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3 years ago

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The distance from the Earth to the Sun is known as an astronomical unit (AU). This distance is also equal to 1.496e+011 m. Neptu

ne has an orbit with an average diameter of 8.80e+009 km. What is the distance from the Sun to Neptune in AU?
A. 24.5 AU
B. 36.3 AU
C. 29.4 AU
D. 68.5 AU
E. 55.8 AU

1 answer:

Lerok [7]3 years ago

7 0

Answer:

The distance from the Sun to Neptune is 29,41 AU.

Explanation:

We know, from the sentence, that the orbit of Neptune has an average diameter around 8.80*10⁹km.

Now, we can calculate the radius of this orbit, which is equivalent to the distance from thsi planet to the Sun. Let's recall tha the radius is the half of the diameter.

$R=\frac{8.80\cdot 10^{9}}{2}=4.4\cdot 10^{9}km=4.4\cdot 10^{12}m$

Ok, we know that 1.496*10¹¹m is an AU, therefore we have:

$4.4\cdot 10^{12}m\cdot \frac{1 AU}{1.496\cdot 10^{11}}=29,41 AU$

Finally, the distance R is 29,41 AU.

I hope it helps you! :)

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Answer:

Al's mass is 102.92 kg

Explanation:

As there are no external forces in the horizontal direction, the horizontal net force must be zero:

$F_{net} = 0$

As the force is the derivative in time of the momentum, this means that the horizontal momentum is constant:

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$p_{horizontal_i }= p_{horizontal_f}$

where the suffix i and f means initial and final respectively.

The initial momentum will be:

$p_{horizontal_}i = m_{Al} \ v_{Al_i} + m_{Jo} \ v_{Jo_i}$

But, as they are at rest, initially

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So, this means:

$p_{horizontal_f} = m_{Al} \ v_{Al_f} + m_{Jo} \ v_{Jo_f} = 0$

We know that the have an combined mass of 195 kg:

$m_{total} = m_{Al} + m_{Jo} = 195 \ kg$ .

so:

$m_{Jo} = 195 \ kg - m_{Al}$ .

$m_{Al} \ v_{Al_f} + (195 \ kg - m_{Al}) \ v_{Jo_f} = 0$

$m_{Al} \ v_{Al_f} - m_{Al} \ v_{Jo_f}= - 195 \ kg \ v_{Jo_f}$

$m_{Al} \ (v_{Al_f} - v_{Jo_f})= - 195 \ kg \ v_{Jo_f}$

$m_{Al} = \frac{ - 195 \ kg \ v_{Jo_f} } { v_{Al_f} - v_{Jo_f} }$

$m_{Al} = \frac{195 \ kg \ v_{Jo_f} } { v_{Jo_f} - v_{Al_f} }$

Now, we can use the values:

$v_{Al_f}= 10.2 \frac{m}{s}$

$v_{Jo_f}= - 11.4 \frac{m}{s}$

where the minus sign appears as they are moving at opposite directions

$m_{Al} = \frac{195 \ kg ( - 11.4 \frac{m}{s} ) } { (- 11.4 \frac{m}{s}) - 10.2 \frac{m}{s} }$

$m_{Al} = 102.92 \ kg$

and this is the Al's mass.

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TRANSFORMA: 765 mm Hg a atm

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Presión

765

=

ATMÓSFERA

1,00658

Answer:

Explanation:

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3 years ago

A car speeds over a hill past point A, as shown in the figure. What is the maximum speed the car can have at point A such that i

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Answer:

see explanations below

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Answer: at 11.8 m/s (26.4 mph) car will leave the track.

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Answer:

Frequency change by a factor of 2.

(b) is correct option.

Explanation:

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Put the value into the formula

$f_{1}=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{16}}$

$f_{1}=\dfrac{1}{2\pi}\dfrac{\sqrt{k}}{4}}$ ....(I)

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Answer:

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