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Butoxors [25]
2 years ago
6

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

Physics
2 answers:
salantis [7]2 years ago
8 0

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

For more information on this visit

brainly.com/question/22271063?referrer=searchResults

Katyanochek1 [597]2 years ago
7 0

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 $4m/sec$.

Learn more about velocity here:

brainly.com/question/18026124

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Homer Agin leads the Varsity team in home runs. In a recent game, Homer hit a 34.5 m/s sinking curve ball head on, sending it of
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Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 34.5 m/s

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Acceleration is simply defined as the rate of change of velocity with time. Mathematically, it is expressed as:

Acceleration = (final velocity – Initial velocity) /time

a = (v – u) / t

With the above formula, we can obtain acceleration of the ball as follow:

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Time (t) = 0.00075 s

Acceleration (a) =?

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a = –77867 m/s/s

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