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3 years ago

6

________ occurs when an object in the outer reaches of the solar system passes between earth and a far distant star, temporarily

blocking light from the star.

1 answer:

murzikaleks [220]3 years ago

6 0

<span>Occulation occurs when an object in outer space reaches the solar system and passes between earth and a far distant star, temporarily blocking light from the star. Occulation is an astronomy term referring to when an object in the foreground blocks the view of an object in the background.</span>

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A thermistor is placed in a 100 °C environment and its resistance measured as 20,000 Ω. The material constant, β, for this therm

Karo-lina-s [1.5K]

Answer:

the thermistor temperature = $325.68 \ ^0 \ C$

Explanation:

Given that:

A thermistor is placed in a 100 °C environment and its resistance measured as 20,000 Ω.

i.e Temperature

$T_1 = 100^0C\\T_1 = (100+273)K\\\\T_1 = 373\ K$

Resistance of the thermistor $R_1 =$ 20,000 ohms

Material constant $\beta$ = 3650

Resistance of the thermistor $R_2$ = 500 ohms

Using the equation :

$R_1 = R_2 \ e^{\beta} (\frac{1}{T_1}- \frac{1}{T_2})$

$\frac{R_1}{ R_2} = \ e^{\beta} (\frac{1}{T_1}- \frac{1}{T_2})$

Taking log of both sides

$In \ \frac{R_1}{ R_2} = In \ \ e^{\beta} (\frac{1}{T_1}- \frac{1}{T_2})$

$In \ \frac{R_1}{ R_2} = {\beta} (\frac{1}{T_1}- \frac{1}{T_2})$

$\frac{ In \ \frac{R_1}{ R_2}}{ {\beta}} = (\frac{1}{T_1}- \frac{1}{T_2})$

$\frac{1}{T_2} = \frac{1}{T_1} - \frac{ In \ \frac{R_1}{ R_2}}{ {\beta}}$

${T_2} = \frac{\beta T_1}{\beta - In (\frac{R_1}{R_2})T}$

Replacing our values into the above equation :

${T_2} = \frac{3650*373}{3650 - In (\frac{20000}{500})373}$

${T_2} = \frac{1361450}{3650 - 3.6888*373}$

${T_2} = \frac{1361450}{3650 - 1375.92}$

${T_2} = \frac{1361450}{2274.08}$

${T_2} = 598.68 \ K$

${T_2} = 325.68 \ ^0 \ C$

Thus, the thermistor temperature = $325.68 \ ^0 \ C$

4 0

4 years ago

The ice cap at the North Pole may be 1000 m thick, with a density of 920 kg/m3. Find the pressure at the bottom and the correspo

bija089 [108]

<span>Pice=920kg/m^3 deltaP=PgH=920kg/m^3 X 9.80665m/s^2 X 1000m = 9022118 Pa P=Po + deltaP=101.325 + 9022 = 9123kPa</span>

7 0

3 years ago

A tow truck is pulling 25,000 kg van with a tow hook that meets the van at an angle of 30° with the ground. How much force must

Law Incorporation [45]

Answer:

86605.08 N

Explanation:

The equation to calculate the force is:

Force = mass * acceleration

The force and the acceleration does not have the same direction in this case, so we need to decompose the force into its horizontal component, which is the force that will generate the horizontal acceleration:

Force_x = Force * cos(30)

Then, we have that:

Force_x = mass * acceleration

Force * cos(30) = 25000 * 3

Force * 0.866 = 75000

Force = 75000 / 0.866 = 86605.08 N

3 0

4 years ago

Both the lens and the cornea of the eye have a primary function of

3241004551 [841]

Answer: B. bending light

Explanation:

The phenomenom of vision in human eye is thanks to refraction (when light changes its direction as it passes through one medium to another), and this is what the cornea and the lens do.

When the ray of light encounters the eye, the first thing it finds is the <u>cornea</u>, which<u> bends this ray and begins to form an image</u>, then light passes through the <u>pupil</u>, which is in charge of regulating the amount of light that enters in the eye.

After light travels through pupil it passes through the <u>lens</u>, where <u>the rays of light change the direction again in order to focus the formed image on the retina. </u>

At this point it is important to note the formed image is downward, then the retina transforms light into electrical impulses that are sent to the brain through the optic nerve and finally the brain interprets these messages, and forms a right upward image.

In the image attached these parts can be seen.

6 0

3 years ago

50POINTS! Find the orbital speed of a satellite in a circular orbit 1700km above the surface of the Earth. M_earth 5.97e24kg, r_

ivolga24 [154]

<h2>Answer: 7020.117 m/s</h2>

Explanation:

The velocity of a satellite describing a circular orbit is<u> constant</u> and defined by the following expression:

$V=\sqrt{G\frac{M}{R}}$ (1)

Where:

$G=6.674({10}^{-11})\frac{N{m}^{2}}{{kg}^{2}}$ is the gravity constant

$M_{Earth}=5.97{10}^{24}kg$ the mass of the massive body around which the satellite is orbiting, in this case, the Earth .

$R=r_{Earth}+h=8080000m$ the radius of the orbit (measured from the center of the planet to the satellite).

This means the radius of the orbit is equal to <u>the sum</u> of the average radius of the Earth $r_{Earth}$ and the altitude of the satellite above the Earth's surface $h$ .

Note this orbital speed, as well as orbital period, does not depend on the mass of the satellite. It depends on the mass of the massive body (the Earth).

Now, rewriting equation (1) with the known values:

$V=\sqrt{(6.674({10}^{-11})\frac{N{m}^{2}}{{kg}^{2}})\frac{5.97{10}^{24}kg}{8080000m}}$

$V=7020.117\frac{m}{s}$

6 0

4 years ago

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