Physics, 04.10.2021 06:50 tannerroland

The Event Horizon Telescope (EHT) recently observed the shadow of M 87* (the supermassive black hole, of about 6 billion solar masses, at the center of the giant elliptical galaxy M87, which is about 55 Mly away). If the observations were made at the radio wavelength of 1.3 mm & were at the diffraction limit for angular resolution at about 50 micro-arcseconds, about what was the distance between EHT’s component telescopes?

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Answer from: Quest

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Answer from: Quest

Then the answer would be d then

Answer from: Quest

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

$a = 32 ft/s^2$

so as we know that

$a = \frac{v^2}{r}$

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

$a_c = 32 ft/s^2$

$a_t = 0$

now radius of curvature is given as

$r = \frac{v^2}{a_c}$

$r = \frac{150^2}{32}$

$r = 703.125ft$

so the radius of curvature just after it is dropped is 703.125 ft

Answer from: Quest

a. carbon monoxide from car exhaust

b.particulate matter, such as soot

c.water vapor from a factory smokestack

d.acid rain

these are the answer choices im on that question right now

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The Event Horizon Telescope (EHT) recently observed the shadow of M 87* (the supermassive black hole...

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