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3 years ago

7

What does Newtons first law of motion say about objects at rest and objects in motion?

1 answer:

Alborosie3 years ago

6 0

Answer:

Explanation:

an object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force.

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Why are the orbits of planets only nearly circular and not perfectly circular?

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Explanation:

this is my answer this is helpful for you

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2 years ago

1.imagine you were able to throw a ball in a frictionless environment such as outer space.once you let go of the ball,what will

Scorpion4ik [409]

Imagine you were able to throw a ball in a frictionless environment
such as outer space. Once you let go of the ball, it will travel forever
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3 years ago

Which of the following was not a famous basketball player for the NBA? Select one: a. Wilt Chamberlain b. Lady Bird Johnson c. M

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Answer:

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3 years ago

What happens to a radioactive isotope as it decays?

kupik [55]

It becomes a different element

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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