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2 years ago

What are the names of neptunes moons

Physics

1 answer:

alexandr402 [8]2 years ago

5 0

Answer:

1. naiad

2.Thalassa

3 Despina

4 Galatea

5 Larissa

6 Hippocamp

7 Proteus ˈ

8 Triton

9 Nereid

10 Halimede

11 Sao

12 Laomedeia

13 Psamathe

14 Neso

Explanation:

hope this helped

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2 years ago

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A particle moves in a straight line with the velocity function v ( t ) = sin ( w t ) cos 3 ( w t ) . find its position function

Sunny_sXe [5.5K]

Integrating the velocity equation, we will see that the position equation is:

$$f(t)=\frac{\cos ^3(\omega t)-1}{3}$

<h3>How to get the position equation of the particle?</h3>

Let the velocity of the particle is:

$$v(t)=\sin (\omega t) * \cos ^2(\omega t)$

To get the position equation we just need to integrate the above equation:

$$f(t)=\int \sin (\omega t) * \cos ^2(\omega t) d t$

$$\mathrm{u}=\cos (\omega \mathrm{t})$

Then:

$$d u=-\sin (\omega t) d t$

$\Rightarrow d t=-d u / \sin (\omega t)$

Replacing that in our integral we get:

$\int \sin (\omega t) * \cos ^2(\omega t) d t$

$-\int \frac{\sin (\omega t) * u^2 d u}{\sin (\omega t)}-\int u^2 d t=-\frac{u^3}{3}+c$

Where C is a constant of integration.

Now we remember that $u=\cos (\omega t)$

Then we have:

$$f(t)=\frac{\cos ^3(\omega t)}{3}+C$

To find the value of C, we use the fact that f(0) = 0.

$$f(t)=\frac{\cos ^3(\omega * 0)}{3}+C=\frac{1}{3}+C=0$

C = -1 / 3

Then the position function is:

$$f(t)=\frac{\cos ^3(\omega t)-1}{3}$

Integrating the velocity equation, we will see that the position equation is:

$$f(t)=\frac{\cos ^3(\omega t)-1}{3}$

To learn more about motion equations, refer to:

brainly.com/question/19365526

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1 year ago

A 38.2 kg wagon is towed up a hill inclined at 17.5 ◦ with respect to the horizontal. The tow rope is parallel to the incline an

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Answer:

v = 8.57 m/s

Explanation:

As we know that the wagon is pulled up by string system

So the net force on the wagon along the inclined is due to tension in the rope and component of weight along the inclined plane

So as per work energy theorem we know that

work done by tension force + work done by force of gravity = change in kinetic energy

$F_t . d - (mgsin\theta)(d) = \frac{1}{2}mv^2 - 0$

so we have

$F_t = 129 N$

$\theta = 17.5^o$

m = 38.2 kg

d = 85.4 m

so now we have

$129(85.4) - (38.2)9.8sin17.5 (85.4) = \frac{1}{2}(38.2) v^2$

$v = 8.57 m/s$

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2 years ago

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2 years ago

The combination of all forces that act on an object are (2 points)

qaws [65]

Answer:

A) The net force

Explanation:

If two forces of equal strength act on an object in opposite directions, the forces will cancel, resulting in a net force of zero and no movement.

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3 years ago