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frosja888 [35]
3 years ago
13

Which of these is exhibiting kinetic energy?

Physics
2 answers:
Stella [2.4K]3 years ago
8 0

Answer:

A.

Explanation:

It's the only moving thing.

pychu [463]3 years ago
3 0

Answer:a

Explanation:

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The distance between two consecutive nodesof a standing wave is 20.9cm.Thehandgen-erating the pulses moves up and down throughac
irina1246 [14]

Answer:

Velocity, v = 0.239 m/s

Explanation:

Given that,

The distance between two consecutive nodes of a standing wave is 20.9 cm = 0.209 m

The hand generating the pulses moves up and down through a complete cycle 2.57 times every 4.47 s.

For a standing wave, the distance between two consecutive nodes is equal to half of the wavelength.

\dfrac{\lambda}{2}=0.209\ m\\\\\lambda=0.418\ m

Frequency is number of cycles per unit time.

f=\dfrac{2.57}{4.47}\\\\f=0.574\ Hz

Now we can find the velocity of the wave.

Velocity = frequency × wavelength

v = 0.574 × 0.418

v = 0.239 m/s

So, the velocity of the wave is 0.239 m/s.

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3 years ago
A factor that determines the inertia of an object is
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Height, weight, size
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What is voltage and current?
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3 years ago
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Atoms with the same atomic number but different atomic mass are called
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Assume the average value of the vertical component of Earth's magnetic field is 42 μT (downward) in some region that has an area
Oliga [24]

Answer:

The net magnetic flux through the rest of Earth's surface is  5.782\times10^{-2}\ T

Explanation:

Given that,

Magnetic field = 42 μT

Area A=3.71\times10^{5}\ km^2

A=3.71\times10^{11}\ m^2

We need to calculate the flux per unit area

flux\ per\ unit\ area=\dfrac{42\times10^{-6}}{3.71\times10^{11}}

flux\ per\ unit\ area=1.132\times10^{-16}\ T/m^2

We need to calculate the total earth's surface area

A'=4\pi r^2

A'=4\times\pi\times(6.3781\times10^{6})^2

A'=5.1120\times10^{14}\ m^2

We need to calculate the rest of earth's area

A''=A-A'

Put the value into the formula

A''=5.1120\times10^{14}-3.71\times10^{11}

A''=5.10829\times10^{14}\ m^2

We need to calculate the net magnetic flux through the rest of Earth's surface

B'=5.10829\times10^{14}\times1.132\times10^{-16}

B'=5.782\times10^{-2}\ T

Hence, The net magnetic flux through the rest of Earth's surface is  5.782\times10^{-2}\ T

3 0
4 years ago
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