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Rainbow [258]
3 years ago
15

The moon Ariel orbits Uranus at a distance of 1.91 x 108 m once every 2.52 days. Use that data to calculate the mass of Uranus.

[?] x 10kg​
Physics
1 answer:
PIT_PIT [208]3 years ago
4 0

Mu = 8.66 × 10^25 kg

Explanation:

centripetal force = gravitational force

m \frac{ {v}^{2} }{r}  = (grav.const) \frac{m \times mu}{ {r}^{2} }

where

m = mass of moon Ariel

mu = mass of Uranus

r = radius of Ariel's orbit

v = Ariel's velocity around Uranus

To find the velocity, we need to find the circumference of the no orbit and then divide it by the period (2.52 days):

circumference = 2πr = 2π×(1.91 × 10^8 m)

= 1.2 × 10^9 m

period = 2.52 days × (24 h/1 day)×(3600 s/1 hr)

= 2.18 × 10^5 s

v = (1.2 × 10^9 m)/(2.18 × 10^5 s)

= 5.5 × 10^3 m/s

(5.5 × 10^3 m/s)^2/(1.91 × 10^8 m) = (6.67 × 10^-11 m^3/kg-s^2)Mu/(1.91 × 10^8 m)^2

Solving Mu,

Mu = 8.66 × 10^25 kg

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Explain in your own words how the Doppler Effect is also applicable in our study of light.
arlik [135]

well in my own words, i'd saw the the doppler effect is similar to light because sound has a speed, and light does too.

so my theory is if you go fast enough everything would just become black, or maybe white? idk its hard to explain

but what my point is, is taht the doppler effect works in the same way, like if a car is moving towards you the sound is being emitted from the car and being pushed by the speed of the car making it have a much higher pitch, when the car is going away however it drops to a lower pitch due the the sound waves being DRAGGED by the car.

there hoped this helped I guess

8 0
3 years ago
Fluorine has 7 valence electrons. Which charge is its ion likely to have?(1 point)
7nadin3 [17]

Answer:

1–

Explanation:

The fluorine is the element with biggest electronegativity in the periodic table, so it usually always take an electron and gets charge 1–

5 0
3 years ago
A steel beam that is 5.50 m long weighs 332 N. It rests on two supports, 3.00 m apart, with equal amounts of the beam extending
ElenaW [278]

Explanation:

The given data is as follows.

    Length of beam, (L) = 5.50 m

    Weight of the beam, (W_{b}) = 332 N

     Weight of the Suki, (W_{s}) = 505 N

After crossing the left support of the beam by the suki then at some overhang distance the beam starts o tip. And, this is the maximum distance we need to calculate. Therefore, at the left support we will set up the moment and equate it to zero.

                 \sum M_{o} = 0

     -W_{s} \times x + W_{b} \times 1.5 = 0

                x = \frac{W_{b} \times 1.5}{W_{s}}

                   = \frac{332 N \times 1.5}{505 N}

                   = 0.986 m

Hence, the suki can come (2 - 0.986) m = 1.014 from the end before the beam begins to tip.

Thus, we can conclude that suki can come 1.014 m close to the end before the beam begins to tip.

8 0
4 years ago
A wheel free to rotate about its axis that is not frictionless is initially at rest. A constant external torque of +46 N·m is ap
Rom4ik [11]

Answer:

Explanation:

Let Torque due to friction be

F  

Net torque

= 46 - F

Angular impulse = change in angular momentum

=(  46 - F ) x 17  = I X 580

When external torque is removed , only friction creates torque reducing its speed to zero in 120 s so

Angular impulse = change in angular momentum

F  x 120 = I X 580

(  46 - F ) x 17 = F  x 120

137 F = 46 x 17

F = 5.7 Nm

b )

Putting this value in first equation

5.7 x 120 = I x 580

I = 1.18 kg m²

8 0
3 years ago
A car tire rotates with an average angular speed of 32 rad/s. In what time interval will the tire rotate 3.5 times? Answer in un
stiks02 [169]

Answer:

\Delta t = 0.687\ s

Explanation:

given,

Angular speed of the tire = 32 rad/s

Displacement of the wheel = 3.5 rev

Δ θ = 3.5 x 2 π

        = 7 π rad

now,

Time interval of the car to rotate 7π rad

using equation

\omega_{avg} =\dfrac{\Delta \theta}{\Delta t}

\Delta t=\dfrac{7\pi}{32}

\Delta t = 0.687\ s

Time taken to rotate 3.5 times is equal to 0.687 s.

3 0
3 years ago
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