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2 years ago

What is the Gravitational strength of the planet Eris

Physics

1 answer:

BabaBlast [244]2 years ago

3 0

<h3>Answer:</h3>

0.82 m/s²

Though its gravity is weak compared to earth, Eris has the strongest gravity out of all the dwarf planets: 0.82 m/s².

<h3>How Is Eris Important?</h3>

Eris is the <u>largest</u> dwarf planet in the Solar System, and the <u>ninth largest body orbiting our Sun.</u> The discovery of Eris was so important because it was a celestial body larger than Pluto, which forced astronomers to consider, for the first time in history, what the definition of a planet truly is!

Hope this helps you!! :D

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If a series circuit contains a 12-V battery, a 6-ohm resistor, and a 4-ohm resistor, what is the current in the circuit?

Shalnov [3]

In a series circuit the total current is the same throughout resistors and so:

$I_{total}=I_1=I_2$

The voltage is distributed throughout the resistors and so:

$V_{total}=V_1+V_2$

and the total resistance can be calculated by adding up the resistors resistance:

$R_{total}=R_1+R_2$

First thing is to calculate the total resistance and so:

$R_{total}=6\Omega + 4\Omega = 10\Omega$

And by Omh's law V=IR we have:

$V_{total}=I_{total}R_{total}\\\\I_{total}=\frac{V_{total}}{R_{total}}= \frac{12V}{10\Omega} =1.2A$

And so the total current of the circuit is 1.2 amps i.e. 1.2 A.

6 0

3 years ago

Read 2 more answers

Galois drove 60.0 kilometers due west in 5.00 hours and then drove 43.0 kilometers due north in 3.00 hours.

IceJOKER [234]

Answer:

Explanation:

See the attachment for the details. A right triangle is formed to find the hypotenuse of the two legs consisting of the actual driving distances and times. The hypotenuse gives the vector information for the displacement at the end of 8 hours of driving.

The individual driving times and distances are summed to provide:

(<u>a) How far did he travel?</u>

103 km

<u>(b) What was his average speed?</u>

12.88 km/h

<u>(c) What was his displacement?</u>

73.82 km

<u>(d) What was his average velocity?</u>

9.228 km/h

8 0

3 years ago

Heather and Jerry are standing on a bridge 49 mm above a river. Heather throws a rock straight down with a speed of 17 m/sm/s .

lara [203]

Answer:

3.467 s

Explanation:

given,

distance , d = 49 mm = 0.049 m

initial speed of the of the rock, v = 17 m/s

time taken by the Heather rock to reach water

using equation of motion

$s = ut +\dfrac{1}{2}at^2$

taking downward as negative

$-0.049 = -17 t -\dfrac{1}{2}\times 9.8\times t^2$

4.9 t² + 17 t - 0.049 = 0

now,

$t_1 = \dfrac{-(17)\pm \sqrt{17^2 - 4\times 4.9 \times (-0.049)}}{2\times 4.9}$

t₁ = -3.47 s , 0.0028 s

rejecting negative values

t₁ = 0.0028 s

now, time taken by the ball of Jerry

using equation of motion

$s = ut +\dfrac{1}{2}at^2$

taking downward as negative

$-0.049 = 17 t -\dfrac{1}{2}\times 9.8\times t^2$

4.9 t² - 17 t - 0.049 = 0

now,

$t_2 = \dfrac{-(-17)\pm \sqrt{17^2 - 4\times 4.9 \times (-0.049)}}{2\times 4.9}$

t₂ = 3.47 s ,-0.0028 s

rejecting negative values

t₂ = 3.47 s

now, time elapsed is = t₂ - t₁ = 3.47 - 0.0028 = 3.467 s

5 0

3 years ago

Lightweight, vertically suspended spiral spring with a spring constant of 8.6 N / m is fitted with 64 g weight. The weight shall

alexandr1967 [171]

Answer:

Explanation:

Given that

Force constant k=8.6N/m

Weight =64g=64/1000=0.064kg

Extension is 45mm=45/1000= 0.045m

It will have it highest spend when the Potential energy is zero

Therefore energy in spring =change in kinetic energy

Ux=∆K.e

½ke² = ½mVf² — ½mVi²

Initial velocity is 0, Vi=0m/s

½ke² = ½mVf²

½ ×8.6 × 0.045² = ½ ×0.064 ×Vf²

0.0087075 = 0.032 Vf²

Then, Vf² = 0.0087075/0.032

Vf² = 0.2721

Vf=√0.2721

Vf= 0.522m/s

The time it will have this maximum velocity?

Using equation of motion

Vf= Vi + gr

0.522= 0+9.81t

t=0.522/9.81

t= 0.0532sec

t= 53.2 milliseconds

5 0

3 years ago

A particle moves horizontally in uniform circular motion, over a horizontal xy plane. At one instant, it moves through the point

Alexus [3.1K]

Answer:

(2.5,0)

Explanation:

The particle can be described by the following equations:

$x=Rsin(-\omega t)+2.5\\y=Rcos(-\omega t)\\\frac{dx}{dt}=-\omega Rcos(-\omega t)\\\frac{dy}{dt}=\omega Rsin(-\omega t)\\\frac{d^2x}{dt^2}=-\omega^2Rsin(-\omega t)\\\frac{d^2y}{dt^2}=-\omega^2Rcos(-\omega t)$

For R = 2.5, ω = 2 and t = 0:

$x=2.5\\y=2.5\\\\\frac{dx}{dt}=-5\\ \frac{d^2y}{dt^2}=-10$

The center of the circle would be at point (2.5,0)

5 0

3 years ago