The cells are called insibatss, I think I spelled that right but
... the angular tilt of the Earth's position on its axis relative to the sun
A uniform solid sphere rolls down an incline without slipping<span>. </span>If the linear acceleration of the center of mass of the sphere is 0.19g<span>, </span>then what is the angle the incline makes with the horizontal<span>?</span>
Answer:
a) dh/dt = -44.56*10⁻⁴ cm/s
b) dr/dt = -17.82*10⁻⁴ cm/s
Explanation:
Given:
Q = dV/dt = -35 cm³/s
R = 1.00 m
H = 2.50 m
if h = 125 cm
a) dh/dt = ?
b) dr/dt = ?
We know that
V = π*r²*h/3
and
tan ∅ = H/R = 2.5m / 1m = 2.5 ⇒ h/r = 2.5
⇒ h = (5/2)*r
⇒ r = (2/5)*h
If we apply
Q = dV/dt = -35 = d(π*r²*h/3)*dt
⇒ d(r²*h)/dt = 3*35/π = 105/π ⇒ d(r²*h)/dt = -105/π
a) if r = (2/5)*h
⇒ d(r²*h)/dt = d(((2/5)*h)²*h)/dt = (4/25)*d(h³)/dt = -105/π
⇒ (4/25)(3*h²)(dh/dt) = -105/π
⇒ dh/dt = -875/(4π*h²)
b) if h = (5/2)*r
Q = dV/dt = -35 = d(π*r²*h/3)*dt
⇒ d(r²*h)/dt = d(r²*(5/2)*r)/dt = (5/2)*d(r³)/dt = -105/π
⇒ (5/2)*(3*r²)(dr/dt) = -105/π
⇒ dr/dt = -14/(π*r²)
Now, using h = 125 cm
dh/dt = -875/(4π*h²) = -875/(4π*(125)²)
⇒ dh/dt = -44.56*10⁻⁴ cm/s
then
h = 125 cm ⇒ r = (2/5)*h = (2/5)*(125 cm)
⇒ r = 50 cm
⇒ dr/dt = -14/(π*r²) = - 14/(π*(50)²)
⇒ dr/dt = -17.82*10⁻⁴ cm/s
It is vital to take measurements at the same time everyday to determine the natural cycle, this will show the numbers on a regular basis. It is important because this will be basis of any action that needs to be taken. If it's not taken at the same time, it may disrupt the correct result or change the expected outcome.
