The acceleration of gravity is 9.8 m/s² . That means that a falling object
is always falling 9.8 m/s faster than it was falling 1 second earlier.
If an object is not slowed by air resistance, and has far enough to go
so that it's still falling after three whole seconds, then at the end of
three seconds it's falling at
(9.8 m/s²) x (3 sec) = 29.4 m/s
Molecules, The movemnet of molecules
Question: A loader sack of total mass
is l000 grams falls down from
the floor of a lorry 200 cm high
Calculate the workdone by the
gravity of the load.
Answer:
19.6 Joules
Explanation:
Applying
W = mgh........................ Equation 1
Where W = Workdone by gravity on the load, m = mass of the loader sack, h = height, g = acceleration due to gravity
From the question,
Given: m = 1000 grams = (1000/1000) kilogram = 1 kg, h = 200 cm = 2 m
Constant: g = 9.8 m/s²
Substitute these values into equation 1
W = (1×2×9.8)
W = 19.6 Joules
Hence the work done by gravity on the load is 19.6 Joules
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;
