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2 years ago

Why is the orbit of the telescope stable nonetheless? What otherforces need to be considered?

1 answer:

7 0

The orbit of the telescope stable because of effect of gravity due to the Sun and another factor maybe position of telescope.

<h3>Effect of gravitational pull of the sun</h3>

The sun exerts a great gravitational force on every object within its radius, thereby creating a stability of the object of planet.

The acceleration due to gravity of the sun is about 275 m/s². This value shows how great the force of attraction of the sun is, since force of attraction is the product of mass and acceleration.

F = ma

where;

m is mass
a is acceleration

Thus, the orbit of the telescope stable because of effect of gravity due to the Sun and another factor maybe position of telescope.

Learn more about acceleration due to gravity here: brainly.com/question/88039

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Which energy resource is renewable?

Alexandra [31]

The answer is B. natural gas

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3 years ago

A supply plane needs to drop a package of food to scientists working on a glacier in greenland. the plane flies 100 m above the

pishuonlain [190]

The package should be dropped <u>678 m</u> short of the target.

A package dropped from a plane which is moving at a speed v, has a horizontal velocity equal to the horizontal velocity of the plane. It has a parabolic trajectory, traversing a horizontal range x while it falls through a vertical height y.

The package has no initial vertical velocity, and it falls through a height y under the action of the acceleration due to gravity g.

Use the equation,

$y=\frac{1}{2} gt^2$

Write an expression for t, the time taken for the package to fall through y.

$t^2=\frac{2y}{g}$

Substitute 100 m for y and 9.81m/s² for g.

$t^2=\frac{2y}{g}\\ =\frac{2(100m)}{9.81m/s^2} \\ =20.39s^2\\ t=4.52s$

In the time t the package travels a horizontal distance x. The horizontal velocity of the package remains constant, since no force acts along the horizontal direction.

Therefore, the horizontal distance traveled by the package is given by,

$x=vt\\ =(150m/s)(4.52s)\\ =678m$

If the package is released <u>678m</u> before the target, the package would reach the scientists working in Greenland.

6 0

4 years ago

How much heat is needed to melt a 3 kg block of cheddar cheese initially at room tmperature of 24 degrees C? (assume the melting

Mnenie [13.5K]

Cheese I'm so sorry for not having an answer I failed you

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4 years ago

Three ideal polarizing filters are stacked, with the polarizing axis of the second and third filters at 21 degrees and 61 degree

kvv77 [185]

Answer:

When second polarizer is removed the intensity after it passes through the stack is

$I_f_3 = 27.57 W/cm^2$

2 When third polarizer is removed the intensity after it passes through the stack is

$I_f_2 = 102.24 W/cm^2$

Explanation:

From the question we are told that

The angle of the second polarizing to the first is $\theta_2 = 21^o$

The angle of the third polarizing to the first is $\theta_3 = 61^o$

The unpolarized light after it pass through the polarizing stack $I_u = 60 W/cm^2$

Let the initial intensity of the beam of light before polarization be $I_p$

Generally when the unpolarized light passes through the first polarizing filter the intensity of light that emerges is mathematically evaluated as

$I_1 = \frac{I_p}{2}$

Now according to Malus’ law the intensity of light that would emerge from the second polarizing filter is mathematically represented as

$I_2 = I_1 cos^2 \theta_1$

$= \frac{I_p}{2} cos ^2 \theta_1$

The intensity of light that will emerge from the third filter is mathematically represented as

$I_3 = I_2 cos^2(\theta_2 - \theta_1 )$

$I_3= \frac{I_p}{2}(cos^2 \theta_1)[cos^2(\theta_2 - \theta_1)]$

making $I_p$ the subject of the formula

$I_p = \frac{2L_3}{(cos^2 \theta [cos^2 (\theta_2 - \theta_1)])}$

Note that $I_u = I_3$ as $I_3$ is the last emerging intensity of light after it has pass through the polarizing stack

Substituting values

$I_p = \frac{2 * 60 }{(cos^2(21) [cos^2 (61-21)])}$

$I_p = \frac{2 * 60 }{(cos^2(21) [cos^2 (40)])}$

$=234.622W/cm^2$

When the second is removed the third polarizer becomes the second and final polarizer so the intensity of light would be mathematically evaluated as

$I_f_3 = \frac{I_p}{2} cos ^2 \theta_2$

$I_f_3$ is the intensity of the light emerging from the stack

substituting values

$I_f_3 = \frac{234.622}{2} * cos^2(61)$

$I_f_3 = 27.57 W/cm^2$

When the third polarizer is removed the second polarizer becomes the

the final polarizer and the intensity of light emerging from the stack would be

$I_f_2 = \frac{I_p}{2} cos ^2 \theta_1$

$I_f_2$ is the intensity of the light emerging from the stack

Substituting values

$I_f_2 = \frac{234.622}{2} cos^2 (21)$

$I_f_2 = 102.24 W/cm^2$

7 0

3 years ago

An object has an initial velocity v and uniform acceleration a. If it covers a distance d, then its final velocity, v, is given

viktelen [127]

Answer:

Explanation:

5 0

3 years ago