the formula

*Kelvin = Celsius + 273*

Put value of C = -50

Kelvin = (−50) + 273 =

Therefore value in Kelvin = 273- 50 = 223 K

There were some of the solved problems based on temperature conversions especially Celsius and Kelvin Temperature conversions . For more updates and study material , stay connected with us .If you want any assistance on a specific topic or if you want us to solve any tricky questions , do comment and we will try to answer your queries as soon as possible .

## More : KCL Solved Examples

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]]>The post Solved Numericals based on SONAR Formula | Physics solved assignments appeared first on Study Material Online.

]]>Practicing these couple of examples with help you to perform better during exams and boot your confidence .These SONAR solved examples cover some of the tricky question also , these questions seem tricky but we have explained them in quite an easy way and you will get them in one run .Let us start with the basic SONAR formula which we are going to use for all the solved problems .

SONAR :

**Formula used : V= 2D/T where**

**V= Speed of sound**

**2D= Total distance traveled**

**T= Time taken**

An ocean vessel is using SONAR for detecting ocean’s bottom. The speed of sound waves that travel through the water is 1450 m/s and the bottom of ocean is 1630 m from the surface .Calculate the time taken by waves to get reflected back to the surface.

**Ans :**

Given

*2D= Total distance traveled = 1630 * 2 m ( Two way distance ) *

*V= Speed of sound = 1450 m/s*

* V= 2D/T therefore T= 2D / V *

T= (1630) (2)/ 1450 = 2.25 s

A ship sends a sound wave to the bottom of the sea and receives an echo after 0.3 sec .Assume speed of sound in water to be 1500 m /s . Calculate the depth of the sea .

**Ans :**

2D= Total distance traveled = to be calculated

Therefore depth of sea will be D ( half of the total )

V= Speed of sound in water = 1500 m /s

T= 0.3

V= 2D/T therefore 2D= T * V = 1500 * 0.3 = 450 m

depth of sea will be D = 225 m

There is a sonar device on a submarine and it sends out a signal .The echo is received exactly after 5 s . If the distance between the submarine and object is 3,625 m. Determine speed of sound waves in water .

**Ans. **

Given,

2D= Total distance traveled = 3,625 m* 2 m ( Two way distance )

T= 5 s

V= Speed of sound in water =

V= 2D/T = (2×3625)/5 = 1450 ms–1

A guy dropped a stone from the tower which is 500 m in height .The stone falls into a pond of water at the base of the tower. After how much time the splash will be heard at the top ? The value of g is 10 ms–2 and speed of sound is 340 ms–1.

**Ans. **

Given

First we will solve the amount of time taken by stone to reach the tower base .

Distance = s = 500 m, u = 0 ms–1, g = 10 ms–2

Now use the formula s = u t + 1/2gt2

By solving , we get value of t = 10 sec

It shows that te amount of time taken by stone to reach the pond is 10 s

Second step = Use SONAR formula to calculate time taken to cover 500 m .

Give : Speed of sound = 340 ms–1

Distance = 500 m

Time taken sound to D = 500/340 = 1.5sec.

Total time taken = (10 + 1.5) s = 11.5 s

:. The splash will be heard at the top after 11.5 s

In this case we will calculate velocity instead of time .A stone is dropped into a well 44.1 m deep, and splash sound is heard after 3.13 s. (Given, g = 9.8 ms–2)

**Ans: **

** **u = 0, distance = s = 44.1 m;

g = 9.8 ms–2

Time taken by stone to reach 44.1m deep will be

Put the values in equation = s = u t + 1/2gt2

By solving , we will get t = 3 sec

The sound takes the remaining time i.e., 3.13 – 3 = 0.13 s to go to up of well through 44.1 m

Speed=d/t= 44.1 m/0.13 = = 339.23 ms–1

A ship sends out ultrasound that returns exactly after 3.42 s. Consider the speed of ultrasound in water to be 1531 m/s, calculate the distance between ship and the seabed .

**Ans. **

Given

2D= Total distance traveled = to be calculated

Therefore depth of sea will be D ( half of the total )

V= Speed of sound in water = 1531 m /s

T= 3.42 sec

V= 2D/T therefore 2D= T * V = 1531 m/s × 3.42 s = 5236 m

D= 5236 m/2 = 2618 m or 2.62 km

A submarine enters a sonar pulse, and it returns from an under water cliff in 1.02 sec. Assume speed of sound in water is 1531 m/s, determine its distance from the cliff .

**Ans : **

2D= Total distance traveled = to be calculated

Therefore depth of sea will be D ( half of the total )

V= Speed of sound in water = 1531 m /s

T= 1.02 sec

V= 2D/T therefore 2D= T * V = 2x1531x1.02 = 1561.62 m

D= 5236 m/2 = 2618 m or 2.62 km

Distance of the cliff = d/2= 1561.62/2 = 780.81 m

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]]>The post KCL Solved Examples and solution | Electric current 12th Physics Concepts appeared first on Study Material Online.

]]>Now here are some solved problems on KCL and examples on properties of current source and we will also discuss about current division method for calculating current in the circuit.

Ques:1Using KCL law find the unknown current in the Figure shown below:Solution: Given that

to find

Apply KCL At Node 3

for i4

for i7

for i9

Ques:2 Find the current ‘‘ through the resister ohm.

Solution: Let us consider the voltage ‘V1’at node A.

Apply KCL at node A ————(1)

Calculate the voltage value at Node

—————————(2)

therefore Ans.

Ques:3 Find the current I through Resistor R When 5 Ampere current source is in:

- same direction to the 10 Ampere current source
- opposite direction to the 10 Ampere current source.
Solution:

In case a: using KCL

In case b: using KCL

Now we are going to discuss current division method for calculating branch current in parallel circuit (divider circuit). In series circuit it is not applicable (voltage division method is applicable). When current flows through more than one impedance path i.e in divider circuit,then each branch shares some portion of total current depending on the impedance on that branch, the formula is given below:

Current to be calculating= Impedance through which current is not calculate

You can make it clear by solving a question on this method.

Ques:4 Find the current through the resistor in the following circuit.

Solution: First calculate the total impedance of the circuit then calculate the current delivered by the voltage source

Let

; ; ;

Now we will calculate Current delivered by the Voltage source using Ohm’s law:

For calculate the current through Resistor,we have to first calculate the branch current.Let Current through R2 be I1 and through R5 be I2 branches can be calculate by using Current division Rule :

Now calculate the current through resistor

Hope you understand the concept through these **KCL solved examples**. If you find any problem or any queries or if you also want to discuss some other problems you have please comment below ,we will surely entertain you. In case, any specific topics needs to be discussed on our website, we welcome & give our audience the first priority. Hit the like button and share the post with your friends who need help with this topic.

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]]>The post What is Electric Current and what is current Effect | Physics concept appeared first on Study Material Online.

]]>**Introduction**: Electric current is all about electric charge i.e rate of change of electric charge. Charged particles can be electrons, protons, ions or holes. Movement of these particles effectively produce current in a closed loop. For example a metallic cylindrical wire having length** ‘l’, **cross-sectional area ‘**a’**** ,**let the ‘**n**‘ number of charged particles are electron having charge on it is denoted as **‘e’**. Place a battery across this metallic wire. Electron will drift with the drift velocity ‘**v’**. Due to the movement of electrons from one place to another per unit of time current will produce in the opposite direction of flow of electrons.

Current measures in Ampere i.e Amp which is equal to flow of one coulomb per second. According to Ohm’s law current I is flowing through the conductor with voltage V and resistance R, V=IR or I=V/R. You can find solved numerical on Ohm’s Law.

German physicist, Gustav Kirchhoff developed two laws based on conservation of energy and charge which are very helpful to calculate the voltage and current values in the complex circuits like T-network or bridge circuits which are difficult to calculate by using ohm’s law.

**Kirchhoff’s current law:** KCL is first law of Kirchhoff which is based on the law conservation of charge. It says algebraic sum of entering current and leaving current at node is equal to Zero. This is all dependent on the **direction of current** flowing in the circuit.

