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Marysya12 [62]
3 years ago
15

Can you think of a scenario when the kinetic and gravitational potential energy could both be zero ? Describe or draw how this c

ould be possible below
Physics
1 answer:
Inga [223]3 years ago
7 0

Both kinetic and gravitational potential energy can become zero at infinite distance from the Earth.

Consider an object  of mass <em>m </em>projected from the surface of the Earth with a velocity <em>v. </em>

The total energy of the body on the surface of the Earth is the sum of its kinetic energy \frac{1}{2} mv^2and gravitational potential energy -\frac{GMm}{R^2}.

here, <em>M</em> is the mass of the Earth, <em>R</em> is the radius of Earth and <em>G</em> is the universal gravitational constant.

The gravitational potential energy of the object is negative since it is in an attractive field, which is the gravitational field of the Earth.

The energy of the object on the surface of the earth is given by,

E_i=\frac{1}{2} mv^2-\frac{GMm}{R^2}

As the object rises upwards, it experiences deceleration due to the gravitational force of the Earth. Its velocity decreases and hence its kinetic energy decreases.

The decrease in kinetic energy is manifested as  an equal increase in potential energy. The potential energy becomes less and less negative as more and more kinetic energy is converted into potential energy.

At a height <em>h</em> from the surface of the Earth, the energy of the object is given by,

E_h=\frac{1}{2} mv_h^2-\frac{GMm}{(R+h)^2}

The velocity v_h is less than <em>v</em>.

When h =∞, the gravitational potential energy increases from a negative value to zero.

If the velocity of projection is adjusted in such a manner that the velocity decreases to zero at infinite distance from the earth, the object's kinetic energy also becomes equal to zero.

Thus, it is possible for both kinetic and potential energies to be zero at infinite distance from the Earth. In this case, kinetic energy decreases from a positive value to zero and the gravitational potential energy increases from  a negative value to zero.


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A car has a mass of 1200kg what is it’s acceleration when the engine exerts a force of 600 n
fredd [130]

<u>Answer:</u>

0.5 ms−2

<u>Explanation</u>:

Newton's second law of motion is mathematically expressed as

F = ma ; where F is the force exerted on an object, m is the mass of the object, and a is the acceleration of that object.

In this case, the mass of the car and the force exerted on it are provided as information.

Let's plug in the values into the formula

F = ma :

⇒600

N =1200

kg × a

Let's express N

using SI base units:

⇒600 kg m s−2 = 1200 kg × a

⇒600 kg m s−2 = 1200 kg = 1200 kg

1200 kg × a

⇒1 on 2 ms −2 = 1 × a

∴ a = 0.5 ms−2

Therefore, the acceleration of the car is

0.5 ms−2

when its engine exerts a force of 600N

4 0
3 years ago
In a Little League baseball game, the 145 g ball reaches the batter with a speed of 15.0 m/s. The batter hits the ball, and it l
Oduvanchick [21]

Answer: 5.075Ns

Explanation:

Given the following :

Mass of ball = 145g

Initial Speed of ball = 15m/s

Final speed of ball when hit by the batter = - 20m/s ( Opposite direction)

The impulse of a body is represented using the relation:

Force(f) * time(t) = mass (m) * (final Velocity(V) - initial velocity(u))

Therefore, using:

m(v - u) = impulse

Mass of ball = 145 / 1000 = 0.145kg

Impulse = 0.145(- 20 - 15)

Impulse = 0.145(-35)

Impulse = 5.075Ns

3 0
3 years ago
An unstable particle is created in the upper atmosphere from a cosmic ray and travels straight down toward the surface of the ea
german

Answer:

Q1)  Time taking by particle to travel the 40 km wrt. earth = 1.34\times10^{-6} sec.

Q2) The distance traveled by particle where the particle created to surface of earth wrt. particle's frame = 3.84 km.

Q3) The time taking by particle to travel from where it is created to the surface of the earth = 1.285\times10^{-5} sec.

Explanation:

Given :

Speed of particle wrt. earth v=0.99537c

Distance between where particle is created and earth surface = 40 km

we know that,

⇒       v = \frac{x}{t}

Where x = 40\times10^{3} m, v = 0.99537c, we know speed of light c = 3 \times10^{8}

∴      t = \frac{x}{v}

         = \frac{40 \times10^{3} }{0.99537\times3\times10^{8} }

      t = 1.34\times10^{-6} sec

∴ Thus, time taking by particle to travel the 40 km wrt. earth t = 1.34\times10^{-6} sec

According to the lorentz transformation,

⇒    l = l_{o} \sqrt{1-\frac{v^2}{c^2} }

Where l = improper length, l_{o} =proper length (distance measured wrt. rest frame) = 40 km

     l = 40 \sqrt{1-\frac{v^2}{c^2 }

     l = 40 \times 0.096

     l = 3.84 km

∴ Thus, the distance traveled by particle where the particle created to surface of earth wrt. particle's frame = 3.84 km.

According to the time dilation,

   \Delta t = \frac{\Delta t_{o} }{\sqrt{1-\frac{v^2}{c^2} } }

Where \Delta t = improper time (wrt. earth frame time) =1.34\times10^{-6} sec ,  \Delta t _{o} = proper time (wrt. particle frame).

 1.34\times10^{-6} = \frac{ \Delta t_{o}}{0.096}

 \Delta t_{o} = 1.285 \times10^{-5} sec

Thus, the time taking by particle to travel from where it is created to the surface of the earth = 1.285 \times10^{-5} sec.

6 0
3 years ago
Two elements in Period 4 that are most similar to Cobalt?
lisabon 2012 [21]
Iron and nickel are the elements most similar to cobalt
5 0
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Studying neutrinos helped to explain how our Sun works but led to changes in theories of particle physics, how is this process c
julsineya [31]

Explanation:

Yes,  evidently, the process is consistent with the scientific process because in scientific process falsification and modification are two very important traits. So this new concept have modified the existing theories.

Through the modification a theory is adjusted without undermining other discovering made through the theory's prediction.

There are many other evidences that prove this fact for example Einstien's Theory of relativity also changes the existing concepts.

3 0
3 years ago
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