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3 years ago

_____ is the kinetic energy of an object, proportional to the object's moment of inertia and the square of its angular velocity.

1 answer:

7 0

Rotational kinetic energy <span>is the kinetic energy of an object, proportional to the object's moment of inertia and the square of its angular velocity.</span>

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What is the distance to a star whose parallex is 0.1 sec?

Arte-miy333 [17]

Answer:

$30.86\times 10^{13} km$

Explanation:

Given the parallex of the star is 0.1 sec.

The distance is inversely related with the parallex of the star. Mathematically,

$d=\frac{1}{P}$

Here, d is the distance to a star which is measured in parsecs, and P is the parallex which is measured in arc seconds.

Now,

$d=\frac{1}{0.1}\\d=10 parsec$

And also know that,

$1 parsec=3.086\times 10^{13} km$

Therefore the distance of the star is $30.86\times 10^{13} km$ away.

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3 years ago

Vectors have which two properties?

stiks02 [169]

Answer:

D. magnitude and direction

Explanation:

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3 years ago

Describe all the characteristics of Earth that make it ideal for living creatures.

Lady_Fox [76]

Earths atmosphere, Temperature, Plant life and Oxygen I guess

8 0

3 years ago

Read 2 more answers

An ice rink is 200 feet long and 85 feet wide. The four corners of the rink are rounded to a radius of 24 feet. What is the surf

Nastasia [14]

Answer:

16,506 ft²

Explanation:

There are different ways you can divide the area using rectangles and circles. One way is to find the area of the entire width and length, then subtract the empty areas in the corners.

If we take the empty areas and put them together, we find their area is the area of a square minus the area of a circle.

A = (2r)² − πr²

A = 4r² − πr²

A = (4 − π) r²

So the area of the rink is:

A = WL − (4 − π) r²

A = (85)(200) − (4 − π) (24)²

A ≈ 16,506 ft²

5 0

2 years ago

One recently discovered extrasolar planet, or exoplanet, orbits a star whose mass is 0.70 times the mass of our sun. This planet

Stels [109]

0.078 times the orbital radius r of the earth around our sun is the exoplanet's orbital radius around its sun.

Answer: Option B

<u>Explanation:</u>

Given that planet is revolving around the earth so from the statement of centrifugal force, we know that any

$\frac{G M m}{r^{2}}=m \omega^{2} r$

The orbit’s period is given by,

$T=\sqrt{\frac{2 \pi}{\omega r^{2}}}=\sqrt{\frac{r^{3}}{G M}}$

Where,

$T_{e}$ = Earth’s period

$T_{p}$ = planet’s period

$M_{s}$ = sun’s mass

$r_{e}$ = earth’s radius

Now,

$T_{e}=\sqrt{\frac{r_{e}^{3}}{G M_{s}}}$

As, planet mass is equal to 0.7 times the sun mass, so

$T_{p}=\sqrt{\frac{r_{p}^{3}}{0.7 G M_{s}}}$

Taking the ratios of both equation, we get,

$\frac{T_{e}}{T_{p}}=\frac{\sqrt{\frac{r_{e}^{3}}{G M_{s}}}}{\sqrt{\frac{r_{p}^{3}}{0.7 G M_{s}}}}$

$\frac{T_{e}}{T_{p}}=\sqrt{\frac{0.7 \times r_{e}^{3}}{r_{p}^{3}}}$

$\left(\frac{T_{e}}{T_{p}}\right)^{2}=\frac{0.7 \times r_{e}^{3}}{r_{p}^{3}}$

$\left(\frac{T_{e}}{T_{p}}\right)^{2} \times \frac{1}{0.7}=\frac{r_{e}^{3}}{r_{p}^{3}}$

$\frac{r_{e}}{r_{p}}=\left(\left(\frac{T_{e}}{T_{p}}\right)^{2} \times \frac{1}{0.7}\right)^{\frac{1}{3}}$

Given $T_{p}=9.5 \text { days }$ and $T_{e}=365 \text { days }$

$\frac{r_{e}}{r_{p}}=\left(\left(\frac{365}{9.5}\right)^{2} \times \frac{1}{0.7}\right)^{\frac{1}{3}}=\left(\frac{133225}{90.25} \times \frac{1}{0.7}\right)^{\frac{1}{3}}=(2108.82)^{\frac{1}{3}}$

$r_{p}=\left(\frac{1}{(2108.82)^{\frac{1}{3}}}\right) r_{e}=\left(\frac{1}{12.82}\right) r_{e}=0.078 r_{e}$

7 0

2 years ago