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1 year ago

5

How high would a projectile go if it was launched from ground level with an initial speed of 26 m/s at an angle of 30 degrees ab

ove the horizontal?

1 answer:

tigry1 [53]1 year ago

4 0

Answer:

Vy = 26 m/s sin 30 = 13 m/s vertical speed

t = Vy / a = 13 m/s / 9.80 m/s^2 = 1.33 sec time to reach Vy = 0

H = Vy t + 1/2 g t^2

H = 13 m/s * 1.33 sec - 1.33^2 * 9.8 / 2 m = 8.62 m

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The answer is C. Pollution

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Hoochie [10]

I have a strange hunch that there's some more material or previous work
that goes along with this question, which you haven't included here.

I can't easily find the dates of Mercury's extremes, but here's some of the
other data you're looking for:

Distance at Aphelion (point in it's orbit that's farthest from the sun):
<span><span><span><span><span>69,816,900 km
0. 466 697 AU</span>

</span> </span> </span> <span> Distance at Perihelion (</span></span><span>point in it's orbit that's closest to the sun):</span>
<span><span><span><span>46,001,200 km
0.307 499 AU</span> </span>

Perihelion and aphelion are always directly opposite each other in
the orbit, so the time between them is 1/2 of the orbital period.

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1/2 (50%) of that is 43.9845 Earth days

The average of the aphelion and perihelion distances is

1/2 ( 69,816,900 + 46,001,200 ) = 57,909,050 km
or
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4 years ago

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3 years ago

M84, M87, and NGC 4258 all have accretion disks around their central black holes for which the rotational velocities have been m

givi [52]

Answer:

<u>For M84:</u>

M = 590.7 * 10³⁶ kg

<u>For M87:</u>

M = 2307.46 * 10³⁶ kg

Explanation:

1 parsec, pc = 3.08 * 10¹⁶ m

The equation of the orbit speed can be used to calculate the doppler velocity:

$v = \sqrt{\frac{GM}{r} }$

making m the subject of the formula in the equation above to calculate the mass of the black hole:

$M = \frac{v^{2} r}{G}$ .............(1)

<u>For M84:</u>

r = 8 pc = 8 * 3.08 * 10¹⁶

r = 24.64 * 10¹⁶ m

v = 400 km/s = 4 * 10⁵ m/s

G = 6.674 * 10⁻¹¹ m³/kgs²

Substituting these values into equation (1)

$M = \frac{( 4*10^{5}) ^{2} *24.64* 10^{16} }{6.674 * 10^{-11} }$

M = 590.7 * 10³⁶ kg

<u>For M87:</u>

r = 20 pc = 20 * 3.08 * 10¹⁶

r = 61.6* 10¹⁶ m

v = 500 km/s = 5 * 10⁵ m/s

G = 6.674 * 10⁻¹¹ m³/kgs²

Substituting these values into equation (1)

$M = \frac{( 5*10^{5}) ^{2} *61.6* 10^{16} }{6.674 * 10^{-11} }$

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The assumption made is that the intrinsic velocity dispersion is needed to match the line widths that are observed.

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3 years ago