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3 years ago

According to the doppler effect What appears to happen when a light source moves closer to an observer

1 answer:

7 1

When the distance between the source and the observer is DEcreasing, the observer sees light that has a<em> shorter wavelength</em> than the light actually had when it left the source.

It doesn't matter how the distance between them is decreasing. The source could be moving toward the observer, the observer could be moving toward the source, or they could both be moving in a way that makes the separation between them decrease as time goes on. The result is the same.

(When this is used in Astronomy and Astrophysics, the big trick is: How does the observer know what the wavelength actually WAS when the light left the source ?)

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If a roller coaster has 50,000 J of potential energy at the top of the first hill, how much kinetic energy does it have at the l

Fed [463]

Answer:

50,000 J because of the transformation of energy

Explanation:

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3 years ago

A standard inverting op-amp circuit has an R1 of 10 kΩ and an Rf of 220 kΩ. If the offset current is 100 nA the output offset vo

kiruha [24]

Answer:

The value is $V_{os} = 0.001 \ V$

Explanation:

From the question we are told that

The circuit resistance is $R_1 = 10 \ k \Omega$

The feedback resistance is $R_f = 220 \ k \Omega$

The offset current is $I_{os } = 100 \ nA = 100 * 1)^{-9} \ A$

Generally the offset voltage is mathematically reparented as

$V_{os} = R_f * I_{os}$

=> $V_{os} = 10 *10^{3}* 100 *10^{-9}$

=> $V_{os} = 0.001 \ V$

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3 years ago

a motorcycle accelerates from 15 m/s to 20 m/s over a distance of 50 meters. what is its average acceleration?

devlian [24]

For this, you need the v-squared equation, which is v(final)² = v(initial)² + 2aΔx The averate acceleration is thus a = (v(final)² - v(initial)²) / 2Δx = (20² - 15²) / 2(50) = 175 / 100 = 1.75 m/s² So the average acceleration is 1.75 m/s²

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4 years ago

How much charge does a 9.0 v battery transfer from the negative to the positive terminal while doing 39 j of work?

sdas [7]

The work done by the battery is equal to the charge transferred during the process times the potential difference between the two terminals of the battery:
$W=q \Delta V$
where q is the charge and $\Delta V$ is the potential difference.

In our problem, the work done is W=39 J while the potential difference of the battery is $\Delta V = 9.0 V$ , so we can find the charge transferred by the battery:
$q= \frac{W}{\Delta V}= \frac{39 J}{9.0 V}=4.33 C$

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4 years ago

You push on a 30 kg box with a force of 120 N. What is the acceleration of the box2

kumpel [21]

Answer:

Explanation:

120/30=6

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3 years ago