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Lubov Fominskaja [6]
1 year ago
14

A spaceship which is 50,000 kilometers from the center of Earth has a mass of 3,000 kilograms. What is the magnitude of the forc

e of gravity acting on the spaceship? (The value of G is 6.673 × 10-11 newton meter2/kilogram2. The mass of Earth is 5.98 × 1024 kilograms.) 400 newtons 478 newtons 500 newtons 1595 newtons
Physics
2 answers:
Anna11 [10]1 year ago
6 0
Fg=Gx(M1M2/r^2)
Fg=478N
vovangra [49]1 year ago
5 0

Answer:

The correct option is 478N

Explanation:

Between two objects of a certain mass exists a force call the gravitational force. This force is the ''attraction'' force between the objects.

The equation to calculate this force is :

F_{G}=\frac{G.m_{1}.m_{2}}{d^{2}} (I)

Where F_{G} is the gravitational force.

Where G is the gravitational constant.

m_{1} and m_{2} are the masses of each object.

And d is the distance between the objects (In fact is the distance between the mass centroid of each object).

In order to calculate the gravitational force, we need to replace the data in the equation.

The distance 50000km is equal to :

50,000km.(\frac{1000m}{1km})=50,000,000m

Now, if we replace in the equation (I) all the data :

F_{G}=\frac{(6.673).(10)^{-11}\frac{Nm^{2}}{kg^{2}}.3000kg.(5.98).10^{24}kg}{(50,000,000m)^{2}}=478.854N

F_{G}=478.854N ≅ 478 N

We find that the magnitude of the force of gravity acting on the spaceship is 478 N.

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Define couple and give 2 examples
Elodia [21]

Answer:

Two equal and opposite parallel forces not acting along the same line, form a couple. A couple is always needed to produce the rotation.

For example, turning a key in a lock and turning a steering wheel.

3 0
1 year ago
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In March 2006, two small satellites were discovered orbiting Pluto, one at a distance of 48,000 km and the other at 64,000 km. P
tatiyna

Answer:

Time period for first satellites 24.46 days and for second satellites 37.67 days

Explanation:

Given :

Distance of first satellites r_{sat1} = 48000 \times 10^{3} m

Distance of second satellites r _{sat2} = 64000 \times 10^{3} m

Distance of charon r_{c} = 19600 \times 10^{3} m

Time period of charon T_{c} = 6.39 days

From the kepler's third law,

Square of the time period is proportional to the cube of the semi major axis.

   T^{2} = r^{3}

   \frac{T}{r^{\frac{3}{2} } } = constant

For first satellites,

  \frac{T_{c} }{r_{c} ^{\frac{3}{2} }  }  = \frac{T_{sat1} }{r_{sat1} ^{\frac{3}{2} }  }

{T_{sat1} } = 6.39 \times \frac{(48000 \times 10^{3} )^{\frac{3}{2} } }{(19600\times 10^{3} )^{\frac{3}{2} }}

T_{sat1} = 24.46 days

For second satellites,

   \frac{T_{c} }{r_{c} ^{\frac{3}{2} }  }  = \frac{T_{sat2} }{r_{sat2} ^{\frac{3}{2} }  }

{T_{sat2} } = 6.39 \times \frac{(64000 \times 10^{3} )^{\frac{3}{2} } }{(19600\times 10^{3} )^{\frac{3}{2} }}

T_{sat2} = 37.67 days

Therefore, time period for first satellites = 24.46 days and for second satellites 37.67 days

8 0
2 years ago
A light ray strikes a plane mirror at an angle of 23° to the normal. What is the angle between the reflected ray and the mirror?
tiny-mole [99]

Answer:67^{\circ}

Explanation:

Given

Ray of light incident on plane making an angle of 23^{\circ} with normal

Thus incident angle i=23^{\circ}

So according to the law of incidence is equal to angle of reflection

\angle i=\angle r

thus \angle r=23^{\circ}

and angle between mirror and reflected ray =90^{\circ}-23^{\circ}

=67^{\circ}

4 0
2 years ago
At an oceanside nuclear power plant, seawater is used as part of the cooling system. This raises the temperature of the water th
Mice21 [21]
Answer is 30. just took it
8 0
2 years ago
Your film idea is about drones that take over the world. In the script, two drones are flying horizontally at the same speed and
Stella [2.4K]

Answer:

vₓ = 20 m/s,    v_{y}  = -15 m / s

Explanation:

This is a conservation of moment problem, since it is a vector quantity we can work each axis independently

The system is formed by the two drones, so the forces during the crash are internal and the moment is conserved

X axis

Initial moment. Before the crash

         p₀ = m₁ v₀ₓ + m₂ v₀ₓ

Final moment. After the crash

       p_{fx} = (m₁ + m₂) vₓ

      p₀ₓ = p_{fx}

      m₁ v₀ₓ + m₂ v₀ₓ = (m₁ + m₂) vₓ

       vₓ = (m₁ + m₂) v₀ₓ / (m₁ + m₂)

       vₓ = v₀ₓ  = 20 m/s

Y Axis

Initial

         p_{oy} = m₁ v_{oy}

Final

         p_{fy} = (m₁ + m₂) v_{y}

         p_{oy} = p_{fy}

the drom rises and when it falls it has the same speed because there is no friction    v_{oy} = -60 m/s          

 

           m₁ v_{oy} = (m₁ + m₂) v_{y}

            v_{y} = m₁ / (m₁ + m₂) v_{oy}

            v_{y}  = 1/4    60

            v_{y}  = -15 m / s

Vertical speed is down

5 0
2 years ago
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