What happens when an object is dropped?
1 answer:
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Answer:
23376 days
Explanation:
The problem can be solved using Kepler's third law of planetary motion which states that the square of the period T of a planet round the sun is directly proportional to the cube of its mean distance R from the sun.
where k is a constant.
From equation (1) we can deduce that the ratio of the square of the period of a planet to the cube of its mean distance from the sun is a constant.
Let the orbital period of the earth be and its mean distance of from the sun be
.
Also let the orbital period of the planet be and its mean distance from the sun be
.
Equation (2) therefore implies the following;
We make the period of the planet the subject of formula as follows;
But recall that from the problem stated, the mean distance of the planet from the sun is 16 times that of the earth, so therefore
Substituting equation (5) into (4), we obtain the following;
cancels out and we are left with the following;
Recall that the orbital period of the earth is about 365.25 days, hence;
Answer:
Explanation:
a )
Each blade is in the form of rod with axis near one end of the rod
Moment of inertia of one blade
= 1/3 x m l²
where m is mass of the blade
l is length of each blade.
Total moment of moment of 3 blades
= 3 x x m l²
ml²
2 )
Given
m = 5500 kg
l = 45 m
Putting these values we get
moment of inertia of one blade
= 1/3 x 5500 x 45 x 45
= 37.125 x 10⁵ kg.m²
Moment of inertia of 3 blades
= 3 x 37.125 x 10⁵ kg.m²
= 111 .375 x 10⁵ kg.m²
c )
Angular momentum
= I x ω
I is moment of inertia of turbine
ω is angular velocity
ω = 2π f
f is frequency of rotation of blade
d )
I = 111 .375 x 10⁵ kg.m² ( Calculated )
f = 11 rpm ( revolution per minute )
= 11 / 60 revolution per second
ω = 2π f
= 2π x 11 / 60 rad / s
Angular momentum
= I x ω
111 .375 x 10⁵ kg.m² x 2π x 11 / 60 rad / s
= 128.23 x 10⁵ kgm² s⁻¹ .