Answer:
Hey
Your answer would be D (Aristotle stated that heavy objects fall faster than light objects). He thought be comparing a feather falling to a rock falling that you could see that heavier objects fall faster than light ones. of course now we know that all objects fall at the same speed and that weight does not affect that. only air drag affects how fast objects fall.
Answer: The energy incident on the solar panel during that day is
.
Explanation:
Given: Mass = 250 kg
Initial temperature = 
Final temperature = 
Specific heat capacity = 4200 
Formula used to calculate the energy is as follows.

where,
q = heat energy
m = mass of substance
C = specific heat capacity
= initial temperature
= final temperature
Substitute the values into above formula as follows.

As it is given that water absorbs 25% of the energy incident on the solar panel. Hence, energy incident on the solar panel can be calculated as follows.

Thus, we can conclude that the energy incident on the solar panel during that day is
.
<h2>a) Average velocity in first 4 seconds is 64 ft/s upward</h2><h2>b) Average velocity in second 4 seconds is 63.5 ft/s downward</h2>
Explanation:
a) Given S(t) = 76 + 128t − 16t²
s(0) = 76 + 128 x 0 − 16 x 0² = 76 ft
s(4) = 76 + 128 x 4 − 16 x 4² = 332 ft
Displacement in 4 seconds = 332 - 76 = 256 ft
Time = 4 - 0 = 4 s

Average velocity in first 4 seconds is 64 ft/s upward
a) Given S(t) = 76 + 128t − 16t²
s(4) = 76 + 128 x 4 − 16 x 4² = 332 ft
s(8) = 76 + 128 x 8 − 16 x 8² = 78 ft
Displacement in 4 seconds = 78 - 332 = -254 ft
Time = 4 - 0 = 4 s

Average velocity in second 4 seconds is 63.5 ft/s downward
The work done by the force is 47.1 J
Explanation:
The work done by a force in moving an object is given by
(1)
where
F is the magnitude of the force
d is the distance covered by the object
is the angle between the direction of the force and the motion of the object
In this problem, the force applied to the object is
F = 3.0 N
This force is always tangential to the track: this means that at every instant, the force is parallel to the motion of the object, so

And the distance covered is equal to the circumference of the circle, which is:

where r = 2.5 m is the radius.
Now we can substitute into eq.(1) to find the work done:

Learn more about work:
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