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2 years ago

9

Whose data did Kepler use to describe the motion of the planets?

2 answers:

lesya [120]2 years ago

8 0

Answer:

A. Tycho Brahe

Explanation:

the person above is correct :)

Levart [38]2 years ago

7 0

Answer:

Tycho Brahe

Explanation:

Tycho Brahe's accurate observations of planetary positions provided the data used by Johannes Kepler to derive his three fundamental laws of planetary motion.

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2 (a) What is the distance from the Sun to Earth in terms of solar radii? Earth radii?

ohaa [14]

a) Distance Earth-Sun is 215.5 solar radii and 23,548 Earth radii

b)

Mercury: 1.7 days

Mars: 6.5 days

Jupiter: 21.7 days

Uranus: 86.8 days

Neptune: 130.2 days

c) $1.3\cdot 10^6$ Earths can fit inside the Sun

Explanation:

a)

The distance between the Sun and the Earth is 150 millions km, so

$d=150\cdot 10^6 km = 1.50\cdot 10^{11} m$

The solar radius is

$r_s = 6.96\cdot 10^5 km = 6.96\cdot 10^8 m$

Therefore the distance Earth-Sun in solar radii is

$d_s = \frac{d}{r_s}=\frac{1.50\cdot 10^{11}}{6.96\cdot 10^8}=215.5$

The Earth radius is

$r_e = 6.37\cdot 10^6 m$

Therefore the distance Earth-Sun in Earth radii is

$d_e=\frac{d}{r_e}=\frac{1.50\cdot 10^{11}}{6.37\cdot 10^6}=23,548$

b)

The speed of the solar wind is

$v=400 km/s = 4\cdot 10^5 m/s$

The value of 1 AU (Astronomical Unit) is

$1 AU = 1.50\cdot 10^{11}m$ (distance Earth-Sun)

The distance between the Sun and Mercury is:

$d=0.4 AU \cdot 1.50\cdot 10^{11}=6.0\cdot 10^{10} m$

So the time taken by a parcel of solar wind to reach Mercury is:

$t=\frac{d}{v}=\frac{6.0\cdot 10^{10}}{4.0\cdot 10^5}=150,000 s$

Converting into days ( $1d=86400 s$ ),

$t=\frac{150,000}{86400}=1.7 d$

The distance between the Sun and Mars is:

$d=1.5 AU \cdot 1.50\cdot 10^{11}=2.25\cdot 10^{11} m$

So the time taken by a parcel of solar wind to reach Mars is:

$t=\frac{d}{v}=\frac{2.25\cdot 10^{11}}{4.0\cdot 10^5}=562,500 s$

Converting into days ( $1d=86400 s$ ),

$t=\frac{562,500}{86400}=6.5 d$

The distance between the Sun and Jupiter is:

$d=5 AU \cdot 1.50\cdot 10^{11}=7.5\cdot 10^{11} m$

So the time taken by a parcel of solar wind to reach Jupiter is:

$t=\frac{d}{v}=\frac{7.5\cdot 10^{11}}{4.0\cdot 10^5}=1.88 \cdot 10^6 s$

Converting into days ( $1d=86400 s$ ),

$t=\frac{1.88\cdot 10^6}{86400}=21.7 d$

The distance between the Sun and Uranus is:

$d=20 AU \cdot 1.50\cdot 10^{11}=3.0\cdot 10^{12} m$

So the time taken by a parcel of solar wind to reach Uranus is:

$t=\frac{d}{v}=\frac{3.0\cdot 10^{12}}{4.0\cdot 10^5}=7.5 \cdot 10^6 s$

Converting into days ( $1d=86400 s$ ),

$t=\frac{7.5\cdot 10^6}{86400}=86.8 d$

The distance between the Sun and Neptune is:

$d=30 AU \cdot 1.50\cdot 10^{11}=4.5\cdot 10^{12} m$

So the time taken by a parcel of solar wind to reach Neptune is:

$t=\frac{d}{v}=\frac{4.5\cdot 10^{12}}{4.0\cdot 10^5}=11.3 \cdot 10^6 s$

Converting into days ( $1d=86400 s$ ),

$t=\frac{11.3\cdot 10^6}{86400}=130.2 d$

c)

