Why is water soaked cloth used on forehead during fever?please can you tell me answer
1 answer:
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Answer:
1) At the highest point of the building.
2) The same amount of energy.
3) The kinetic energy is the greatest.
4) Potential energy = 784.8[J]
5) True
Explanation:
Question 1
The moment when it has more potential energy is when the ball is at the highest point in the building, that is when the ball is at a height of 40 meters from the ground. It is taken as a point of reference of potential energy, the level of the soil, at this point of reference the potential energy is zero.
Question 2)
The potential energy as the ball falls becomes kinetic energy, in order to be able to check this question we can calculate both energies with the input data.
And the kinetic energy will be:
Therefore it is the ball has the same potential energy and kinetic energy as it is half way through its fall.
Question 3)
As the ball drops all potential energy is transformed into kinetic energy, therefore being close to the ground, the ball will have its maximum kinetic energy.
Question 4)
It can be easily calculated using the following equation
Question 5)
True
The potential energy at 20[m] is:
Answer:
a) It takes the stone 0.7743 s to reach a height of 11 m for the first time on its way up and 2.899 s to reach again that height on its way down.
b) At t = 0.7743 s the velocity is 10.41 m/s and at t = 2.899 s the velocity is -10.41 m/s.
c) There are two answers because the stone reaches the height of 11 m one time on its way up and one more time again on its way down.
Please, see the attached figures and the explanation for a description of the figures.
Explanation:
Hi there!
The equations for the height and velocity of the stone are as follows:
y = y0 + v0 · t + 1/2 · g · t²
v = v0 + g · t
Where:
y = height
y0 = initial height
v0 = initial velocity
t = time
g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive)
v = velocity at time t
a) Let´s calculate the time it takes the stone to reach a height of 11 m. The origin of the frame of reference is at the throwing point so that y0 = 0:
y = y0 + v0 · t + 1/2 · g · t²
11 m = 18 m/s · t - 1/2 · 9.8 m/s² · t²
0 = -4.9 m/s² · t² + 18 m/s · t - 11 m
Solving the quadratic equation:
t = 0.7743 s and t = 2.899 s
(Notice that I have used more significant figures to avoid error by rounding)
The stone will be two times at a height of 11 m, one on its way up (at 0.7743 s) and one on its way down (at 2.899 s). Then, it takes the stone 0.7743 s to reach a height of 11 m for the first time.
b) Let´s use the equation of velocity:
v = v0 + g · t
at t = 0.77443 s
v = 18 m/s - 9.8 m/s² · 0.77443 s
v = 10.41 m/s
at t = 2.899 s
v = 18 m/s - 9.8 m/s² · 2.899 s
v = - 10.41 m/s
(Both velocities have to be of the same magnitude but of different sign, that´s why I haven´t rounded the time.)
c) There are two answers because the stone reaches the height of 11 m one time on its way up and one more time again on its way down. On its way up, the velocity is 10.41 m/s and on its way down it is -10.41 m/s.
Figures
The functions to plot are the following:
height in function of time (figure 1, x-axis: time. y-axis: height)
y = -4.9t² + 18t
velocity in function of time (figure 2, x-axis: time. y-axis velocity)
v = -9.8t + 18
Acceleration in function of time (figure 3, x-axis: time. y-axis: acceleration)
a = -9.8
