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tia_tia [17]
3 years ago
5

Differences in ocean-surface height can be measured by _____.\

Physics
2 answers:
kakasveta [241]3 years ago
6 0
The answer that best fits the blank is the word SATELLITES. It is the satellites that are used to determine the measurement of the ocean-surface height differences. This is used in Ocean Surface Topography wherein the sea surface is affected by both ocean circulation and gravity. 
bonufazy [111]3 years ago
3 0
<h3><u>Answer;</u></h3>

Satellite

Differences in ocean-surface height can be measured by<u> Satellite</u>

<h3><u>Explanation;</u></h3>
  • The topography of the ocean or the height of the ocean surface relative to a level of no motion provides the information on tides, and the distribution of heat and mass in the Earths's oceans.
  • <em><u>The ocean topography is measured using satellites altimeter. Satellites use radar altimeters that are specially made to measure the height of the ocean surface.  The satellites measure the height of the ocean surface with an accuracy of 3 cm relative to the center of the earth.</u></em>
  • Satellite altimeter combines precise orbit determination with accurate ranging by a microwave altimeter of ocean distance to the satellite.
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Due to radiation and absorption, what is the maximum flame temperature in a fire?
ludmilkaskok [199]

The maximum flame temperature in a fire is 1500-1600K due to radiation and absorption.

<h3>What do you mean by radiation and absorption?</h3>

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1 year ago
Two loudspeakers emit sound waves of the same frequency along the x-axis. The amplitude of each wave is a. The sound intensity i
leonid [27]

Answer:

Explanation:

To find the amplitude of the sound, we must first determine the wavelength and the phase difference between the two speakers.

For the wavelength;

Recall that, the separation between two successive max. and min. intensity points are \dfrac{\lambda}{2}

Thus; for both speakers; the wavelength of the sound is:

\dfrac{\lambda}{2} = (10+30) cm

\dfrac{\lambda}{2} = (40) cm

λ = 80 cm

The relation between the path difference(Δx) and the phase difference(Δ∅) is:

\Delta \phi = \dfrac{2 \pi}{\lambda}\Delta x + \Delta \phi_o

where;

Δx = 10 cm

λ = 80 cm

Δ∅ = π rad

∴

\Delta \phi = \dfrac{2 \pi}{\lambda}\Delta x + \Delta \phi_o

\pi \ rad  = \dfrac{2 \pi}{80 \ cm}(10 \ cm) + \Delta \phi_o

\pi \ rad  = \dfrac{2 \pi}{8}+ \Delta \phi_o

\pi \ rad  = \dfrac{ \pi}{4}+ \Delta \phi_o

\Delta \phi_o  =  \pi -\dfrac{ \pi}{4}

\Delta \phi_o  = \dfrac{ 4\pi - \pi}{4}

\Delta \phi_o  = \dfrac{ 3\pi}{4} \ rad

Suppose both speakers are placed side-by-side, then the path difference between the two speakers is: Δx = 0 cm

Thus, we have:

\Delta \phi = \dfrac{2 \pi}{\lambda}\Delta x + \Delta \phi_o

\Delta \phi = \dfrac{2 \pi}{\lambda}(0 \ cm ) + \dfrac{3 \pi}{4} \ rad

\Delta \phi = \dfrac{3 \pi}{4} \ rad

∴

The amplitude of the sound wave if the two speakers are placed side-by-side is:

A = 2a \ cos \bigg (\dfrac{\Delta \phi }{2} \bigg)

A = 2a \ cos \bigg (\dfrac{\dfrac{3 \pi}{4} }{2} \bigg)

A = 2a \ cos \bigg ({\dfrac{3 \pi}{8} } \bigg)

A = 0.765a

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Answer:

Convex lens and convex mirrors are similar that

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Answer:

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