Answer:
I(x) = 1444×k ×
I(y) = 1444×k ×
I(o) = 3888×k ×
Explanation:
Given data
function = x^2 + y^2 ≤ 36
function = x^2 + y^2 ≤ 6^2
to find out
the moments of inertia Ix, Iy, Io
solution
first we consider the polar coordinate (a,θ)
and polar is directly proportional to a²
so p = k × a²
so that
x = a cosθ
y = a sinθ
dA = adθda
so
I(x) = ∫y²pdA
take limit 0 to 6 for a and o to
for θ
I(x) =
y²p dA
I(x) =
(a sinθ)²(k × a²) adθda
I(x) = k
da ×
(sin²θ)dθ
I(x) = k
da ×
(1-cos2θ)/2 dθ
I(x) = k
×
I(x) = k ×
× (
I(x) = k ×
×
I(x) = 1444×k ×
.....................1
and we can say I(x) = I(y) by the symmetry rule
and here I(o) will be I(x) + I(y) i.e
I(o) = 2 × 1444×k ×
I(o) = 3888×k ×
......................2
Answer:
wouldn't it be 25 miles?? yeah
Explanation:
Momentum = (mass) x (speed)
Momentum = (70 kg) x (10 m/s)
<em>Momentum = 700 kg-m/s</em>
Answer:
138.3 days
Explanation:
Given that a Planet Ayanna has a radius of 6.2 X 10%m and orbits the star named Dayli in 98 days. A new neighboring planet Clayton J-21 has been discovered and has a radius of 7.8 X 10 meters.
The period of time for Clayton J-21 to orbit Dayli can be calculated by using Kepler law.
T^2 is proportional to r^3
That is,
T^2/r^3 = constant
98^2 / 62^3 = T^2 / 78^3
Make T^2 the subject of formula.
T^2 = 98^2 / 62^3 × 78^3
T^2 = 19123.2
T = sqrt ( 19123.2 )
T = 138.2867 days
Therefore, the period of time for Clayton J-21 to orbit Dayli is 138.3 days approximately.