Physics, asked by gsreenivasulu4490, 10 months ago

The speed of sound in air is v. The velocity of a source of sound whose frequency appears to be doubled to a stationary observer is

Answers

Answered by Anonymous
1

Answer:

When the source moves towards a stationary observer, the apparent frequency of sound increases by Doppler effect and is given by

n 1

=n⋅ v−v sv

where v is the velocity of sound, v s

is the velocity of the source of sound and n is the original frequency of sound emitted by the source.

hope help u

Answered by Theopekaaleader
1

Explanation:

</p><p></p><p>Solve the following by both substitution and elimination methods.</p><p></p><p>2x - \w</p><p>3x</p><p>	</p><p> −y=1</p><p></p><p>Solve the following by both substitution and elimination methods.</p><p></p><p>2x - \sqrt{2} y = 02x− </p><p>2</p><p>	</p><p> y=0</p><p></p><p>and</p><p></p><p>\frac{3x}{ \sqrt{2} } - y = 1 </p><p>2</p><p>	</p><p> </p><p>3x</p><p>	</p><p> −y=1</p><p></p><p>Solve the following by both substitution and elimination methods.</p><p></p><p>2x - \sqrt{2} y = 02x− </p><p>2</p><p>	</p><p> y=0</p><p></p><p>and</p><p></p><p>\frac{3x}{ \sqrt{2} } - y = 1 </p><p>2</p><p>	</p><p> </p><p>3x</p><p>	</p><p> −y=1</p><p></p><p>Solve the following by both substitution and elimination methods.</p><p></p><p>2x - \sqrt{2} y = 02x− </p><p>2</p><p>	</p><p> y=0</p><p></p><p>and</p><p></p><p>\frac{3x}{ \sqrt{2} } - y = 1 </p><p>2</p><p>	</p><p> </p><p>3x</p><p>	</p><p> −y=1</p><p></p><p>Solve the following by both substitution and elimination methods.</p><p></p><p>2x - \sqrt{2} y = 02x− </p><p>2</p><p>	</p><p> y=0</p><p></p><p>and</p><p></p><p>\frac{3x}{ \sqrt{2} } - y = 1 </p><p>2</p><p>	</p><p> </p><p>3x</p><p>	</p><p> −y=1</p><p></p><p>Solve the following by both substitution and elimination m elimination methods.</p><p></p><p>2x - \sqrt{2} y = 02x− </p><p>2</p><p>	</p><p> y=0</p><p></p><p>and</p><p></p><p>\frac{3x}{ \sqrt{2} } - y = 1 </p><p>2</p><p>	</p><p> </p><p>3x</p><p>	</p><p> −y=1</p><p></p><p>Solve the following by both substitution and elimination methods.</p><p></p><p>2x - \sqrt{2} y = 02x− </p><p>2</p><p>	</p><p> y=0</p><p></p><p>and</p><p></p><p>\frac{3x}{ \sqrt{2} } - y = 1 </p><p>2</p><p>	</p><p> </p><p>3x</p><p>	</p><p> −y=1</p><p></p><p>Solve the following by both substitution and elimination methods.</p><p></p><p>2x - \sqrt{2} y = 02x− </p><p>2</p><p>	</p><p> y=0</p><p></p><p>and</p><p></p><p>\frac{3x}{ \sqrt{2} } - y = 1 </p><p>2</p><p>	</p><p> </p><p>3x</p><p>	</p><p> −y=1</p><p></p><p>Solve the following by both substitution and elimination methods.</p><p></p><p>2x - \sqrt{2} y = 02x− </p><p>2</p><p>	</p><p> y=0</p><p></p><p>and</p><p></p><p>\frac{3x}{ \sqrt{2} } - y = 1 </p><p>2</p><p>	</p><p> </p><p>3x</p><p>	</p><p> −y=1</p><p></p><p>Solve the following by both substitution and elimination methods.</p><p></p><p>2x - \sqrt{2} y = 02x− </p><p>2</p><p>	</p><p> y=0</p><p></p><p>and</p><p></p><p>\frac{3x}{ \sqrt{2} } - y = 1 </p><p>2</p><p>	</p><p> </p><p>3x</p><p>	</p><p> −y=1

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