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Kisachek [45]
1 year ago
5

A comet of mass 1.20 × 10¹⁰kg moves in an elliptical orbit around the Sun. Its distance from the Sun ranges between 0.500 AU and

50AU . (a) What is the eccentricity of its orbit?
Physics
1 answer:
irina [24]1 year ago
4 0

The eccentricity of its orbit is $$U=-2.13 \times 10^{17} \mathrm{~J} .\end{aligned}$$

<h3>What is mass?</h3>
  • Mass is a physical body's total amount of matter. It also serves as a gauge for the body's inertia or resistance to acceleration (change in velocity) in the presence of a net force. The strength of an object's gravitational pull to other bodies is also influenced by its mass.
  • The kilogram is the SI unit of mass (kg). In science and technology, a body's weight in a given reference frame is the force that causes it to accelerate at a rate equal to the local acceleration of free fall in that frame.
  • For instance, a kilogram mass weighs around 2.2 pounds at the surface of the planet. However, the same kilogram mass would weigh just about 0.8 pounds on Mars and about 5.5 pounds on Jupiter.
  • An object's mass is a crucial indicator of how much stuff it contains. Weight is a measurement of an object's gravitational pull. It is influenced by the object's location in addition to its mass. As a result, weight is a measurement of force.

The length of the semi-major axis is calculated as follows:

where, $G=6.67 \times 10^{-1} \mathrm{~m}^3 / \mathrm{kgs}$

$M=1.99 \times 10^{30} \mathrm{~kg}=$mass of sur

$m=1.20 \times 10^{10} \mathrm{~kg}$ - a mass of the comet

$$\begin{aligned}\therefore \quad \text { At aphelion, } r &=50 \times U \\&=50 \times 1.496 \times 10^{11} \mathrm{~m} . \\U=-\frac{6.67 \times 10^{-11} \times \mathrm{m} 1.99 \times 10^{30} \times 1.20 \times 10^{10}}{50 \times 1.496 \times 10^{11}} \\U=-2.13 \times 10^{17} \mathrm{~J} .\end{aligned}$$

$$U=-2.13 \times 10^{17} \mathrm{~J} .\end{aligned}$$

To learn more about mass, refer to:

brainly.com/question/3187640

#SPJ4

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Answer:

Acceleration of gravity on Noveria = 4.4 m/s²

Explanation:

Commander Shepard, an N7 spectre for Earth, weighs 799 N on the Earth's surface.

We have weight, W = mg

Acceleration due to gravity, g = 9.81m/s²    

799 = m x 9.81

Mass of Shepard, m = 81.45 kg            

She lands on Noveria, a distant planet in our galaxy, she weighs 356 N.

We have weight, W = mg'

                 356 = 81.45 xg'

Acceleration of gravity on Noveria, g' = 4.4 m/s²

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7 0
2 years ago
A pet-store supply truck moves at 25.0 m/s north along a highway. inside, a dog moves at 1.75 m/s at an angle of 35.0° east of
koban [17]

<u>Answer:</u>

 Velocity of the dog relative to the road = 26.04 m/s 3.15⁰ north of east.

<u>Explanation:</u>

  Let the east point towards positive X-axis and north point towards positive Y-axis.

  Speed of truck = 25 m/s north = 25 j m/s

  Speed of dog = 1.75 m/s at an angle of 35.0° east of north = (1.75 cos 35 i + 1.75 sin 35 j)m/s

                          = (1.43 i + 1.00 j) m/s

    Velocity of the dog relative to the road = 25 j + 1.43 i + 1.00 j = 1.43 i + 26.00 j

    Magnitude of velocity = 26.04 m/s

    Angle from positive horizontal axis = 86.85⁰

 So Velocity of the dog relative to the road = 26.04 m/s 86.85⁰ east of north = 26.04 m/s 3.15⁰ north of east.

4 0
3 years ago
A car accelerates from rest at 1.0 m/s2 for 20.0 second along a straight road. It then moves at a constant speed for half an hou
Whitepunk [10]

Total distance = 36500 m

The average velocity = 19.73 m/s

<h3>Further explanation</h3>

Given

vo=initial velocity=0(from rest)

a=acceleration= 1 m/s²

t₁ = 20 s

t₂ = 0.5 hr = 1800 s

t₃= 30 s

Required

Total distance

Solution

State 1 : acceleration

\tt d=vo.t+\dfrac{1}{2}at^2\\\\d=\dfrac{1}{2}\times 1\times 20^2\rightarrow vo=0\\\\d=200~m

\tt vt=vo+at\\\\vt=at\rightarrow vo=0\\\\vt=1\times 20\\\\vt=20~m/s

State 2 : constant speed

\tt d=v\times t\\\\d=20\times 1800\\\\d=36000~m

State 3 : deceleration

\tt vt=vo+at\rightarrow vt=0(stop)\\\\vo=-at\\\\20=-a.30~s\\\\a=-\dfrac{2}{3}m/s^2(negative=deceleration)

\tt d=vot+\dfrac{1}{2}at^2\\\\d=20.30-\dfrac{1}{2}.\dfrac{2}{3}.30^2\\\\d=300~m

Total distance : state 1+ state 2+state 3

\tt 200 + 36000 + 300=36500~m

the average velocity = total distance : total time

\tt avg~velocity=\dfrac{36500}{20~s+1800~s+30~s}=19.73~m/s

4 0
2 years ago
A rotating object starts from rest at t = 0 s and has a constant angular acceleration. At a time of t = 2.5 s the object has an
Evgesh-ka [11]

Answer:

52 rad

Explanation:

Using

Ф = ω't +1/2αt²................... Equation 1

Where Ф = angular displacement of the object, t = time, ω' = initial angular velocity, α = angular acceleration.

Since the object states from rest, ω' = 0 rad/s.

Therefore,

Ф = 1/2αt²................ Equation 2

make α the subject of the equation

α = 2Ф/t².................. Equation 3

Given: Ф = 13 rad, t = 2.5 s

Substitute into equation 3

α = 2(13)/2.5²

α = 26/2.5

α = 4.16 rad/s².

using equation 2,

Ф = 1/2αt²

Given: t = 5 s, α = 4.16 rad/s²

Substitute into equation 2

Ф = 1/2(4.16)(5²)

Ф = 52 rad.

6 0
3 years ago
Read 2 more answers
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