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Firdavs [7]
3 years ago
14

in this model, the velocity of the spacecraft at position 2 is A.) equal to B.) greater than C.) less than the velocity of the c

raft at position 4. At position 1, the direction of the spacecraft changes because of A.) the gravitational force between earth and the spacecraft B.) the momentum of the spacecraft C.) secondary fuel tanks. Position 3 represents the A.) final destination of the spacecraft B.) gravitational force of the spacecraft on Earth C.) orbital path or velocity of Earth

Physics
2 answers:
slava [35]3 years ago
8 0

Answer:

1. Right answer: the velocity of the spacecraft at position 2 is greater than the velocity of the craft at position 4.

This is due the gravity field of the planet (The Earth in this case) is used to accelerate the craft. This is true when in a specific point the direction of the movement of the craft is the same direction of the movement of the planet.

In this case the craft will be “catched” by the Earth’s gravitational field, making the craft  to enter a circular orbit.

2. Right answer: At position 1, the direction of the spacecraft changes because of the gravitational force between Earth and the spacecraft.

As explained in the prior answer, this is the exact and correct point where the trajectory of the spacecraft enters into a circular orbit because of the attraction due gravity of the Earth and therefore changes its direction.

3. Right answer: Position 3 represents the orbital path or velocity of Earth

Being this the orbital path of the Earth and considering the trajectory of the craft, the condition of accelerating the craft is accomplished.

If the orbital path of the Earth were the opposite from the shown in the figure, the effect on the craft would be braking.

Note all of these is related to the gravitational assistance.

Gravitational assistance is the maneuver in which the energy of the gravitational field of a planet or satellite is used to obtain an acceleration or braking of the probe changing its trajectory.

This maneuver is also called slingshot effect, swing-by or gravity assist. It is a common technique in space for the outer Solar System missions , in order to save costs in the launch rocket and thrusters.

finlep [7]3 years ago
6 0
<h2>1. Right answer: the velocity of the spacecraft at position 2 is <u>greater than</u> the velocity of the craft at position 4. </h2>

This is due the gravity field of the planet (The Earth in this case) is used to accelerate the craft. This is true when in a specific point the direction of the movement of the craft is the same direction of the movement of the planet.

In this case the craft will be “catched” by the Earth’s gravitational field, making the craft  to enter a circular orbit.

<h2>2. Right answer: At position 1, the direction of the spacecraft changes because of <u>the gravitational force between Earth and the spacecraft. </u></h2>

As explained in the prior answer, this is the exact and correct point where the trajectory of the spacecraft enters into a circular orbit because of the attraction due gravity of the Earth and therefore changes its direction.


<h2>3. Right answer: Position 3 represents <u>the orbital path or velocity of Earth </u></h2>

Being this the orbital path of the Earth and considering the trajectory of the craft, the condition of accelerating the craft is accomplished.

If the orbital path of the Earth were the opposite from the shown in the figure, the effect on the craft would be braking.

Note all of these is related to the <u>gravitational assistance. </u>

<u>Gravitational assistance</u> is the maneuver in which the energy of the gravitational field of a planet or satellite is used to obtain an acceleration or braking of the probe changing its trajectory.

This maneuver is also called <em>slingshot effect, swing-by</em> or <em>gravity assist</em>. It is a common technique in space for the outer Solar System missions , in order to save costs in the launch rocket and thrusters.


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A small electric motor produces a force of 5 N that moves a remote-control car 5 m every second. How much power does the motor p
kati45 [8]

Answer:

Explanation:

Given:

Force, f = 5 N

Velocity, v = 5 m/s

Power, p = energy/time

Energy = mass × acceleration × distance

Poer, p = force × velocity

= 5 × 5

= 25 W.

Note 1 watt = 0.00134 horsepower

But 25 watt,

0.00134 hp/1 watt × 25 watt

= 0.0335 hp.

