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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 $f\; {\rm Hz}$ .

6 0

3 years ago

Help on matching please, struggling on it. Even just 1 or 2 would help

n200080 [17]

Answer:

7. free fall -- h. 9.8m/s^2

3. Velocity -- x. 60 km/hr west

6. Acceleration -- d. change in velocity/time

8. Centrifugal -- s. towards the centre

13. Work done --w. Force * displacement

5. Uniform circular motion --j. spin cycle in washer

18. Power -- r. kW an hour

7. g -- a. 10N

hope this helps

6 0

3 years ago

Heat transfer occurs

Archy [21]

It depends on the tempature of something or an object,
The Object has to go from hot to cold, that is a primary example of how heat transfer occurs

8 0

3 years ago

Bill and Ted are standing on a bridge 40 ft above a river. Bill drops a stone, while Ted decides to throw a stone downward at 10

Y_Kistochka [10]

Answer:

D) - 0.72 secs

Explanation:

Parameters given:

Height of bridge = 40ft = 12.19 m

Initial velocity of Bill's stone = 0m/s

Initial velocity of Ted's stone = 10m/s

We find the time it take Bill's stone to bit the river and the time it takes Ted's stone to hit the river. Then we find the time difference.

Using one of the equations of motion:

For Bill:

S = ut + ½gt²

Where g = 9.8 m/s

12.19 = 0 + ½*9.8*t²

t² = 12.19/4.9 = 2.49

t = 1.58 secs

For Ted:

S = uT + ½gT²

12.19 = 10*T + ½*9.8*T²

=> 4.9T² + 10T - 12.19 = 0

Using quadratic formula and retaining only the positive value, we get that:

T = 0.86 secs

Time difference between Bill's throw and Ted's throw is:

0.86 - 1.58 = - 0.72 secs

In reality, this means that Ted must throw his stone 0.72 secs before Bill throws his for both stones to land the same time.

6 0

4 years ago

Johannes Kepler used math to show that the planets move in perfect circles around the sun.

svet-max [94.6K]

Kepler actually showed that the planets move around the sun in ellipses, not circles. So the answer is false.

4 0

3 years ago

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