Question

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

Answer 1

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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Answer 2

Answer:

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

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:

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

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

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

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

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

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

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

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

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

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