Answer:

Explanation:
The apparent brightness follows an inverse square law, therefore we can write:

where I is the apparent brightness and r is the distance from the Sun.
We can also rewrite the law as
(1)
where in this problem, we have:
apparent brightness at a distance
, where
million km
We want to estimate the apparent brightness at
, where
is ten times
, so

Re-arranging eq.(1), we find
:

Lindsay has to fly this plane towards this direction [W 12.5° S] to get to Hamilton.
From this question, the plane is still up in the air.
We have wind blowing in [W 60° N ]
To solve the problem we have to make use of the sine rule

We put the values in the equation, we have:
50/Sinθ = 200/sin60°
The next step is to cross multiply
50 x sin60° = 200Sinθ
50 x 0.8660 = 200sinθ
We make Sin θ the subject
Sine θ = 43.30/200
sine θ = 0.2165
we find the value of θ
θ = sine⁻¹(0.2165)
θ = 12.50
So Lindsay has to fly this plane towards this direction
[W 12.5° S]
Here is a similar question brainly.com/question/13338067?referrer=searchResults
Here we will the speed of seagull which is v = 9 m/s
this is the speed of seagull when there is no effect of wind on it
now in part a)
if effect of wind is in opposite direction then it travels 6 km in 20 min
so the average speed is given by the ratio of total distance and total time


now since effect of wind is in opposite direction then we can say



Part b)
now if bird travels in the same direction of wind then we will have


now we can find the time to go back



Part c)
Total time of round trip when wind is present


now when there is no wind total time is given by


So due to wind time will be more
Its a thermometer . i hope this helps you
Explanation:
Let us first calculate long does it take to go 12m at 30m/s( assumed speed)
12/30 = 0.4 seconds
horizontal distance the ball drop in that time
H= (0)(0.4)+1/2(-9.8)(0.4)2
H= -0.78m
negative sign shows that the height of the ball at the net from the top.
Height of the ball at the net and from the ground= H1-H=2.4-0.78=1.62m
As 1.62m>0.9m so the ball will clear the net.
H_1= V0y t’ + ½ g t’^2
-2.4= (0)t’ + ½ (-9.8) t’^2
t’= 0.69s
X’=V0x t’
X’=(30)(0.96)
X’= 20.7m