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:
g = 8.61 m/s²
Explanation:
distance of the International Space Station form earth is 200 Km
mass of the object = 1 Kg
acceleration due to gravity on earth = 9.8 m/s²
mass of earth = 5.972 x 10²⁴ Kg
acceleration due to gravity = ?
r = 6400 + 200 = 6800 Km = 6.8 x 10⁶ n
using formula


g = 8.61 m/s²
Answer:
I have no clue what's really going on I'm just here to get answer maybe I will just try to get an answer but I have no clue I'm sorry I am confused and dint really know what to do here.
It lets the viewer know it's something to do with underwater.
Answer:
a) u = 30.29 m/s
b) t = 2.09 s
Explanation:
given,
velocity = 45 m/s
angle (θ) = 50°
horizontal velocity = 45 cos 50°
time taken to reach 150 m.
times = 
t = 5.19 s
a) height of arrow



s = 46.78 m
v² - u² = 2 g s
u² = 2 × 9.81 × 46.78
u = 30.29 m/s
b) time taken by the apple = 
= 3.09 s
time after which it has to be thrown = 5.19-3.09 = 2.1 s