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3 years ago

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One consequence of Newton's third law of motion is that __________. A. every object that has mass has inertia B. a force acting

upon an object increases that objects acceleration C. all actions have equal and opposite reactions D. none of the above

2 answers:

bazaltina [42]3 years ago

7 0

Answer:

C. All actions have an equal and opposite reaction.

Explanation:

Arlecino [84]3 years ago

7 0

Answer: the answer is C

Explanation: just took test made 100%

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When blueshift occurs,the preceived frequency of the wave would be?

LiRa [457]

Answer:

When blueshift happens, the perceived frequency of the wave would be higher than the actual frequency.

Explanation:

As the name suggests, when blueshift happens to electromagnetic waves, the frequency of the observed wave would shift towards the blue (high-frequency) end of the visible spectrum. Hence, there would be an increase to the apparent frequency of the wave.

Blueshifts happens when the source of the wave and the observer are moving closer towards one another.

Assume that the wave is of frequency $f\; {\rm Hz}$ at the source. In other words, the source of the wave sends out a peak after every $(1/f)\; {\text{seconds}}$ .

Assume that the distance between the observer and the source of the wave is fixed. It would then take a fixed amount of time for each peak from the source to reach the observer.

The source of this wave sends out a peak after each period of $(1/f)\; {\text{seconds}}$ . It would appear to the observer that consecutive peaks arrive every $(1/f)\; {\text{seconds}}\!$ . That would correspond to a frequency of $f\; {\rm Hz}$ .

On the other hand, for a blueshift to be observed, the source of the wave needs to move towards the observer. Assume that the two are moving towards one another at a constant speed of $v \; {\rm m \cdot s^{-1}}$ .

Again, the source of this wave would send out a peak after each period of $(1/f)\; {\text{seconds}}$ . However, by the time the source sends out the second peak, the source would have been $v \cdot (1 / f) \; { \rm m}= (v / f)\; {\rm m}$ closer to the observer then when the source sent out the first peak.

When compared to the first peak, the second peak would need to travel a slightly shorter distance before it reach the observer. Hence, from the perspective of the observer, the time difference between the first and the second peak would be shorter than $(1/f)\; {\text{seconds}}$ . The observed frequency of this wave would be larger than the original $f\; {\rm Hz}$ .

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3 years ago

Which method would be best to separate a mixture of iron and lead pellets (of the same approximate diameter)? magnetism screenin

KATRIN_1 [288]

Separation by density filtration because lead slightly interacts with magnets also.

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3 years ago

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Cells are visualized using light microscopes and electron microscopes. Which statement about microscopes is false. Group of answ

denpristay [2]

Answer:

Light microscopes use light and glass objectives to illuminate and magnify objects

Explanation:

Light microscopes and electron microscopes are used to study cells. The electron microscope has many times more resolving power than the ordinary light microscope. A light microscope contains an objective lens and an eyepiece through which the final image is seen.

Both light and electron microscopes magnifies the image of the object. The magnifying power of an electron microscope is many times that of the light microscope.

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3 years ago

The number ocean waves that pass a buoy in one second is _ of the wave

mr_godi [17]

The number of ocean waves that pass a buoy in one second is the frequency of the <span>wave. The crest of a transverse wave is its highest point. </span>

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3 years ago

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A light source shines light consisting of two wavelengths, λ1 = 540 nm (green) and λ2 = 450 nm (blue), on two slits separated by

densk [106]

Answer:

The maximun distance is $z_1 = z_2 = 0.0138m$

Explanation:

From the question we are told that

The wavelength are $\lambda _ 1 = 540nm (green) = 540 *10^{-9}m$

$\lambda_2 = 450nm(blue) = 450 *10^{-9}m$

The distance of seperation of the two slit is $d = 0.180mm = 0.180 *10^{-3}m$

The distance from the screen is $D = 1.53m$

Generally the distance of the bright fringe to the center of the screen is mathematically represented as

$z = \frac{m \lambda D}{d}$

Where m is the order of the fringe

For the first wavelength we have

$z_1 = \frac{m_1 (549 *10^{-9} * (1.53))}{0.180*10^{-3}}$

$z_1=0.00459m_1 m$

$z_1= 4.6*10^{-3}m_1 m ----(1)$

For the second wavelength we have

$z_2 = m_2 \frac{450*10^{-9} * 1.53 }{0.180*10^{-3}}$

$z_2 = 0.003825m_2$

$z_2 = 3.825 *10^{-3} m_2 m$ ----(2)

From the question we are told that the two sides coincides with one another so

$zy_1 =z_2$

$4.6*10^{-3}m_1 m = 3.825 *10^{-3} m_2 m$

$\frac{m_1}{m_2} = \frac{3.825 *10^{-3}}{4.6*10^{-3}}$

Hence for this equation to be solved

$m_1 = 3$

and $m_2 = 4$

Substituting this into the equation

$z_1 = z_2 = 3 * 4.6*10^{-3} = 4* 3.825*10^{-3}$

Hence $z_1 = z_2 = 0.0138m$

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3 years ago