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3 years ago

A train moves from rest to a speed of 25 m/s in 30.0 seconds. What is its acceleration?

Physics

1 answer:

iris [78.8K]3 years ago

4 0

Answer:

Explanation:

The equation for acceleration is

$a=\frac{v_f-v_0}{t}$ where vf is the final velocity and v0 is the initial velocity. For us, the final velocity is 25 m/s and the initial is 0 because the train started from resting position. Filling in and solving for a:

$a=\frac{25-0}{30.0}$ so

a = .83 m/s/s

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The James Webb Space Telescope is positioned around 1.5 million kilometres from the Earth on the side facing away from the Sun.

Bad White [126]

The angular velocity depends on the length of the orbit and the orbital

speed of the telescope.

Response:

First question:

The angular velocity of the telescope is approximately <u>0.199 rad/s</u>

Second question:

The telescope should accelerates away by approximately F = <u>0.0005·m </u>

Third question:

<u>The pulling force between the Earth and the satellite</u>

<h3>What equations can be used to calculate the velocity and forces acting on the telescope?</h3>

The distance of the James Webb telescope from the Sun = 1.5 million kilometers from Earth on the side facing away from the Sun

The orbital velocity of the telescope = The Earth's orbital velocity

First question:

$Angular \ velocity = \mathbf{\dfrac{Angle \ turned}{Time \ taken}}$

The orbital velocity of the Earth = 29.8 km/s

The distance between the Earth and the Sun = 148.27 million km

The radius of the orbit of the telescope = 148.27 + 1.5 = 149.77

Radius of the orbit, r = 149.77 million kilometer from the Sun

The length of the orbit of the James Webb telescope = 2 × π × r

Which gives;

r = 2 × π × 149.77 million kilometers ≈ 941.03 million kilometers

Therefore;

$Angular \ velocity = \dfrac{29.8}{941.03}\times 2 \times \pi \approx 0.199$

The angular velocity of the telescope, ω ≈ <u>0.199 rad/s</u>

Second question:

Centrifugal force force, $F_{\omega}$ = m·ω²·r

Which gives;

$F_{\omega} = m \cdot \dfrac{28,500^2 \, m^2/s^2}{149.77 \times 10^9 \, m} \approx 0.0054233 \cdot m$

$Gravitational \ force, F_G = \mathbf{G \cdot \dfrac{m_{1} \cdot m_{2}}{r^{2}}}$

Universal gravitational constant, G = 6.67408 × 10⁻¹¹ m³·kg⁻¹·s⁻²

Mass of the Sun = 1.989 × 10³⁰ kg

Which gives;

$F_G = 6.67408 \times 10^{-11} \times \dfrac{1.989 \times 10^{30} \times m}{149.77 \times 10^9} \approx 0.00592 \cdot m$

Which gives;

$F_{\omega}$ < $F_G$ , therefore, the James Webb telescope has to accelerate away from the Sun

F = $\mathbf{F_{\omega}}$ - $\mathbf{F_G}$

The amount by which the telescope accelerates away is approximately 0.00592·m - 0.0054233·m ≈ <u>0.0005·m (away from the Sun)</u>

Third part:

Other forces include;

<u>The force of attraction between the Earth and the telescope </u>which can contribute to the the telescope having a stable orbit at the given speed.

Learn more about orbital motion here:

brainly.com/question/11069817

3 0

3 years ago

Why are different constellations<br> of stars seen during different<br> seasons?

slamgirl [31]

Actually, they're not. There's a group of stars and constellations arranged
around the pole of the sky that's visible at any time of any dark, clear night,
all year around. And any star or constellation in the rest of the sky is visible
for roughly 11 out of every 12 months ... at SOME time of the night.

Constellations appear to change drastically from one season to the next,
and even from one month to the next, only if you do your stargazing around
the same time every night.

Why does the night sky change at various times of the year ? Here's how to
think about it:

The Earth spins once a day. You spin along with the Earth, and your clock is
built to follow the sun . "Noon" is the time when the sun is directly over your
head, and "Midnight" is the time when the sun is directly beneath your feet.

Let's say that you go out and look at the stars tonight at midnight, when you're
facing directly away from the sun.

In 6 months from now, when you and the Earth are halfway around on the other
side of the sun, where are those same stars ? Now they're straight in the
direction of the sun. So they're directly overhead at Noon, not at Midnight.

THAT's why stars and constellations appear to be in a different part of the sky,
at the same time of night on different dates.

5 0

4 years ago

Read 2 more answers

In 1666 at the age of 23, what scientist

scoray [572]

Answer:

<h3>B - Isaac Newton</h3>

Explanation:

He first thought of his system of gravitation which he hit upon by observing an apple fall from a tree,

The incident occurring in the late summer of 1666.

3 0

3 years ago

The given function represents the position of a particle traveling along a horizontal line. s(t) = 2t3 − 3t2 − 36t + 6 for t ≥ 0

igor_vitrenko [27]

1) The velocity of the particle is given by the derivative of the position. So, if we derive s(t), we get the velocity of the particle as a function of the time:

$v(t)=s'(t)=(2t^3-3t^2-36t+6)'=6t^2-6t-36$

2) The acceleration of the particle is given by the derivative of the velocity. So, if we derive v(t), we get the acceleration of the particle as a function of the time:

$a(t)=v'(t)=(6t^2-6t-36)'=12t-6$

8 0

3 years ago

The __________ energy in food is changed into mechanical energy by your muscles.

Orlov [11]

Potential energy which is the stored energy an object has waiting to be used

8 0

3 years ago