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Nikitich [7]
4 years ago
11

What is one major difference between the attributes of Mercury and Jupiter? A. Mercury is the planet that is closest to the Sun,

while Jupiter is the farthest. B. Mercury is the coldest planet in the solar system, while Jupiter is the hottest. C. Mercury has no orbiting satellites, while Jupiter has many satellites circling it. D. Mercury has bands and rings visible from Earth, while Jupiter has no such bands. E. Mercury has a rich supply of methane, while Jupiter has a rich supply of hydrogen.
Physics
2 answers:
Sergeu [11.5K]4 years ago
4 0

Answer:

C. Mercury has no orbiting satellites, while Jupiter has many satellites circling it.

Explanation:

Mercury is the smallest and closest planet to sun. It is a rocky planet. It does not have any natural satellites. On the other hand, Jupiter is the largest planet and is made up of gases. It has bands. Jupiter has more than 60 natural satellites revolving about it. It is mostly composed of hydrogen and helium.

Thus, the major difference between Mercury and Jupiter is C. Mercury has no orbiting satellites, while Jupiter has many satellites circling it.

sergeinik [125]4 years ago
3 0

The major difference is: Mercury has no orbiting satellites, while Jupiter has many satellites that surround it, so the correct answer is C

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If Dwight races his Chevy towards Chatham and travels 2460 meters in 60 seconds. what is his velocity?
labwork [276]

Answer:

divide

Explanation:

whenever looking for velocity.just devide

8 0
3 years ago
A coyote can locate a sound source with good accuracy by comparing the arrival times of a sound wave at its two ears. Suppose a
il63 [147K]

Answer:

0.0000109261200583 s

0.0109261200583

Explanation:

d_2 = Distance from right ear = 3 m

s = Distance between ears = 15 cm

v = Speed of sound in air = 343 m/s

Distance between the left ear and the bird

d_1=\sqrt{s^2+d_2^2}\\\Rightarrow d_1=\sqrt{0.15^2+3^2}\\\Rightarrow d_1=3.00374765918\ m=3.004\ m

Time

t=\dfrac{Distance}{Speed}

Time difference would be

\Delta T=\dfrac{d_1}{v}-\dfrac{d_2}{v}\\\Rightarrow \Delta T=\dfrac{3.00374765918}{343}-\dfrac{3}{343}\\\Rightarrow \Delta T=0.0000109261200583\ s

The time difference is 0.0000109261200583 s

Time period is given by

T=\dfrac{1}{f}\\\Rightarrow T=\dfrac{1}{1000}\\\Rightarrow T=10^{-3}\ s

The ratio is

\dfrac{\Delta T}{T}=\dfrac{0.0000109261200583}{10^{-3}}\\\Rightarrow \dfrac{\Delta T}{T}=0.0109261200583

The ratio is 0.0109261200583

7 0
3 years ago
A gyroscope flywheel of radius 3.25 cm is accelerated from rest at 11.6 rad/s2 until its angular speed is 1820 rev/min. (a) What
kati45 [8]

Explanation:

The given data is as follows.

           radius (r) = 3.25 cm,    \alpha = 11.6 rad/s^{2}

Now, we will calculate the tangential acceleration as follows.

          a_{tangential} = \alpha \times r

Putting the given values into the above formula as follows.

         a_{tangential} = \alpha \times r

                      = 11.6 rad/s^{2} \times 3.25 cm

                      = 37.7 rad cm/s^{2}

Thus, we can conclude that the tangential acceleration of a point on the rim of the flywheel during this spin-up process is 37.7 rad cm/s^{2}.

6 0
3 years ago
A sound wave can cause the particles in a medium to vibrate slowly or quickly what is this rate of vibration called?
stealth61 [152]

Answer:

FREQUENCY

Explanation:

3 0
3 years ago
Read 2 more answers
Earth is 149.6 million meters from the Sun and takes 365 days to make one complete revolution around the Sun. Mars is 227.9 mill
luda_lava [24]

Answer:

\frac{a_{r,earth}}{a_{r,mars}} = 2.325

Explanation:

The distance of Earth from the Sun is 149.6\times 10^{9}\,m and of Mars from the Sun is 227.9\times 10^{9}\,m. Let assume that both planets have circular orbits. The centripetal accelaration can be found by using the following expression:

a_{r} = \frac{v^{2}}{R}

Since planet has translation at constant speed, this formula is applied to compute corresponding speeds:

v=\frac{2\pi\cdot r}{\Delta t}

Earth:

v_{earth} = \frac{2\pi\cdot (149.6\times 10^{9}\,m)}{(365\,days)\cdot(\frac{24\,hours}{1\,day} )\cdot(\frac{3600\,s}{1\,h} )}

v_{earth}=29806.079\,\frac{m}{s}

Mars:

v_{mars} = \frac{2\pi\cdot (227.9\times 10^{9}\,m)}{(687\,days)\cdot(\frac{24\,hours}{1\,day} )\cdot(\frac{3600\,s}{1\,h} )}

v_{mars}=24124.244\,\frac{m}{s}

Now, centripetal accelarations can be found:

Earth:

a_{r,earth} = \frac{(29806.079\,\frac{m}{s} )^{2}}{149.6\times 10^{9}\,m}

a_{r,earth} = 5.939\times 10^{-3}\,\frac{m}{s^{2}}

Mars:

a_{r,mars} = \frac{(24124.244\,\frac{m}{s} )^{2}}{227.9\times 10^{9}\,m}

a_{r,mars} = 2.554\times 10^{-3}\,\frac{m}{s^{2}}

The ratio of Earth's centripetal acceleration to Mars's centripetal acceleration is:

\frac{a_{r,earth}}{a_{r,mars}} = \frac{5.939}{2.554}

\frac{a_{r,earth}}{a_{r,mars}} = 2.325

7 0
3 years ago
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