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ddd [48]
3 years ago
9

explain how climatic shifts that are naturally occuring and not the result of some castrophic events like asteroid impact happen

?
Physics
1 answer:
77julia77 [94]3 years ago
6 0

Answer:

Climatic shift occur due to changes in climate systems caused by varying  energy balances happening on earth.Asteroids can trigger increase in kinetic energy with global effect but can not really cause a change in climate systems

Explanation:

Climatic shift could be caused by increased levels of greenhouse gases in the atmosphere,changes in  sun strength, shift of ocean currents, movement of tectonic plates and changes in levels of carbondioxide in the ocean.All these factors and others are natural occurring and should balance to create a specific climatic condition.Changes in one of these will led to addition on another resulting to climatic shift.However, for the asteroid, an impact on earth could result to dust and aerosols resulting to pollution, but not really a climatic shift.

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A constant electric field with magnitude 1.50 ✕ 103 N/C is pointing in the positive x-direction. An electron is fired from x = −
romanna [79]

Answer:

The speed of electron is 1.5\times10^{7}\ m/s

Explanation:

Given that,

Electric field E=1.50\times10^{3}\ N/C

Distance = -0.0200

The electron's speed has fallen by half when it reaches x = 0.190 m.

Potential energy P.E=5.04\times10^{-17}\ J

Change in potential energy \Delta P.E=-9.60\times10^{-17}\ J as it goes x = 0.190 m to x = -0.210 m

We need to calculate the work done by the electric field

Using formula of work done

W=-eE\Delta x

Put the value into the formula

W=-1.6\times10^{-19}\times1.50\times10^{3}\times(0.190-(-0.0200))

W=-5.04\times10^{-17}\ J

We need to calculate the initial velocity

Using change in kinetic energy,

\Delta K.E = \dfrac{1}{2}m(\dfrac{v}{2})^2+\dfrac{1}{2}mv^2

\Delta K.E=\dfrac{-3mv^2}{8}

Now, using work energy theorem

\Delta K.E=W

\Delta K.E=\Delta U

So, \Delta U=W

Put the value in the equation

\dfrac{-3mv^2}{8}=-5.04\times10^{-17}

v^2=\dfrac{8\times(5.04\times10^{-17})}{3m}

Put the value of m

v=\sqrt{\dfrac{8\times(5.04\times10^{-17})}{3\times9.1\times10^{-31}}}

v=1.21\times10^{7}\ m/s

We need to calculate the change in potential energy

Using given potential energy

\Delta U=-9.60\times10^{-17}-(-5.04\times10^{-17})

\Delta U=-4.56\times10^{-17}\ J

We need to calculate the speed of electron

Using change in energy

\Delta U=-W=-\Delta K.E

\Delta K.E=\Delta U

\dfrac{1}{2}m(v_{f}^2-v_{i}^2)=4.56\times10^{-17}

Put the value into the formula

v_{f}=\sqrt{\dfrac{2\times4.56\times10^{-17}}{9.1\times10^{-31}}+(1.21\times10^{7})^2}

v_{f}=1.5\times10^{7}\ m/s

Hence, The speed of electron is 1.5\times10^{7}\ m/s

4 0
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Answer:

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