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3 years ago

10

What is the formula for displacement?

2 answers:

qwelly [4]3 years ago

7 0

Displacement = final - initial
The formula most closely resembling that is delta = xf - xi

Montano1993 [528]3 years ago

3 0

D = xf - xi , so it would be the first choice.

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A speed boat travels from the dock to the first buoy a distance of 20 meters in 18 seconds it began the trip at a speed of 0 m/s

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1.11 m/s

Explanation:

The motion of the boat is an example of accelerated motion, since the velocity is not constant. However, we don't need to find the acceleration, because we are only interested in the average velocity of the boat, which is given by:

$v=\frac{d}{t}$

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3 years ago

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4 years ago

The______ of a sound wave is defined as the amount of energy passing through a unit area of the wave front in a unit of time

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4 years ago

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Solution :

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$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

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Acceleration due to gravity at 2000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-2000) \times 5.5 \times 10^3\right)$

$= 673552 \times 10^{-8}$

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= 6.73 m/s

Acceleration due to gravity at 3000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-3000) \times 5.5 \times 10^3\right)$

$= 3371 \times 153.86 \times 10^{-8}$

= 5.18 m/s

Acceleration due to gravity at 4000 km depths is :

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$= 153.84 \times 2371 \times 10^{-8}$

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= 3.64 m/s

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3 years ago