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ser-zykov [4K]
3 years ago
12

An object has 1200 J of gravitational potential energy when it is dropped. Right when the object hits the ground how much kineti

c energy does it have?
Physics
1 answer:
Fittoniya [83]3 years ago
5 0

Answer:

The kinetic energy is 1200 J

Explanation:

The Principle of Conservation of energy states that "energy is neither created nor destroyed, it is transformed".

This means that energy can be transformed from one form to another, but the total amount of energy always remains constant, that is, the total energy is the same before and after each transformation.

The mechanical energy of a body or a physical system is the sum of its kinetic energy and the potential energy. According to the Principle of Conservation of Energy for mechanical energy, the total mechanical energy that a body possesses is constant at every instant of time.

Since mechanical energy is equal to the sum of kinetic energy and gravitational potential energy that a body possesses, the only way to stay constant is that:

  • when the kinetic energy increases the gravitational potential energy decreases,
  • when gravitational potential energy increases, kinetic energy decreases.

Due to the Principle of Conservation of Energy you can say that the gravitational potential energy is converted to kinetic energy.  So Gravitational potential energy at the top = kinetic energy at the bottom

<u><em>The kinetic energy is 1200 J</em></u>

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A block slides down a rough ramp with a 30-degree incline as shown.
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Answer:

Image 4 ?

Explanation:

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4 years ago
uniform disk with mass 40.0 kg and radius 0.200 m is pivoted at its center about a horizontal, frictionless axle that is station
Alex787 [66]

Answer:

The magnitude of the tangential velocity is v= 0.868 m/s

The magnitude of the resultant acceleration at that point is  a = 4.057 m/s^2

Explanation:

From the question we are told that

      The mass of the uniform disk is m_d = 40.0kg

       The radius of the uniform disk is R_d = 0.200m

       The force applied on the disk is F_d = 30.0N

Generally the angular speed i mathematically represented as

             w = \sqrt{2 \alpha  \theta}

Where \theta is the angular displacement given from the question as

           \theta  = 0.2000 rev = 0.2000 rev * \frac{2 \pi \ rad }{1 rev}

                 =1.257\  rad

   \alpha is the angular acceleration which is mathematically represented as

                    \alpha = \frac{torque }{moment \ of  \ inertia}  = \frac{F_d * R_d}{I}

    The moment of inertial is mathematically represented as

                     I = \frac{1}{2} m_dR^2_d

Substituting values

                    I = 0.5 * 40 * 0.200^2

                        = 0.8kg \cdot m^2

Considering the equation for angular acceleration

               \alpha = \frac{torque }{moment \ of  \ inertia}  = \frac{F_d * R_d}{I}

Substituting values

               \alph\alpha = \frac{(30.0)(0.200)}{0.8}

                   = 7.5 rad/s^2

Considering the equation for angular velocity

    w = \sqrt{2 \alpha  \theta}

Substituting values

     w =\sqrt{2 * (7.5) * 1.257}

         = 4.34 \ rad/s

The tangential velocity of a given point on the rim is mathematically represented as

                 v = R_d w

Substituting values

                    = (0.200)(4.34)

                     v= 0.868 m/s

The radial acceleration at hat point  is mathematically represented as

            \alpha_r = \frac{v^2}{R}

                  = \frac{0.868^2}{0.200^2}

                 = 3.7699 \ m/s^2

The tangential acceleration at that point is mathematically represented as

               \alpha _t = R \alpha

Substituting values

           \alpha _t = (0.200) (7.5)

                 = 1.5 m/s^2

The magnitude of resultant acceleration at that point is

                 a = \sqrt{\alpha_r ^2+ \alpha_t^2 }

Substituting values

                a = \sqrt{(3.7699)^2 + (1.5)^2}

                   a = 4.057 m/s^2

         

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Within the theory of G relativity what, exactly, is meant by " the speed of light WITHIN A VACUUM" ? &amp; what does that have t
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seconds \times  \frac{meters}{seconds} =meters
This is why we can talk about lightyears.  It's not a unit of time, but distance light travels in a year.  We are now free to define distance as a unit of time because we have a way to convert them.  
As it turns out light is not special in that it gets to travel faster than anything else.  Firstly, other things travel that fast too (gravity and information to name two).  But NO events or information can travel faster than this.  Not because they are not allowed to beat light to the finish line---remember my claim that light has nothing to do with it.  It's because this speed (called "c") converts space and time.  A speed greater than c isn't unobtainable---it simply does not exist.  Period.  Just like I can't travel 10 meters without actually moving 10 meters, I cannot travel 10 meters without also "traveling" at least about 33 nanoseconds (about the time it takes light to get 10 meters)  There is simply no way to get there in less time, anymore than there is a way to walk 10 meters by only walking 5.  
We don't see this in our daily life because it is not obvious that space and time are intertwined this way.  This is a result of our lives spent at such slow speeds relative to the things around us.
This is the fundamental part to the Special Theory of Relativity (what you called the "FIRST" part of the theory)  Here is where Einstein laid out the idea of spacetime and the idea that events (information) itself propagates at a fixed speed that, unlike light, does not slow down in any medium.  The idea that what is happening "now" for you is not the same thing as what is "now" for distant observers or observers that are moving relative to you.  It's also where he proposed of a conversion factor between space and time, which turned out to be the speed of light in vacuum.
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Which explanation best describes a wave?
Gekata [30.6K]

Answer:

C. a disturbance that travels through a medium with a transfer of energy and without a transfer of matter

Explanation:

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