As we said in part a), we have:

$r_e = 6.37\cdot 10^6 m$ (radius of the Earth)

$r_s=6.96\cdot 10^8 m$ (radius of the Sun)

So the volume of the Earth can be calculated as:

$V_e=\frac{4}{3}\pi r_e^3 = \frac{4}{3}\pi (6.37\cdot 10^6)^3=1.08\cdot 10^{21} m^3$

While the volume of the Sun is

$V_s=\frac{4}{3}\pi r_s^3 = \frac{4}{3}\pi (6.96\cdot 10^8)^3=1.41\cdot 10^{27} m^3$

Therefore, the number of Earths that could fit inside the Sun is:

$\frac{V_s}{V_e}=\frac{1.41\cdot 10^{27}}{1.08\cdot 10^{21}}=1.3\cdot 10^6$

Learn more about the Solar System:

brainly.com/question/2887352

brainly.com/question/10934170

#LearnwithBrainly

6 0

3 years ago

A hoop of mass 2 kg, radius 0.5 m is rotating about its center with an angular speed of 3 rad's. A force of 10N is applied tange

Degger [83]

Answer:

The rotational kinetic energy of the hoop and the instantaneous change rate of the kinetic energy are 2.25 J and 15 J.

Explanation:

Given that,

Mass = 2 kg

Radius = 0.5 m

Angular speed = 3 rad/s

Force = 10 N

(I). We need to calculate the rotational kinetic energy

Using formula of kinetic energy

$K.E =\dfrac{1}{2}\timesI\omega^2$

$K.E=\dfrac{1}{2}\times mr^2\times\omega^2$

$K.E=\dfrac{1}{2}\times2\times(0.5)^2\times(3)^2$

$K.E=2.25\ J$

(II). We need to calculate the instantaneous change rate of the kinetic energy

Using formula of kinetic energy

$K.E=\dfrac{1}{2}mv^2$

On differentiating

$\dfrac{K.E}{dt}=\dfrac{1}{2}m\times2v\times\dfrac{dv}{dt}$

$\dfrac{K.T}{dt}=mva$ ....(I)

Using newton's second law

$F = ma$

$a= \dfrac{F}{m}$

$a=\dfrac{10}{2}$

$a=5 m/s^2$

Put the value of a in equation (I)

$\dfrac{K.E}{dt}=mva$

$\dfrac{K.E}{dt}=mr\omega a$

$\dfrac{K.E}{dt}=2\times0.5\times3\times5$

$\dfrac{K.E}{dt}=15\ J/s$

Hence, The rotational kinetic energy of the hoop and the instantaneous change rate of the kinetic energy are 2.25 J and 15 J.

3 0

2 years ago

In each cycle, a heat engine an input of 1940 J of heat and exhausts 1480 J of heat. What is the thermal efficiency?

AleksAgata [21]

Answer:

0.237 (23.7 %)

Explanation:

The thermal efficiency of an engine is given by:

$\eta=\frac{W}{Q_{in}}$

where

W is the useful work output of the engine

$Q_{in}$ is the heat in input

Here we have:

$Q_{in}=1940 J$

and the work done is the total heat in input minus the heat exhausted:

$W=1940 J - 1480 J=460 J$

So, the efficiency is

$\eta=\frac{460 J}{1940 J}=0.237$ (23.7 %)

8 0

3 years ago

What is the current through a wire if 240 coulombs of charge pass through the wire in 2.0 minutes?

xenn [34]

Current = charge/time = 240/2x120 =1A.
Current flowing is 1 ampere.

4 0

3 years ago

Read 2 more answers

You kick a soccer ball with a mass of 2 kg. The ball leaves your foot with a speed of 30 m/s. How much kinetic energy does the b

Lyrx [107]

Answer:

KE= 900 (I think)

Explanation:

KE=½mv²

KE= ½(2)(30)²

KE=½(60)²

KE=30²

KE=900

Hope this helps!

7 0

2 years ago

Read 2 more answers