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3 years ago

Winding roads are made on hills, why?

1 answer:

3 0

Answer.

The road on the hills goes upward with the increasing length of the road.That's why we make a winding roads because it reduces the allover surface inclination in every portion of the road which is suitable for driving the vehicles.

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3 years ago

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 0.199 rad/s

Second question:

The telescope should accelerates away by approximately F = 0.0005·m

Third question:

The pulling force between the Earth and the satellite

<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 ≈ 0.199 rad/s

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 ≈ 0.0005·m (away from the Sun)

Third part:

Other forces include;

The force of attraction between the Earth and the telescope 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

Winding roads are made on hills, why?​

Winding roads are made on hills, why?