Answer: D. it has been demonstrated to be without exception under certain stated conditions.
Explanation:
A <u>Law</u> is an affirmation (something established) based on repeated long-term observation of a phenomenon that has been studied and verified.
That is: A law is present in all known theories and therefore is considered universal. In addition, a law can not be refuted, nor changed, because its precepts have been proven through various studies.
<u>Unlike theory</u>, which is the set of rules and principles that describe and explain a particular phenomenon and <u>is subject to changes as new evidence emerges that gives meaning to it. </u>
Then, based on what is explained above, the law of universal gravitation is a statement that exists because it was rigorously tested and verified, therefore it can not be refuted.
<span>The tides on earth are caused mainly by earths gravitational interactions with the sun and the moon.</span>
We know that:
d=vt
d=32mph*5h
d=160mi
Answer: The spring constant is K=392.4N/m
Explanation:
According to hook's law the applied force F will be directly proportional to the extension e produced provided the spring is not distorted
The force F=ke
Where k=spring constant
e= Extention produced
h=2m
Given that
e=20cm to meter 20/100= 0.2m
m=100g to kg m=100/1000= 0.1kg
But F=mg
Ignoring air resistance
assuming g=9.81m/s²
Since the compression causes the plastic ball to poses potential energy hence energy stored in the spring
E=1/2ke²=mgh
Substituting our values to find k
First we make k subject of formula
k=2mgh/e²
k=2*0.1*9.81*2/0.1²
K=3.921/0.01
K=392.4N/m
Answer:
W = ½ m v²
Explanation:
In this exercise we must solve it in parts, in a first part we use the conservation of the moment to find the speed after the separation
We define the system formed by the two parts of the rocket, therefore the forces during internal separation and the moment are conserved
initial instant. before separation
p₀ = m v
final attempt. after separation
= m /2 0 + m /2 v_{f}
p₀ = p_{f}
m v = m /2 
v_{f}= 2 v
this is the speed of the second part of the ship
now we can use the relation of work and energy, which establishes that the work is initial to the variation of the kinetic energy of the body
initial energy
K₀ = ½ m v²
final energy
= ½ m/2 0 + ½ m/2 v_{f}²
K_{f} = ¼ m (2v)²
K_{f} = m v²
the expression for work is
W = ΔK = K_{f} - K₀
W = m v² - ½ m v²
W = ½ m v²
