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Westkost [7]
3 years ago
7

On December 26, 2004, a great earthquake occurred off the coast of Sumatra and triggered immense waves (tsunami) that killed som

e 200000 people. Satellites observing these waves from space measured 800 kmkm from one wave crest to the next and a period between waves of 1.0 hour. Part A What was the speed of these waves in m/sm/s
Physics
2 answers:
horrorfan [7]3 years ago
5 0

Answer:

The speed of these waves is 222.22 m/s

Explanation:

A.

Since, we know that the speed of a wave is given by the formula:

v = fλ

where,

v = speed of wave

f = frequency of wave

λ = wavelength

Here,

λ = distance between consecutive wave crests

λ = 800 km = 8 x 10⁵ m

f = 1/Time Period = (1/1 hr)(1 hr/3600 s)

f = 2.77 x 10⁻⁴ s⁻¹

Therefore,

v = (2.77 x 10⁻⁴ s⁻¹)(8 x 10⁵ m)

<u>v = 222.22 m/s</u>

FromTheMoon [43]3 years ago
5 0

Answer:

Explanation:

Given that,

200,000 people where killed due to earth quake in December 2004

Satellite observing measure a wavelength of 800km crest to crest

Crest to crest implies one wavelength

λ = 800km = 800,000m

Period of One hour

T = 1 hour = 3600s

Then, we want to find speed of wave.

Using wave equation

v = fλ

Where

f is frequency in Hz

v is speed of wave in m/s

λ is wavelength in m

Where f = 1 / T

Then, v = λ / T

v = 800,000 / 3600

v = 222.22 m/s

Then, the speed of the wave is 222.22 m/s

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3 years ago
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b) The two bodies have emission in this region, the body of 3000K in the part of rise of the emission and the body to 12000K in the descent of the emission even when this body emits 256 times more than the other, so this body should have the highest broadcast in this area

c) The emission of the hottest 12000K body is mainly in UV

d) The hottest body emits more energy in UV and visible

e) No body has greater emission in all zones

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