4 0
3 years ago
Read 2 more answers
A ball is thrown horizontally from the top of a 60 m building and lands 100 m from the base of the building. How long is the bal
zhannawk [14.2K]

Answer:

The ball is in the air for 3.5 seconds

The initial horizontal component of velocity is 28.6 m/s

The vertical component of the final velocity is 34.3 m/s downward

The final velocity is 44.7 m/s in the direction 50.2° below the horizontal

Explanation:

A ball is thrown horizontally

That means the vertical component of the initial velocity u_{y}=0

The initial velocity is the horizontal component u_{x}

The ball is thrown from the top of a 60 m

That means the vertical displacement component y = 60 m

→ y = u_{y} t + \frac{1}{2} gt²

where g is the acceleration of gravity and t is the time

y = -60 m , g = -9.8 m/s² , u_{y}=0

Substitute these values in the rule

→ -60 = 0 + \frac{1}{2} (-9.8)t²

→ -60 = -4.9t²

Divide both sides by -4.9

→ 12.2449 = t²

Take √ for both sides

∴ t = 3.5 seconds

* <em>The ball is in the air for 3.5 seconds </em>

The initial velocity is the horizontal component u_{x}

The ball lands 100 meter from the base of the building

That means the horizontal displacement x = 100 m

→ x = u_{x} t

→ t = 3.5 s , x = 100 m

Substitute these values in the rule

→ 100 = u_{x} (3.5)

Divide both sides by 3.5

→ u_{x} = 28.57 m/s

<em>The initial horizontal component of velocity is 28.6 m/s</em>

The vertical component of the final velocity is v_{y}

→ v_{y} = u_{y} + gt

→ u_{y} = 0 , g = -9.8 m/s² , t = 3.5 s

Substitute these values in the rule

→ v_{y} = 0 + (-9.8)(3.5)

→ v_{y} = -34.3 m/s

<em>The vertical component of the final velocity is 34.3 m/s downward</em>

The final velocity v is the resultant vector of  v_{x} and v_{y}

→ Its magnetude is v=\sqrt{(v_{x})^{2}+(v_{y})^{2}}

→ Its direction tan^{-1}\frac{v_{y}}{v_{x}}

→ v_{y} = 28.6 , v_{y} = -34.3

Substitute this values in the rules above

→ v=\sqrt{(28.6)^{2}+(-34.3)^{2}}=44.66

→ Its direction tan^{-1}\frac{-34.3}{28.6}=-50.18

The negative sign means the direction is below the horizontal

<em>The final velocity is 44.7 m/s in the direction 50.2° below the horizontal</em>

7 0
3 years ago
It would really help if anyone could answers please and thanks
mojhsa [17]

well it would be A because 55 degrees is going strait well 75 is going literally straight up

4 0
3 years ago
Read 2 more answers
You pull with a force of 295 N on a rope that is attached to a block of mass 22 kg, and the block slides across the floor at a c
Sergeeva-Olga [200]

Answer:

Fnet = 0

Explanation:

  • Since the block slides across the floor at constant speed, this means that it's not accelerated.
  • According Newton's 2nd Law, if the acceleration is zero, the net force on the sliding mass must be zero.
  • This means that there must be a friction force opposing to the horizontal component of the applied force, equal in magnitude to it:

       F_{appx} = F_{app} * cos \theta = 295 N * cos 35 = 242 N  (1)

  • In the vertical direction, the block is not accelerated either, so the sum of the normal force and the vertical component of the applied force, must be equal in magnitude to the force of gravity on the block:

      F_{appy} = F_{app} * cos \theta = 295 N * sin 35 = 169 N  (2)

⇒    169 N + Fn = Fg = 216 N  (3)

  • This means that there must be a normal force equal to the difference between Fappy and Fg, as follows:
  • Fn = 216 N - 169 N = 47  N (4)

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3 years ago
If the object on the right gained mass so that it had as much mass as the object on the left, how would the gravitational force
just olya [345]
The object would stay constant
5 0
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