What is the speed over the ground of a mosquito flying 2 m/s relative to the air against a 2 m/s headwind?

Sagot :

Answer:

The speed of mosquito over the ground is [tex]$4m/sec$[/tex].

Explanation:

• Relative velocity is defined as the velocity of an object B in the rest frame of another object A.

• To find the speed of mosquito over the ground, use this formula:

[tex]${V_{{M \mathord{\left/ {\vphantom {M A}} \right[/tex].

[tex]\kern-\nulldelimiterspace} A}}} = {V_{{M \mathord{\left/ {\vphantom {M G}} \right. \kern-\nulldelimiterspace} G}}} + {V_{{A \mathord{\left/ {\vphantom {A G}} \right. \kern-\nulldelimiterspace} G}}}$[/tex]

Where [tex], $A$[/tex] represents air,[tex]$M$[/tex] represents mosquito and [tex]$G$[/tex] represents ground, thus [tex]${V_{{A \mathord{\left/ {\vphantom {A G}} \right. \kern-\nulldelimiterspace} G}}}$[/tex] represents velocity of air with respect to ground and so on.

[tex]\\${V_{{M \mathord{\left/ {\vphantom {M A}} \right. \kern-\nulldelimiterspace} A}}} = 2m/sec$ \\${V_{{A \mathord{\left/ {\vphantom {A G}} \right. \kern-\nulldelimiterspace} G}}} = - 2m/sec$[/tex]

• Placing the value of the given  in the above formula.

[tex]& {V_{{M \mathord{\left/ {\vphantom {M A}} \right. \kern-\nulldelimiterspace} A}}} = {V_{{M \mathord{\left/ {\vphantom {M G}} \right.[/tex]

[tex]\kern-\nulldelimiterspace} G}}} + {V_{{A \mathord{\left/ {\vphantom {A G}} \right. \kern-\nulldelimiterspace} G}}} \\ & \Rightarrow 2 = {V_{{M \mathord{\left/ {\vphantom {M G}} \right. \kern-\nulldelimiterspace} G}}} - 2 \\[/tex]

 [tex]& \Rightarrow {V_{{M \mathord{\left/ {\vphantom {M G}} \right. \kern-\nulldelimiterspace} G}}} = 4m/sec \\\end{align}\]$[/tex]

• Hence, the speed of mosquito over the ground is [tex]$4m/sec$[/tex].

Learn more about velocity here:

https://brainly.com/question/18026124

We have that the speed over the ground of a mosquito flying 2 m/s relative to the air against a 2 m/s headwind is

X=0m/s

From the question we are told that

mosquito flying 2 m/s

against a 2 m/s headwind

Generally

The speed over the ground is the Flight Speed minus resistance speed

Generally the equation for the  speed over the ground  is mathematically given as

X=Flight Speed-resistance speed

Therefore

X=2-2

X=0m/s

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