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
Question:<em> </em><em>Find, separately, them mass of the balloon and the basket (incidentally, most of the balloon's mass is air)</em>
Answer:
The mass of the balloon is 2295 kg, and the mass of the basket is 301 kg.
Explanation:
Let us call the mass of the balloon 
and the mass of the basket 
, then according to newton's second law:
,
where 
is the upward acceleration, and 
is the net propelling force (counts the gravitational force).
Also, the tension 
in the rope is 79.8 N more than the basket's weight; therefore,
and this tension must equal
Combining equations (2) and (3) we get:
since , we have
Putting this into equation (1) and substituting the numerical values of 
and 
, we get:
Thus, the mass of the balloon and the basket is 2295 kg and 301 kg respectively.
Answer:
4.42 x 10⁷ W/m²
Explanation:
A = energy absorbed = 500 J
η = efficiency = 0.90
E = Total energy
Total energy is given as
E = A/η
E = 500/0.90
E = 555.55 J
t = time = 4.00 s
Power of the beam is given as
P = E /t
P = 555.55/4.00
P = 138.88 Watt
d = diameter of the circular spot = 2.00 mm = 2 x 10⁻³ m
Area of the circular spot is given as
A = (0.25) πd²
A = (0.25) (3.14) (2 x 10⁻³)²
A = 3.14 x 10⁻⁶ m²
Intensity of the beam is given as
I = P /A
I = 138.88 / (3.14 x 10⁻⁶)
I = 4.42 x 10⁷ W/m²
