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VARVARA [1.3K]
3 years ago
7

A small 22 kilogram canoe is floating downriver at a speed of 3 m/s. What is the canoe's kinetic energy?

Physics
2 answers:
tekilochka [14]3 years ago
6 0
Kinetic energy = 1/2 (mass)(speed squared).

According to somebody on the river bank watching the canoe float by, it has 99 joules of kinetic energy.

According to anybody in another boat, floating down the same river, the canoe has no kinetic energy at all.

They're both correct, each in his own frame of reference.
amid [387]3 years ago
4 0
Kinetic energy can be calculated using the following rule:
kinetic energy = 0.5*m*v^2 where:
m is the mass = 22 kg
v is the velocity = 3 m/sec

Substitute with the givens in the above equation to get the kinetic energy as follows:
kinetic energy = 0.5(22)(3)^2
kinetic energy = 99 Joules
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Your science teacher gives you three liquids to pour into a jar. After pouring, the liquids layer as seen here. What property of
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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 = −
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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

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