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3 years ago

At its closest point, Mercury is approximately 46 million kilometers from the sun. What is this distance in AU?

2 answers:

5 0

1 astronomical unit 1 AU = 1.4960 * 10^11 meters
it is the average distance between earth and sun
mercury to sun distance is = 46,000,000 * 1000 meters
= 4.6 * 10^9 meters = 4.6 * 10^9 / 1.4960 * 10^11 AU
= 3.0.74 / 100 = 0.0374 AU

Radda [10]3 years ago

5 0

Explanation :

Astronomical unit is the unit to find the average distance between the earth and the sun.

$1\ AU=1.496\times 10^8\ Km$

So, $1\ Km=\dfrac{1}{1.496\times 10^8}\ AU$

46 million kilometers means, $46\times 10^6\ Km$

$46\times 10^6\ Km=\dfrac{46\times 10^6}{1.496\times 10^8}$

$46\ million\ Km=0.3074\ AU$

Hence, this is the required solution.

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bagirrra123 [75]

Answer:

The marble was moving in a projectile and the speed of the engine was 2.716 m/s

Explanation:

The vertical component of the marble's flight path relative to the train

is given by the equation y(t) = v*t - (4.9)*t^2,

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So with y(1) = v - 4.9 = 0 we have

v = 4.9 m/s.

The marble will reach maximum height after 0.5 seconds, at which the

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This horizontal velocity component of the marble is the same as the

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3 years ago

Read 2 more answers

Ask Your Teacher The height h in feet reached by a dolphin t seconds after breaking the surface of the water is given by h = −16

kirza4 [7]

Answer:

1 second

Explanation:

h = −16t² + 32t

When, h = 16

16 = −16t² + 32t

Divide each of the numbers by 16

1 = -1t² + 2t

Rearrange the equation

1t²-2t+1 = 0

Solving by the quadratic formula, we get

$t=\frac{-(-2)\pm \sqrt{(-2)^2-4\times 1\times 1}}{2\times 1}\\\Rightarrow t=\frac{2\pm 0}{2}\\\Rightarrow t=1\ s$

So, time taken by the dolphin to jump out of the water and touch the trainer's hand is 1 second.

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3 years ago