Answer:
The answer is "False"
Explanation:
The geologic time scale is the "schedule" for occasions in Earth history. It partitions time into named units of unique time called in descending order of duration "eons, eras, periods, epochs, and ages". The specification of those geologic time units depends on stratigraphy, which is the relationship and order of rock layers. The fossil structures that happen in the stones, nonetheless, give the central methods for setting up a geologic time scale, with the circumstance of the development and vanishing of far and wide species from the fossil record being used to outline the beginnings and endings of ages,, periods, and different stretches.
Geologic time is the broad time period involved by the geologic history of Earth. Formal geologic time starts toward the beginning of the Archean Eon (4.0 billion to 2.5 billion years back) and proceeds to the current day.
The cell theory states that:
1) All living organisms are composed of one or more cells.
2) The cell is the basic unit of structure and organization in organisms.
3) Cells arise from pre-existing cells.
Your statement is not present, so the answer would be no.
♧
Answer: elastic potential energy = 20.27 J
Explanation:
Given that the
Mass M = 0.470 kg
Height h = 4.40 m
Spring constant K = 85 N/m
The maximum elastic potential will be equal to the maximum kinetic energy experienced by the block.
But according to conservative of energy, the maximum kinetic energy is equal to the maximum potential energy experienced by the block of mass M.
That is
K .E = P.E = mgh
Where g = 9.8m/s^2
Substitutes all the parameters into the formula
K.E = 0.470 × 9.8 × 4.4
K.E = 20.27 J
Where K.E = maximum elastic potential energy stored in the spring during the motion of the blocks after the collision which is 20.27J.
Answer:
g(h) = g ( 1 - 2(h/R) )
<em>*At first order on h/R*</em>
Explanation:
Hi!
We can derive this expression for distances h small compared to the earth's radius R.
In order to do this, we must expand the newton's law of universal gravitation around r=R
Remember that this law is:
In the present case m1 will be the mass of the earth.
Additionally, if we remember Newton's second law for the mass m2 (with m2 constant):
Therefore, we can see that
With a the acceleration due to the earth's mass.
Now, the taylor series is going to be (at first order in h/R):
a(R) is actually the constant acceleration at sea level
and
Therefore:
Consider that the error in this expresion is quadratic in (h/R), and to consider quadratic correctiosn you must expand the taylor series to the next power:
Vertical force on the box=mg
<span>the component of gravity parallel=mg*SinTheta </span>
<span>the component of gravity normal=mg*CosTheta </span>
<span>frictional force up the plane: mg*cosTheta*mu max, but if it is sitting still, it is equal and opposite to mg*cosTheta (it cannot be greater than this or it would go up the plane).</span>
