<span>The answers are --
a) wind direction
b) wind speed
e) intensity of precipitation
f) location of precipitation</span>
Answer:
F = 4.147 × 10^23
v = 1.31 × 10^4
Explanation:
Given the following :
mass of Jupiter (m1) = 1.9 × 10^27
Mass of sun (m2) = 1.99 × 10^30
Distance between sun and jupiter (r) = 7.8 × 10^11m
Gravitational force (F) :
(Gm1m2) / r^2
Where ; G = 6.673×10^-11 ( Gravitational constant)
F = [(6.673×10^-11) × (1.9 × 10^27) × (1.99 × 10^30)] / (7.8 × 10^11)^2
F = [25.231 × 10^(-11+27+30)] / (60.84 × 10^22)
F = (25.231 × 10^46) / (60.84 × 10^22)
F = 3.235 × 10^(46 - 22)
F = 0.4147 × 10^24
F = 4.147 × 10^23
Speed of Jupiter (v) :
v = √(Fr) / m1
v = √[(4.147 × 10^23) × (7.8 × 10^11) / (1.9 × 10^27)
v = √32.3466 × 10^(23+11) / 1.9 × 10^27
v = √32.3466× 10^34 / 1.9 × 10^27
v = √17. 023 × 10^34-27
v = √17.023 × 10^7
v = 13047.221
v = 1.31 × 10^4
(a) 5.66 m/s
The flow rate of the water in the pipe is given by
where
Q is the flow rate
A is the cross-sectional area of the pipe
v is the speed of the water
Here we have
the radius of the pipe is
r = 0.260 m
So the cross-sectional area is
So we can re-arrange the equation to find the speed of the water:
(b) 0.326 m
The flow rate along the pipe is conserved, so we can write:
where we have
and where 
is the cross-sectional area of the pipe at the second point.
Solving for A2,
And finally we can find the radius of the pipe at that point:
Answer:
The child represented by a star on the outside path.
Explanation:
