An asteroid of mass 2 times 10^23 kg has been discovered orbiting the Sun. If it

Question

An asteroid of mass 2 times 10^23 kg has been discovered orbiting the Sun. If it takes the asteroid 5 Earth years to make 1 complete orbit, what is the radius of its orbit relative to the radius of the Earth's own orbit? A pipe of diameter 1 m is at the top of a 20 m tall hill. Water flows through the pipe with a velocity of v = 5 m/s for a distance of 10 m before the pipe constricts to a radius of .5 m. The water then flows for another 8 m before the pipe bends downwards and travels to the ground. Find the fluid velocity at the ground level. What is the pressure difference between the lower and upper parts of the pipe?

Explanation / Answer

Question 1

we can use Kepler's third law for two objects

Te2= Ks ae3

Ta2= ks aa3

(Ta/Te)2 = (aa/ae)3

(5/1 )2= (aa/ ae)3

aa = 52/3 ae

the distance from the earth to the sun is ae= 1.496 1011m

aa= 52/31.496 1011m = 2.924 1.496 1011

aa = 4.37 1011 m

Result

We give the result of two ways:

the orbit of the asteroid is 52/3ae = 2.9 ae

or as the distance in meters4.37 1011 m

Question 2

data

D1 = 1 m

L = 20 m

v = 5 m/s

x1 = 10 m

R = 0.5 m

x2 = 8 m

Part 1

We write the equation of continuity

A1v1 = A2v2

A= r2

we have a problem in Part 1 of tube diameter and give in Part 2 give the radius and are equal, if this is real fluid velocity is the same in both parts

in case of an error and in part 1 being the radius, the result is

r12 v1 = r22 v2

v2 = (r1/r2)2 v1

v2= (1/0.5)25

v2 = 20 m/s

I will assume this is our case

Part 2

We write the Bernoulli equation

P1 + ½ v12+ g y1 = P2 + ½ v22+ g y2

(P2-P1) = ½ (v12– v22)+ g (y1-y2)

(P2-P1) = ½ 1000 (52 – 202 ) + 1000 9.8 (20 – 0)

(P2-P1) = - 1.876 105 + 1.96 105

(P2-P1) = 8.4 103 Pa

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