An ideal current source has infinite output impedance.It behave like an open circuit. It is because current follows least resistance path. Ideal current source provide the constant current supply with 100% efficiency. Practical current source does not provide constant current supply therefore efficiency of practical current source can be 98% or 99% but can not be 100%.

The V-I (voltage -current) characteristics can be describe by the following relation for the practical current source and ideal current source.

Basically there is four types of current sources :

- DC Current source
- Time varying non-sinusoidal current source
- Sinusoidal current source
- Controlled and dependent current source.

- Two Resistance connected in series have same current.
- Two resistance connected in parallel have different value of current.
- When a voltage source connected in series with current source then value of voltage source can be ignored.
- If two current source connected in parallel than the total current value is sum or difference of the both current values, depending on their directions(for example:If current source I1 and I2 have same direction then I=I1 + I2 and if I1 and I2 are in opposite direction than I = I1 – I2).
- If two current source have different values connected in series than this not obey KCL and this is invalid situation.

Hope you find some good **Electric Current Explanation** here. For more **Electricity** Related Topics, stay in touch with us. If you have any queries related to any such topics, do comment and we will surely entertain you. In case, any specific topics needs to be discussed on our website, we welcome & give our audience the first priority. Hit the like button and share the post with your friends who need help with this topic.

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]]>The post Ohm’s Law V=IR Solved Examples and Solution | Electric Current appeared first on Study Material Online.

]]>Ohm’s Law is one of the fundamental formula which is widely used in Electrical Circuits. Ohm’s Law defines that the Electric Current is directly dependent or proportional to voltage or potential difference across the circuit or the conductor. V=IR Law was first introduced while experimenting the behaviors of electricity conducted by the metals. It was developed by the German Physicist in 1827 (Georg Ohm). Here we will discuss about some of the popular **V=IR Solved Examples.**

Ohm’s Law basically defined the relationship between three parameters i.e resistance, voltage and current.

- where V is the Voltage or potential difference in volts (V)
- I is the current applied in amperes (I)
- R is the resistance in Ω (Omega)

So, solver can rearrange the parameters as per her or her own need. You can get more familiar to ohm’s law by solving the **V=IR Solved Examples****.**

For example if current needs to be calculated than:

if Resistance(R) needs to be calculated than:

Below you can find some types of **V=IR Solved Examples**** and solution**, which you can practice and have the idea.

Ques 1. In a circuit, Voltage across the circuit is 10V. Resistance applied to the circuit is 5 Ohms. What is the Current flowing through the circuit?

Ans. As we know V=IR; So rearranging to calculate Current (I) we will have:

So, 2 Ampere current is flowing in the circuit.

Ques 2. Voltage applied across the circuit is 220 Volts while the current flowing is 5 A. Find the Resistance applied in the circuit.

Ans. Applying V=IR, we get

Ques 3. A circuit with 11 KΩ resistance has 750 μA current flowing inside it. What is the voltage drop across the resistance applied?

Ans. Ohm’s Law states that V=IR so, V = 750 μA * 11 KΩ = 8.25 V.

Ques 4. While measuring current in a particular circuit a voltage drop was noticed. Initially the current was 40 mA operated at 18 V. Later the current was dropped to 20 mA. Calculate the voltage dropped.

Ans. Applying Ohm’s Law V=IR;

Now using this Resistance Value to calculate the Voltage when current was 20 mA.

Voltage Dropped = V1 – V2 = 18 V – 9 V = 9 Volts.

Hope you find some good **V=IR Solved Examples** here. For more **Ohm’s Law Solved Example** and **Ohm’s Law principle**, stay in touch with us. If you have any queries related to any such topics, do comment and we will surely entertain you. In case, any specific topics needs to be discussed on our website, we welcome & give our audience the first priority. Hit the like button and share the post with your friends who need help with this topic.

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]]>The post What is Heisenberg Uncertainty Principle Formula | Quantum Physics appeared first on Study Material Online.

]]>## About Heisenberg Uncertainty Principle definition

**Quantum Physics** explains a lot of concept and practicality around us. This topic of Uncertainty principle is also one of the important topics from the book of quantum mechanics. Uncertainty Principle in quantum mechanics is also referred to as Heisenberg Uncertainty Principle or it is also known as Heisenberg Indeterminacy Principle. It was expressed by one of the German Physicist Werner Heisenberg in the year of 1927. **Heisenberg Uncertainty Principle** explains that for any object moving with velocity ‘v’ and at position ‘x’, both the values that is velocity and position cannot be gathered at the same time. This concept of not getting the exact position and the exact velocity has no reference in the nature.

Indeterminacy in position * Indeterminacy in momentum h/2π

OR

where is Plank’s Constant i.e.

In the real world and visible to eyes this uncertainty principle has no reference.

- Let us take a car moving on a road, we can measure or note both the velocity and the position of the vehicle. So the uncertainty produced with this principle is very small as compared to such large objects.
- This rule mostly refers to the extremely small particles of small mass subatomic particles.
- As per this principle, the product of position and velocity is either greater or equal to the constant (h/2Π) where “h” known as Plank’s Constant with the value h= 6.6 × 10−34 joule-second.
- This product is significant for the atoms with small masses and subatomic particles.

If any experiment is made to check the velocity of particle very precisely, it will knock out the electron in some unpredictable nature and hence it has no significant position at that time.

Every object particle is associated with the wave like properties and behaviors. Hence the particle is mostly located at the intense crust or trough of the wave. A localized wave particle having uncertain velocity and a defined position will have indeterminate wavelength while the wave particle with precise velocity and is not having defined position or is spread out will have well defined wavelength.

Here was the detailed information regarding the Heisenberg Uncertainty Principle. For more information regarding **Heisenberg Uncertainty Formula** and **Heisenberg Uncertainty Equation **Stay in touch . If you have any queries related to any such topics, do comment and we will surely entertain you. In case, any specific topics needs to be discussed on our website, we welcome & give our audience the first priority. Hit the like button and share the post with your friends who need help with this topic.

Also check our latest article on What are signals and different classifications of Signals in Communication

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]]>The post SONAR solved numerical for exams | Physics topics appeared first on Study Material Online.

]]>Here is the list of solved problems (**sonar solved numerical**)based on working of SONAR and how we use to calculate various parameters related to this concept. As we know that in this technology, we make use of sound sound concept for various functions such as communication, navigation and detection of any objects that are underlying below the surface of water. Let us have a look at some of the basic SONAR problems and how these can be solved by using time distance formula.

**Quick Facts :**

**Formula used : V= 2D/T where**

**V= Speed of sound**

**2D= Total distance traveled**

**T= Time taken**

## SONAR solved problem # 1

A sonar device on a submarine sends out a signal and receives an echo 5 s later. Calculate the speed of sound in water if the distance of the object from sub-marine is 3,625 m.

Answer :

Given, d ( distance ) = 3,625 m;

t ( time taken )= 5 s ;

u = ? (to find)

According to SONAR Formula :

d = (v x t)/2 v = 2d/t = (2×362)/5 = 1450 ms^{-1}

## SONAR solved problem # 2

** **A ship sends on a high frequency sound wave and receives an echo after 1 second. What is the depth of the sea? Given is Speed of sound in water : 1500 m/s.

Answer : If the depth of sea is d , then total distance travelled by the sound wave is 2d

Time taken = 1 s

As per question , Speed of the sound = 1500 m/s

Put the values in the formula = 1500 m s^{_1} = 2d/1s

therefore d = (1500 x 1 ) / 2 = 750 m

Sea depth = 750 metres.

These are the two basic examples of sonar numerical.In our other post we have lot more **sonar solved numerical** related to it. Apart from solved problems, there are other important topics such as principle and working of SONAR , SONAR types, echo role ,and many other conceptual questions based on this technique .

If you want to have unsolved problems for practice, do comment and we will surely entertain you. We hope that these solved problems will help in understanding the concept better and help you in solving your assignments. In case, any specific topics needs to be discussed on our website, we do welcome and give our audience the first priority. Hit the like button and share the post with your friends who need help with this topic.

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