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alex41 [277]
3 years ago
11

A 320-kg satellite experiences a gravitational force of 800 N. What is the radius of the satellite’s orbit? What is its altitude

?
Physics
1 answer:
Westkost [7]3 years ago
3 0

Answer:

Radius of orbit = 3.992 × 10^{7} m

Altitude of Satellite =33541.9×  m

Explanation:

Formula for gravitational force for a satellite of mass m  moving in an orbit of radius r around a planet of mass M is given by;

F= G\frac{m M}{r^{2} }

Where G = Gravitational constant = 6.67408 × 10-11 \frac{m^{3} }{Kg sec^{2} }

We are given

F= 800 N

m = 320 Kg

M = 5.972 × 10^{24} Kg

G = 6.67408 × 10-11 \frac{m^{3} }{Kg sec^{2} }

We have to find radius r =?

putting values in formula;

==> 800 =6.67408 ×10^{-11}× 320 × 5.972 × 10^{24} / r^{2}

==> 800= 39.8576 × 10^{13} × 320 / r^{2}

==> 800 = 12754.43 × 10^{13} / r^{2}

==> r^{2} = 12754.43 × 10^{13} /800

==> r^{2}=15.94 × 10^{13}

==> r =  3.992 × 10^{7} m

==> r = 39920×10^{3} m

This is the distance of satellite from center of earth. To find altitude we need distance from surface of earth. So we will subtract radius of earth from this number to find altitude.

Radius of earth =6378.1 km = 6378.1 × 10^{3} m

Altitude = 39920×10^{3} - 6378.1 ×

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An Atwood's machine consists of two different masses, both hanging vertically and connected by an ideal string which passes over
Tasya [4]

Answer:

V₁ = √ (gy / 3)

Explanation:

For this exercise we will use the concepts of mechanical energy, for which we define energy n the initial point and the point of average height and / 2

Starting point

    Em₀ = U₁ + U₂

    Em₀ = m₁ g y₁ + m₂ g y₂

Let's place the reference system at the point where the mass m1 is

     y₁ = 0

    y₂ = y

    Em₀ = m₂ g y = 2 m₁ g y

End point, at height yf = y / 2

    E_{mf} = K₁ + U₁ + K₂ + U₂

    E_{mf} = ½ m₁ v₁² + ½ m₂ v₂² + m₁ g y_{f} + m₂ g y_{f}

Since the masses are joined by a rope, they must have the same speed

     E_{mf} = ½ (m₁ + m₂) v₁² + (m₁ + m₂) g y_{f}

   E_{mf}= ½ (m₁ + 2m₁) v₁² + (m₁ + 2m₁) g y_{f}

How energy is conserved

   Em₀ =  E_{mf}

   2 m₁ g y = ½ (m₁ + 2m₁) v₁² + (m₁ + 2m₁) g y_{f}

   2 m₁ g y = ½ (3m₁) v₁² + (3m₁) g y / 2

   3/2 v₁² = 2 g y -3/2 g y

   3/2 v₁² = ½ g y

   V₁ = √ (gy / 3)

5 0
3 years ago
What fraction of all the electrons in a 25 mg water
mihalych1998 [28]

Answer:

9.11\times 10^{-15}.

Explanation:

The water droplet is initially neutral, it will obtain a 40 nC of charge when a charge of  -40 nC is removed from the water droplet.

The charge on one electron, \rm e=-1.6\times 10^{-19}\ C.

Let the N number of electrons have charge -40 nC, such that,

\rm Ne=-40\ nC\\\Rightarrow N=\dfrac{-40\ nC}{e}=\dfrac{-40\times 10^{-9}\ C}{-1.6\times 10^{-19}\ C}=2.5\times 10^{11}.  

Now, mass of one electron = \rm 9.11\times 10^{-31}\ kg.

Therefore, mass of N electrons = \rm N\times 9.11\times 10^{-31}=2.5\times 10^{11}\times 9.11\times 10^{-31}=2.2775\times 10^{-19}\ kg.

It is the mass of the of the water droplet that must be removed in order to obtain a charge of 40 nC.

Let it is m times the total mass of the droplet which is 25\ \rm mg = 25\times 10^{-6}\ kg.

Then,

\rm m\times (25\times 10^{-3}\ kg) = 2.2775\times 10^{-19}\ kg.\\m=\dfrac{2.2775\times 10^{-19}\ kg}{25\times 10^{-3}\ kg}=9.11\times 10^{-15}.

It is the required fraction of mass of the droplet.

3 0
3 years ago
Which determines the reactivity of an alkali metal?
shutvik [7]

Answer:

A

Explanation:

I'm pretty sure it's A because when you look up the properties of alkali metal it states stuff about its melting points and what to expect. (correct me if I'm wrong but I had this same question on a test and I'm sure this is the answer lol)

8 0
3 years ago
what are the main morphological types of galaxies, and what is the difference in their rotation patterns?
ankoles [38]

The main morphological types of galaxies are elliptical, spiral, and irregular.

Based on their morphology , galaxies have been classified into 3 types namely elliptical, spiral, and irregular.

These galaxies have various sizes and shapes ranging from dwarf galaxies to giant galaxies.

Elliptical Galaxy:

  • The shape of it is generally circular
  • These are the largest among all the types of galaxies because according to astronomers, it is formed by the merger of other small galaxies.
  • Their rotational pattern is symmetric.

Spiral Galaxy:

  • A spiral galaxy consists of a bright nucleus surrounded by a thin outer disk forming a spiral shape.
  • This type of galaxy is the most common in our universe.
  • It is divided into three classes: Spiral a, Spiral b, and Spiral c.
  • Their rotational pattern has circular symmetry.

Irregular Galaxy:

  • These types of galaxies have no central nucleus and irregular arms which are bluish.
  • They don’t have any rotational symmetry.

To know more about "galaxies", refer to the following link:

brainly.com/question/24836631?referrer=searchResults

#SPJ4

3 0
2 years ago
As a cold air mass advances on a warm air mass, what usually comes before it?​
Marina CMI [18]

Answer: A cold front occurs when a cold air mass advances into a region occupied by a warm air mass. If the boundary between the cold and warm air masses doesn't move, it is called a stationary front.

Explanation: Two types of occluded front exist: the warm-type and the cold-type. They’re distinguished by the relative temperatures of the air mass ahead of the occlusion – in other words, the air mass ahead of the original warm front – and the air mass behind the cold front. If the air behind the cold front is colder than the air ahead of the occlusion, it shoves beneath that air (because it’s denser) to form a cold-type occluded front. If the air behind the cold front is warmer than the air ahead, it rides over it to form a warm-type occluded front – which appears to be the more common case. In either situation, the lighter warm air representing the air mass originally between the warm and cold fronts sits above the boundary between the two cooler air masses.

Hope this helps!!

8 0
3 years ago
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