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Vikki [24]
3 years ago
14

You are the pilot of a spaceship (9.1835 × 104 kg) and you enter into elliptical orbit around an uncharted planet, Rosseforp. At

periapsis (closest approach to the planet), your ship is 4.5 × 107 m above the surface of the planet; at apoapsis (most distant) it is 2.257×108 m above the surface. The mass of Rosseforp is found by your ship’s equipment to be 1.96×1025 kg, and the period of your ship’s orbit is 83.37 hours
Find the following:i. the semimajor axis of your ship’s orbit,ii. the radius of planet Rosseforp,iii. the eccentricity of your ship’s orbit,iv. the speed of your ship at perihelion and aphelion,

Physics
1 answer:
Kaylis [27]3 years ago
7 0

Answer:

Check the explanation

Explanation:

The orbital period of a satellite that is given as (T) and the mean distance from the central body (R) are connected by the following equation: representing T as the satellite period, R will be represented as the average radius of orbit for the satellite (which is the distance from center of central planet)

Kindly check the attached image below to see the step by step explanation to the above question.

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Which kind of force do you exert on an object when you pull it toward you?
jarptica [38.1K]

By definition we have to:

Applied force: It is the external force that acts directly on a body.

Therefore, we can say that if you have an object and push it towards yourself, you are exerting an external force on the object.

This external force was not acting on the object previously, therefore, it is a force that you are applying at that moment.

Answer:

you exert an Applied Force on an object when you pull it towards you

A. Applied Force

4 0
3 years ago
Read 2 more answers
A hockey puck is sliding across a frozen pond with an initial speed of 9.3 m/s. It comes to rest after sliding a distance of 42.
kondaur [170]

Answer:

The coefficient of kinetic friction between the puck and the ice is 0.11

Explanation:

Given;

initial speed, u = 9.3 m/s

sliding distance, S = 42 m

From equation of motion we determine the acceleration;

v² = u² + 2as

0 = (9.3)² + (2x42)a

- 84a = 86.49

a = -86.49/84

|a| = 1.0296

F_k = \mu_k N = ma

where;

Fk is the frictional force

μk is the coefficient of kinetic friction

N is the normal reaction = mg

μkmg = ma

μkg = a

μk = a/g

where;

g is the gravitational constant = 9.8 m/s²

μk = a/g

μk = 1.0296/9.8

μk = 0.11

Therefore, the coefficient of kinetic friction between the puck and the ice is 0.11

3 0
3 years ago
Someone help please by providing work and answers please :)
Nastasia [14]
First we gotta use an equation of motion:

d = ut + \frac{1}{2} a {t}^{2}

Our vertical distance d= 100 m, initial vertical speed u = 0 m/s (because velocity is fully horizontal), and vertical acceleration a = 9.8 m/s2 because of gravity. Let's plug it all in!

100 = 0 + \frac{1}{2} (9.8) {t}^{2}

Now we just need to solve for t:

{t}^{2} = \frac{2(100)}{9.8} \\ \\ t = \sqrt{\frac{2(100)}{9.8}}

Hit the calculators, and you'll get 4.5 seconds!
5 0
3 years ago
determine whether the following equations are dimensionally correct if not how can you make them dimensionally correct 1 /2 mv2
serg [7]

Answer:

1 /2 mv^{2} =mgh

Explanation:

The correct equation follows the law of conservation of energy where kinetic energy is all transformed to potential energy, since we know that kinetic energy is expressed as

1 /2 mv^{2} while potential energy is mgh where m is the mass of the object, v is the velocity or speed of the object, g is acceleration due to gravity and h is the vertical height. Therefore, relating the two equations we should have 1 /2 mv^{2} =mgh

6 0
3 years ago
The sun is 150,000,000 km from the earth.
aniked [119]

Answer:

<h3>What is the angular speed of the earth around the sun? </h3>

It takes the Earth approximately 23 hours, 56 minutes and 4.09 seconds to make one complete revolution (360 degrees). This length of time is known as a sidereal day. The Earth rotates at a moderate angular velocity of

7.2921159 × 10 {}^{ - 5}  \:  \:  \frac{radians}{second}

<h3>What is the tangential speed of the earth? </h3>

The earth rotates once every 23 hours, 56 minutes and 4.09053 seconds, called the sidereal period, and its circumference is roughly 40,075 kilometers. Thus, the surface of the earth at the equator moves at a speed of 460 meters per second--or roughly 1,000 miles per hour.

8 0
3 years ago
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