According to Ptolemy's model, he, too, believed in a geocentric Universe and that the planets and stars were perfect spheres, though Earth itself was not.
He further thought that the movements of the planets and stars must be circular since they were perfect and, if the motions were circular, then they could go on forever.
<h3>What is comets and shooting stars?</h3>
Shooting stars are very different from comets, although the two can be related. A Comet is a ball of ice and dirt, orbiting the Sun (usually millions of miles from Earth). ... A shooting star on the other hand, is a grain of dust or rock (see where this is going) that burns up as it enters the Earth's atmosphere.
Learn more about ptolemy's model:
brainly.com/question/12639459
Answer:
ΔS=2*m*Cp*ln((T1+T2)/(2*(T1*T2)^1/2))
Explanation:
The concepts and formulas that I will use to solve this exercise are the integration and the change in the entropy of the universe. To calculate the final temperature of the water the expression for the equilibrium temperature will be used. Similarly, to find the change in entropy from cold to hot water, the equation of the change of entropy will be used. In the attached image is detailed the step by step of the resolution.
Well idk if this helps but the formula to solve acceleration is
a=F/m=(100kg)=1.0m/s 2
Answer: The height above the release point is 2.96 meters.
Explanation:
The acceleration of the ball is the gravitational acceleration in the y axis.
A = (0, -9.8m/s^)
For the velocity we can integrate over time and get:
V(t) = (9.20m/s*cos(69°), -9.8m/s^2*t + 9.20m/s^2*sin(69°))
for the position we can integrate it again over time, but this time we do not have any integration constant because the initial position of the ball will be (0,0)
P(t) = (9.20*cos(69°)*t, -4.9m/s^2*t^2 + 9.20m/s^2*sin(69°)*t)
now, the time at wich the horizontal displacement is 4.22 m will be:
4.22m = 9.20*cos(69°)*t
t = (4.22/ 9.20*cos(69°)) = 1.28s
Now we evaluate the y-position in this time:
h = -4.9m/s^2*(1.28s)^2 + 9.20m/s^2*sin(69°)*1.28s = 2.96m
The height above the release point is 2.96 meters.
From the geometry of the problem, the 20 m-long cable creates
the hypotenuse of a right triangle, with the extended of the other two sides of
size 20 m * cos(30 deg), which is around 17.3 m. Therefore, the ball has increased
by 20 m - 17.3 m = 2.7 m.
The potential energy will have altered by m*g*h, which is 1400 kg * 9.8 m/s^2 *
1.6 m , or about 37044 joules.
