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siniylev [52]
3 years ago
10

A boulder is raised above the ground, so that its potential energy relative to the ground is 200 J. Then it is dropped. Estimate

what its kinetic energy will be just before hitting the ground.
Physics
1 answer:
babymother [125]3 years ago
6 0

Answer:

200 J

Explanation:

In this problem, I assume there is no air resistance, so the  system is isolated (=no external forces).

For an isolated system, the total mechanical energy is constant, and it is given by:

E=KE+PE

where

KE is the kinetic energy

PE is the potential energy

The kinetic energy is the energy due to the motion of the object,  while the potential energy is the energy due to the position of the object relative to the ground.

At the beginning, when the boulder is raised above the ground, its height above the ground is maximum, while its  speed is zero; it means that all its mechanical energy is just potential energy, and it is:

E=PE_{max}=200 J

As the boulder falls  down, its altitude decreases, so its potential energy decreases, while the speed increases, and the kinetic energy increases. Therefore, potential energy is converted into kinetic energy.

Eventually, just before the boulder hits the ground, the height of the object is zero, and the speed is maximum; this means that all the energy has now converted into kinetic energy, and we have

E=KE_{max}=200 J

Therefore, the kinetic energy just before hitting the ground is 200 J.

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A rifle fires a 2.90 x 10-2-kg pellet straight upward, because the pellet rests on a compressed spring that is released when the
iren2701 [21]

Answer: 586.60N/m

Explanation:

In this scenario, the elastic potential energy of the spring is converted into potential energy.

0.5*K*x^2 = mgh

Thus K = 2mgh/x^2

=(2*2.90*10^-2*9.8*7.23)/(8.37*10^-2)^2

=586.599

Therefore K = 586.60N/m

7 0
3 years ago
Read 2 more answers
What change in entropy occurs when a 0.15 kg ice cube at -18 °C is transformed into steam at 120 °c 4.
Studentka2010 [4]

<u>Answer:</u> The change in entropy of the given process is 1324.8 J/K

<u>Explanation:</u>

The processes involved in the given problem are:

1.)H_2O(s)(-18^oC,255K)\rightarrow H_2O(s)(0^oC,273K)\\2.)H_2O(s)(0^oC,273K)\rightarrow H_2O(l)(0^oC,273K)\\3.)H_2O(l)(0^oC,273K)\rightarrow H_2O(l)(100^oC,373K)\\4.)H_2O(l)(100^oC,373K)\rightarrow H_2O(g)(100^oC,373K)\\5.)H_2O(g)(100^oC,373K)\rightarrow H_2O(g)(120^oC,393K)

Pressure is taken as constant.

To calculate the entropy change for same phase at different temperature, we use the equation:

\Delta S=m\times C_{p,m}\times \ln (\frac{T_2}{T_1})      .......(1)

where,

\Delta S = Entropy change

C_{p,m} = specific heat capacity of medium

m = mass of ice = 0.15 kg = 150 g    (Conversion factor: 1 kg = 1000 g)

T_2 = final temperature

T_1 = initial temperature

To calculate the entropy change for different phase at same temperature, we use the equation:

\Delta S=m\times \frac{\Delta H_{f,v}}{T}      .......(2)

where,

\Delta S = Entropy change

m = mass of ice

\Delta H_{f,v} = enthalpy of fusion of vaporization

T = temperature of the system

Calculating the entropy change for each process:

  • <u>For process 1:</u>

We are given:

m=150g\\C_{p,s}=2.06J/gK\\T_1=255K\\T_2=273K

Putting values in equation 1, we get:

\Delta S_1=150g\times 2.06J/g.K\times \ln(\frac{273K}{255K})\\\\\Delta S_1=21.1J/K

  • <u>For process 2:</u>

We are given:

m=150g\\\Delta H_{fusion}=334.16J/g\\T=273K

Putting values in equation 2, we get:

\Delta S_2=\frac{150g\times 334.16J/g}{273K}\\\\\Delta S_2=183.6J/K

  • <u>For process 3:</u>

We are given:

m=150g\\C_{p,l}=4.184J/gK\\T_1=273K\\T_2=373K

Putting values in equation 1, we get:

\Delta S_3=150g\times 4.184J/g.K\times \ln(\frac{373K}{273K})\\\\\Delta S_3=195.9J/K

  • <u>For process 4:</u>

We are given:

m=150g\\\Delta H_{vaporization}=2259J/g\\T=373K

Putting values in equation 2, we get:

\Delta S_2=\frac{150g\times 2259J/g}{373K}\\\\\Delta S_2=908.4J/K

  • <u>For process 5:</u>

We are given:

m=150g\\C_{p,g}=2.02J/gK\\T_1=373K\\T_2=393K

Putting values in equation 1, we get:

\Delta S_5=150g\times 2.02J/g.K\times \ln(\frac{393K}{373K})\\\\\Delta S_5=15.8J/K

Total entropy change for the process = \Delta S_1+\Delta S_2+\Delta S_3+\Delta S_4+\Delta S_5

Total entropy change for the process = [21.1+183.6+195.9+908.4+15.8]J/K=1324.8J/K

Hence, the change in entropy of the given process is 1324.8 J/K

4 0
3 years ago
2 Points
mezya [45]
The advantage is that we do not run out of resources and a disadvantage is that is dangerous when a “human” gets too close and gets sick by the radiation.
5 0
2 years ago
The angle between the axes of two polarizing filters is 45.0^\circ45.0 ​∘ ​​ . By how much does the second filter reduce the int
suter [353]

Answer

given,                                                                      

angle between two polarizing filters = 45°

filter reduce intensity = ?                          

a) I = I₀ Cos² θ                                

here θ = 45⁰                                

I = \dfrac{I_0}{2}                      

intensity of the light is reduced by 0.500

correct answer from the given option D

b) direction of the polarization                    

                        θ = 45°                  

7 0
3 years ago
A force acting over a large area will exert less pressure per square inch than the same force acting over a smaller area.
babunello [35]

Answer:

True

Explanation:

Pressure is defined as:

p=\frac{F}{A}

where

F is the magnitude of the force perpendicular to the surface

A is the surface

Therefore, pressure is inversely proportional to the area of the surface:

p\propto \frac{1}{A}

this means that, assuming that the forces in the two situations (which have same magnitude) are both applied perpendicular to the surface, the force exerted over the smaller area will exert a greater pressure. Hence, the statement"

<em>"A force acting over a large area will exert less pressure per square inch than the same force acting over a smaller area"</em>

is true.

8 0
3 years ago
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