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Nadya [2.5K]
3 years ago
12

Define an astronomical unit. Choose all that apply. a. 8.3 minutes Average distance from Earth to the Sun. b. Distance that ligh

t travels in one year. c. Average distance from any planet to the Sun. d. 93 miles 150 million km
Physics
1 answer:
Naddik [55]3 years ago
6 0

Answer:

a. 8.3 minutes average distance from earth to the sun

d. 93 miles or 150 million km

Explanation:

The distance between the earth and the sun is defined as an astronomical unit (AU). It takes 8.3 minutes to go from earth to the sun at the speed of light. That distance has a length of 150 million Kilometers or 93 miles.  

It is common to see in planet charts that distance to the sun are compared in astronomical units. In the case of Mars is 1.524 AU away from the sun.

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27b. You're in an airplane that flies horizontally with speed 1000 km/h (280 m/s) when an engine falls off. Ignore air resistanc
jonny [76]

Answer:

8400m

Explanation:

The engine that falls off would have the same constant horizontal velocity as the airplane's when if falls off if we ignore air resistance. So it would have a horizontal velocity of 280m/s for 30seconds before it hits the ground.

Therefor the horizontal distance the engine travels during its fall is

280 * 30 = 8400m

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Answer is:

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5 0
2 years ago
Two iron weights, one twice the mass of the other, are dropped from the top of a building. Compared with the lighter weight, the
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The twice as heavy weight will hit the ground with more force, or impact.
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2 years ago
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Include units with all of your answers. A 1.84 kg bucket full of water is attached to a 0.76 m long string.
aleksandrvk [35]
<span>c. What is the magnitude of the tension in the string at the bottom of the circle if you are swinging it at 3.37 m/s? </span>
7 0
3 years ago
The intensity of the Sun's light in the vicinity of the Earth is about 1000W/m^2. Imagine a spacecraft with a mirrored square sa
NeX [460]

Explanation:

Formula to represent thrust is as follows.

             F = \frac{dP}{dt}

                = \frac{2p}{dt}

or,           p = \frac{E}{c}

         \frac{p}{dt} = \frac{W}{c}

                  F = \frac{2IA}{c}

                     = \frac{2 (1000 W/m^{2})(5.5 \times 10^{3} m)^{2}}{3 \times 10^{8} m/s}

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Thus, we can conclude that the thrust is 201.67 N.

8 0
3 years ago
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