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3 years ago

The distance between Jupiter and the Sun is 5.2 AU. What is the distance in millions of kilometers?

Physics

1 answer:

miss Akunina [59]3 years ago

8 0

Answer:

777.9 Mkm

Explanation:

(M = Mega = $10^{6}$ )

AU is the Astronomical Unit for distance (1AU=149,597,870.691 km)

The total distance in km then is:

$5.2 * 149,597,870.691 km = 777908927.593 km$

To give the result in millions, we use the prefix Mega:

$(\frac{1M}{10^{6} } ) * 777908927.593 km = 777.9 Mkm$

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The angle of incidence must equal the angle of reflection is always less than the angle of reflection is always greater than the

saw5 [17]

Answer: The angle of incidence is not always equal to the angle of reflection.

Explanation:

The angle of incidence might not be equal to the angle of reflection. It depends of the type of surface in consideration. If the surface is smooth, the incident ray will reflect out at the same angle the incident ray makes with surface. This is not the same for a rough or irregular surface.

For an irregular surface, the angle of incident is not equal to the angle if reflection because the reflected ray always reflects at different angles to the horizontal.

6 0

2 years ago

Local environmental change have no effect on the global climate true or false

Otrada [13]

Answer:

false

Explanation:

Even small environmental changes have large effects because small things become bigger

5 0

2 years ago

17. Suppose you were standing on a scale in an elevator in free fall. What would the scale read?

lianna [129]

What would the scale read? zero

18 In free fall you are being pulled by a gravity. "Truly" weightless presumably happens in deep space where there is nothing to pull you.

19 coasters accelerate down to simulate weight loss/zeroised. As do NASA planes,

Roller coasters are for fun seekers. NASA is for science

5 0

3 years ago

Read 2 more answers

You stand on a straight desert road at night and observe a vehicle approaching. This vehicle is equipped with two small headligh

spayn [35]

To solve this problem we will apply the concepts related to Reyleigh's criteria. Here the resolution of the eye is defined as 1.22 times the wavelength over the diameter of the eye. Mathematically this is,

$\theta = \frac{1.22 \lambda }{D}$

Here,

D is diameter of the eye

$D = \frac{1.22 (539nm)}{5.11 mm}$

$D= 1.287*10^{-4}m$

The angle that relates the distance between the lights and the distance to the lamp is given by,

$Sin\theta = \frac{d}{L}$

For small angle, $sin\theta = \theta$

$sin \theta = \frac{d}{L}$

Here,

d = Distance between lights

L = Distance from eye to lamp

For small angle $sin \theta = \theta$

Therefore,

$L = \frac{d}{sin\theta}$

$L = \frac{0.691m}{1.287*10^{-4}}$

$L = 5367m$

Therefore the distance is 5.367km.

4 0

3 years ago

At room temperature, sound travels at a speed of about 344 m/s in air. You see a distant flash of lightning and hear the thunder

beks73 [17]

Answer:

$d=1.67mi$

Explanation:

Assuming the light takes essentially no time to reach you, the distance at which the lightning occurred can be calculated by multiplying the speed of sound by the time it takes to hear the thunder:

$d=v_st\\d=344\frac{m}{s}(7.8s)\\d=2683.2m*\frac{1mi}{1609.34m}=1.67mi$

4 0

2 years ago