The sum of the kinetic and potential energies of a system of objects is conserved only when no external force acts on the objects.
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Conservation of mechanical energy</h3>
The principle of conservation of mechanical energy states that the total mechanical energy of an isolated system (absence of external force) is always constant.
M.A = P.E + K.E
where;
P.E is potential energy
K.E is kinetic energy
Thus, the sum of the kinetic and potential energies of a system of objects is conserved only when no external force acts on the objects.
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- The data for the first part of the experiment support the first hypothesis.
- As the force applied to the cart increased, the acceleration of the cart increased.
- Since the increase in the applied force caused the increase in the cart's acceleration, force and acceleration are directly proportional to each other, which is in accordance with Newton's second law.
When we state something about the results on the basis whether the observed data supports the original hypothesis, we say that we are concluding the results.
What is the relationship between force and acceleration based on Newton's 2nd law?
Newton's second law of motion can be formally stated as follows: The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.
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Answer: 1
Explanation: The highest eccentricity an ellipse can have is '1', a straight line.
Why did the narrator use the phrase “fit together like puzzle pieces” to describe how the cells of a developing human embryo form a human face
Answer:
ball hit the ground from her feet is 1.83 m far away
Explanation:
given data
speed = 5.3 m/s
angle = 12°
height = 1 m
to find out
how far from her feet ball hit ground
solution
we consider here x is horizontal component and y is vertical component
so in vertical
velocity will be = v sin12
vertical speed u = 5.3 sin 12 = 1.1 m/s downward
and
in horizontal , velocity we know v = 5.3 m/s
so from motion of equation
s = ut + 0.5×a×t²
s is distance t is time a is 9.8
put all value
1 = 1.1 ( t) + 0.5×9.8×t²
solve it we get t
t = 0.353 s
and
horizontal distance is = vcos12 × t
so horizontal distance = 5.3×cos12 × ( 0.353)
horizontal distance = 1.83 m
so ball hit the ground from her feet is 1.83 m far away
