An undiscovered planet, many lightyears from Earth, has one moon in a periodic orbit. This moon takes 23.0 days on average to co
mplete one nearly circular revolution around the unnamed planet. If the distance from the center of the moon to the surface of the planet is 2.060×105 km and the planet has a radius of 4.000×103 km, calculate the moon's radial acceleration a=_________m/s^2
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Answer:
Explanation:
From the question we are told that:
Height
Length
Mass
Final speed
Generally the equation for Potential Energy P.E is mathematically given by
Therefore
Initial potential energy
Generally the equation for Kinetic Energy K.E is mathematically given by
Therefore
Final kinetic energy
Generally the equation for Work_done is mathematically given by
Therefore
Answer:
= 0.66
Hence, the fraction of the length of the rod above water = = 0.66
and fraction of the length of the rod submerged in water = 1 - = 1 - 0.66 = 0.34
Explanation:
Data given:
Density of the rod = 5/9 of the density of the water.
Let's denote density of Water with w
And density of rod with r
So,
r = 5/9 x w
Required:
Fraction of the length of the rod above water.
Let's denote total length of the rod with L
and length of the rod above with = y
Let's denote the density of rod = r
And density of water = w
So, the required is:
Fraction of the length of the rod above water = y/L
y/L = ?
In order to find this, we first need to find out the all type of forces acting upon the rod.
We know that, a body will come to equilibrium if the net torque acting upon a body is zero.
As, we know
F = ma
Density = m/v
m = Density x volume
Volume = Area x length = X ( L-y)
So, let's say X is the area of the cross section of the rod, so the forces acting upon it are:
F = mg
F = (Density x volume) x g
g = gravitational acceleration
F1 = X(L-y) x w x g (Force on the length of the rod submerged in water)
where,
X (L-y) = volume
w = density of water.
Another force acting upon it is:
F = mg
F2 = X x L x r x g
Now, the torques acting upon the body:
T1 + T2 = 0
F1 ( y + ) g sinФ - F2 x (
) x gsinФ = 0
plug in the equations of F1 and F2 into the above equation and after simplification, we get:
=
. r
where, w is the density of water and r is the density of rod.
As we know that,
r = 5/9 x w
So,
=
. 5/9 x w
Hence,
=
=
Taking common and solving for
, we will get
= 0.66
Hence, the fraction of the length of the rod above water = = 0.66
and fraction of the length of the rod submerged in water = 1 - = 1 - 0.66 = 0.34