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3 years ago

10

explain why earths acceleration is usually very small compared to the acceleration of the object the earth interact with

1 answer:

Marianna [84]3 years ago

6 0

Answer:

F-ma

Explanation:

If you are speaking of objects like satellites, etc. then their mass is much less than that of the Earth. A good approximation is Newton's first law of motion:

Force = Mass × Acceleration

often written:

F = m a

The gravitational force is the same between the Earth and the object - only the mass differs. So the acceleration is inversely proportional to the mass.

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TEA [102]

Answer:

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Explanation:

Because it is impossible for it to show the real depth of the ocean and how deep it is

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3 years ago

if somebody swim to the right for 30 meters then swims to left for 15 meters and then swims back to the right for 50 meters what

zubka84 [21]

50m

Explanation:

Displacement is the length of path traveled which is measured from start to the finishing of the path.

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3 years ago

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3 years ago

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4 0

3 years ago

Chapter 14, Problem 042 A flotation device is in the shape of a right cylinder, with a height of 0.588 m and a face area of 4.19

Alisiya [41]

Answer:

The workdone is $W = 9.28 * 10^{3} J$

Explanation:

From the question we are told that

The height of the cylinder is $h = 0.588\ m$

The face Area is $A = 4.19 \ m^2$

The density of the cylinder is $\rho = 0.346 * \rho_w$

Where $\rho_w$ is the density of freshwater which has a constant value

$\rho_w = 1000 kg/m^3$

Now

Let the final height of the device under the water be $= h_f$

Let the initial volume underwater be $= V_n$

Let the initial height under water be $= h_i$

Let the final volume under water be $= V_f$

According to the rule of floatation

The weight of the cylinder = Upward thrust

This is mathematically represented as

$\rho_c g V_n = \rho_w gV_f$

$\rho_c A h = \rho A h_f$

So $\frac{0.346 \rho_w}{\rho_w} = \frac{h_f}{h}$

=> $\frac{h_f}{h_c} = 0.346$

Now the work done is mathematically represented as

$W = \int\limits^{h_f}_{h} {\rho_w g A (-h)} \, dh$

$= \rho_w g A [\frac{h^2}{2} ] \left | h_f} \atop {h}} \right.$

$= \frac{g A \rho}{2} [h^2 - h_f^2]$

$= \frac{g A \rho}{2} (h^2) [1 - \frac{h_f^2}{h^2} ]$

Substituting values

$W = \frac{(9.8 ) (4.19) (10^3)}{2} (0.588)^2 (1 - 0.346)$

$W = 9.28 * 10^{3} J$

4 0

3 years ago