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3 years ago

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

2 answers:

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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dezoksy [38]

<h2><em>The Answer to Youre Question:</em></h2>

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3 years ago

The distance between two objects is increased by three times the oringinal distance. How will this change the force of attractio

tigry1 [53]

<span>The distance between two objects is increased by three times the oringinal distance. Since they were already separated by one time the original distance,
the additional three times the oringinal distance now puts them four times the original distance apart.

Whether we're talking about the gravitational forces of attraction or
the electrical forces of attraction, either one is inversely proportional
to the square of the distance between the objects.

So changing the distance to four times the original distance causes
the forces to become 1/4</span>² as strong as they were originally.

The forces become 1/16 of their original magnitude.<span>
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8 0

3 years ago

A 10.0 cm3 sample of copper has a mass of 89.6

Romashka-Z-Leto [24]

Density is mass divides by volume, so
89.6g / 10cm^3 =8.96g /cm^3

*cm^3 is a standard unit of volume*

4 0

4 years ago

A billiard ball is moving in the x-direction at 30.0 cm/s and strikes another billiard ball moving in the y-direction at 40.0 cm

ipn [44]

To solve this problem it is necessary to apply the trigonometric ratios of the given velocity components.

If we make a graph of the velocity vectors in their respective velocities according to the given description we will have something similar to the attached graph.

The angle could be obtained from the components of the opposite leg and the adjacent leg so that

$tan\theta = \frac{x}{y}$

$\theta = tan^{-1}(\frac{x}{y})$

The opposite leg value (y) is 40cm / s and the adjacent leg (x) is 30cm / s

$\theta = tan^{-1}(\frac{30}{40})$

$\theta = 36.87\°$

Therefore the final direction that does the first ball is 36.87°

6 0

3 years ago

You are watching an archery tournament when you start wondering how fast an arrow is shot from the bow. Remembering your physics

spayn [35]

Answer:

$v_0 = 3.53~{\rm m/s}$

Explanation:

This is a projectile motion problem. We will first separate the motion into x- and y-components, apply the equations of kinematics separately, then we will combine them to find the initial velocity.

The initial velocity is in the x-direction, and there is no acceleration in the x-direction.

On the other hand, there no initial velocity in the y-component, so the arrow is basically in free-fall.

Applying the equations of kinematics in the x-direction gives

$x - x_0 = v_{x_0} t + \frac{1}{2}a_x t^2\\63 \times 10^{-3} = v_0t + 0\\t = \frac{63\times 10^{-3}}{v_0}$

For the y-direction gives

$v_y = v_{y_0} + a_y t\\v_y = 0 -9.8t\\v_y = -9.8t$

Combining both equation yields the y_component of the final velocity

$v_y = -9.8(\frac{63\times 10^{-3}}{v_0}) = -\frac{0.61}{v_0}$

Since we know the angle between the x- and y-components of the final velocity, which is 180° - 2.8° = 177.2°, we can calculate the initial velocity.

$\tan(\theta) = \frac{v_y}{v_x}\\\tan(177.2^\circ) = -0.0489 = \frac{v_y}{v_0} = \frac{-0.61/v_0}{v_0} = -\frac{0.61}{v_0^2}\\v_0 = 3.53~{\rm m/s}$

6 0

3 years ago