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2 years ago

The orbits of all the planets in the solar system are slightly elliptical with the exception of _______.

Physics

1 answer:

pogonyaev2 years ago

7 0

Except for Pluto, all of the planets in the solar system have slightly elliptical orbits.

<h3>What distinguishes Pluto from other planets in terms of its orbit?</h3>

Pluto, a dwarf planet, the orbit is highly elliptical, with an eccentricity of 0.25, and it travels between 4.4 and 7.4 109 kilometers (2.7-4.6 109 miles) from the Sun on average. As a result, Pluto's orbit occasionally brings it closer to the Sun than Neptune, which implies that sometimes Pluto is much closer to the Sun than other times.

Pluto orbits the Sun at an angle of 17 degrees. This indicates that Pluto occasionally orbits above the other planets and occasionally orbits below them.

Learn more about pluto here:

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You stand on a merry-go-round which is spinning at f = 0.25 revolutions per second. You are R = 200 cm from the center. (a) Find

wariber [46]

Answer:

a) ω = 9.86 rad/s

b) ac = 194. 4 m/s²

c) minimum coefficient of static friction, µs = 19.8

Explanation:

a) angular speed, ω = 2πf, where f is frequency of revolution

1 rps = 6.283 rad/s, π = 3.142

ω = 2 * 3.14 * 0.25 * 6.28

ω = 9.86 rad/s

b) centripetal acceleration, a = rω²

where r is radius in meters; r = 200 cm or 2 m

a = 2 * 9.86²

a = 194. 4 m/s²

c) µs = frictional force/ normal force

frictional force = centripetal force = ma; where a is centripetal acceleration

normal force = mg; where g = 9.8 m/s²

µs = ma/mg = a/g

µs = 194.4 ms⁻²/9.8 ms⁻²

c) minimum coefficient of static friction, µs = 19.8

5 0

3 years ago

What are the scientific mesurments ordered from greatest to least

Sveta_85 [38]

Meters Micrometers centimeters millimeters

5 0

3 years ago

If 2.0 x 10^-4 C charge passes a point in 5.0 x 10^-5 s, what is the rate of current flow?

vampirchik [111]

Current is defined as the rate of charge flowing a point every second. Having a current of 1 Ampere signifies 1 Coulomb is flowing in a circuit every second. It is measured by the use of an ammeter which is positioned in series to the component to be measured. The current in the problem is calculated as follows:

I = 2.0 x 10^-4 C / 5.0 x 10^-5 s
<span>I = 4 A</span>

8 0

4 years ago

Read 2 more answers

Prof. Kopp is working on an experiment located in an abandoned mine near Duluth, MN. The experiment is located approximately 713

MA_775_DIABLO [31]

Answer:

The average speed of the elevator going down in the abandoned mine is 17.722mph.

Explanation:

If the elevator takes 90 seconds to descend a height of 713m, the average speed of the elevator is:

$v_{av}=x_T/t_T=713m/90s=7.922m/s$

And if 1m/s is 2.23694mph, the average speed is:

$v_{av}=7.922m/s=17.722mph$ .

8 0

3 years ago

Your grandmother enjoys creating pottery as a hobby. She uses a potter's wheel, which is a stone disk of radius R-0.520 m and ma

Lesechka [4]

Answer:

0.54454

104.00902 N

Explanation:

m = Mass of wheel = 100 kg

r = Radius = 0.52 m

t = Time taken = 6 seconds

$\omega_f$ = Final angular velocity

$\omega_i$ = Initial angular velocity

$\alpha$ = Angular acceleration

Mass of inertia is given by

$I=\dfrac{mr^2}{2}\\\Rightarrow I=\dfrac{100\times 0.52^2}{2}\\\Rightarrow I=13.52\ kgm^2$

Angular acceleration is given by

$\alpha=\dfrac{\tau}{I}\\\Rightarrow \alpha=\dfrac{\mu fr}{I}\\\Rightarrow \alpha=\dfrac{\mu 50\times 0.52}{13.52}$

Equation of rotational motion

$\omega_f=\omega_i+\alpha t\\\Rightarrow \omega_f=\omega_i+\dfrac{\mu (-50)\times 0.52}{13.52}t\\\Rightarrow 0=60\times \dfrac{2\pi}{60}+\dfrac{\mu (-50)\times 0.52}{13.52}\times 6\\\Rightarrow 0=6.28318-11.53846\mu\\\Rightarrow \mu=\dfrac{6.28318}{11.53846}\\\Rightarrow \mu=0.54454$

The coefficient of friction is 0.54454

At r = 0.25 m

$\omega_f=\omega_i+\dfrac{0.54454 (-50)\times 0.52}{13.52}6\\\Rightarrow 0=60\times \dfrac{2\pi}{60}+\dfrac{0.54454 f\times 0.25}{13.52}6\\\Rightarrow 2\pi=0.06041f\\\Rightarrow f=\dfrac{2\pi}{0.06041}\\\Rightarrow f=104.00902\ N$

The force needed to stop the wheel is 104.00902 N

5 0

3 years ago