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4 years ago

Which of the following is a good example of a physical change?

Physics

1 answer:

den301095 [7]4 years ago

5 0

Candle burning because it’s going from a solid to a liquid

Plz give me a good mark

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Problem 4 A meteoroid is first observed approaching the earth when it is 402,000 km from the center of the earth with a true ano

Nikitich [7]

Answer:

Part a: The eccentricity is 1.086.

Part b: The altitude at closest approach is 5088 km

Part c: The velocity at perigee is 8.516 km/s

Part d: The turn angle is 134.08 while the aiming radius is 5641.28 km

Explanation:

Specific energy is given by

$\epsilon=\frac{v^2}{2}-\frac{\mu}{r}$

Here

ε is the specific energy
v is the velocity which is given as 2.23 km/s
μ is the gravitational constant whose value is 398600
r is the distance between earth and the meteorite which is 402,000 km

$\epsilon=\frac{v^2}{2}-\frac{\mu}{r}\\\epsilon=\frac{2.2^2}{2}-\frac{398600}{402,000}\\\epsilon=1.495 km^2/s^2$

Value of specific energy is also given as

$\epsilon=\frac{\mu}{2a}\\a=\frac{\mu}{2\epsilon}\\a=\frac{398600}{2\times 1.495}\\a=13319 km$

Orbit formula is given as

$r=a(\frac{e^2-1}{1+ecos \theta})\\ae^2-recos\theta-(a+r)=0$

Putting values in this equation and solving for e via the quadratic formula gives

$ae^2-recos\theta-(a+r)=0\\(133319)e^2-(402000)(cos 150) e-(133319+402000)=0\\133319e^2+348142.21 e-535319=0\\\\e=\frac{-348142.21 \pm \sqrt{348142.21^2-4(133319)(535319)}}{2 (133319)}\\\\e=1.086 \, or \, -3.69$

As the value of eccentricity cannot be negative so the eccentricity is 1.086.

The radius of trajectory at perigee is given as

$r_p=a(e-1)\\$

Substituting values gives

$r_p=133319 (1.086-1)\\r_p=11465.4 km$

Now for estimation of altitude z above earth is given as

$z=r_p-R_E\\z=11465.4-6378\\z=5087.434\\z\approx 5088 km$

So the altitude at closest approach is 5088 km

radius of perigee is also given as

$r_p=\frac{h^2}{\mu}\frac{1}{1+e}$

Rearranging this equation gives

$h=\sqrt{r_p\mu(1+e)}\\h=\sqrt{11465.4 \times 3986000 \times (1+1.086)}\\h=97638.489 km^2/s$

Now the velocity at perigee is given as

$v_p=\frac{h}{r_p}\\v_p=\frac{97638.489}{11465.4}\\v_p=8.516 km/s\\$

So the velocity at perigee is 8.516 km/s

Turn angle is given as

$\delta =2 sin^{-1} (\frac{1}{e})$

Substituting value in the equation gives

$\delta =2 sin^{-1} (\frac{1}{e})\\\delta =2 sin^{-1} (\frac{1}{1.086})\\\delta =134.08$

Aiming radius is given as

$\Delta =a \sqrt{e^2-1}$

Substituting value in the equation gives

$\Delta =a \sqrt{e^2-1}\\\Delta =13319 \sqrt{1.086^2-1}\\\Delta=5641.28 km$

So the turn angle is 134.08 while the aiming radius is 5641.28 km

3 0

3 years ago

Please help!!!!<br> Will give brainlest

Evgen [1.6K]

Answer:

Explanation:

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3 years ago

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castortr0y [4]

That is a paradox b/c stationary means not moving

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3 years ago

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kap26 [50]

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the way I found this out because i read the definitions and found the key words in there.

7 0

3 years ago

Answer:

notka56 [123]

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