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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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Astronauts aboard the U.S.S. Burger decide to fire the rocket thrusters for 3.0 seconds to make a course correction. While the t
Cloud [144]

Answer:

340 m/s

Solution:

As per the question;

Time, t = 3.0 s

Total distance moved by the ship, d = 10,500 km

The increase in speed, v = 340 m/s

Now,

To calculate the U.S.S Burger's change in speed, \Delta V:

The final velocity is given by:

v_{f} = 340 + v_{i}                      (1)

where

v_{f} = final\ velocity

v_{i} = initial\ velocity

Also, the change in velocity is given by:

\Delta v = v_{f} - v_{i}                    (2)

Now, from eqn (1) and (2):

\Delta = 340 + v_{i} - v_{i} = 340\ m/s

8 0
3 years ago
An apple is initially sitting on the bottom shelf of a pantry. A hungry physics student picks up the apple to eat it, but change
andre [41]

Answer:

Zero

Explanation:

Work done is given by multiplying force and distance moved. The distance is moved both positive and negative and it's equal distance. Since force used is the same hence work

W=F*d+ (F*-d)=0

Therefore, total work done is zero

8 0
3 years ago
An ideal photo-diode of unit quantum efficiency, at room temperature, is illuminated with 8 mW of radiation at 0.65 µm wavelengt
nadya68 [22]

Answer:

I = 4.189 mA    V = 0.338 V

Explanation:

In order to do this, we need to apply the following expression:

I = Is[exp^(qV/kT) - 1]   (1)

However, as the junction of the diode is illuminated, the above expression changes to:

I = Iopt + Is[exp^(qV/kT) - 1]   (2)

Now, as the shunt resistance becomes infinite while the current becomes zero, we can say that the leakage current is small, and so:

I ≅ Iopt

Therefore:

I ≅ I₀Aλq / hc  (3)

Where:

I₀A = Area of diode (radiation)

λ: wavelength

q: electron charge (1.6x10⁻¹⁹ C)

h: Planck constant (6.62x10⁻³⁴ m² kg/s)

c: speed of light (3x10⁸ m/s)

Replacing all these values, we can get the current:

I = (8x10⁻³) * (0.65x10⁻⁶) * (1.6x10⁻¹⁹) / (6.62x10⁻³⁴) * (3x10⁸)

I = 4.189x10⁻³ A or 4.189 mA

Now that we have the current, we just need to replace this value into the expression (2) and solve for the voltage:

I = Is[exp^(qV/kT) - 1]

k: boltzman constant (1.38x10⁻²³ J/K)

4.189x10⁻³ = 9x10⁻⁹ [exp(1.6x10⁻¹⁹ V / 1.38x10⁻²³ * 300) - 1]

4.189x10⁻³ / 9x10⁻⁹ = [exp(38.65V) - 1]

465,444.44 + 1  = exp(38.65V)

ln(465,445.44) = 38.65V

13.0508 = 38.65V

V = 0.338 V

6 0
3 years ago
A space explorer is moving through space far from any planet or star. He notices a large rock, taken as a specimen from an alien
yaroslaw [1]
He should push it gently.
This is because the forces of resistance this situation are minimal, so the rock will not slow as it would on Earth. Kicking the rock may result in it travelling too fast and hitting something else, causing damage. Moreover, the rock could start rebounding off of surfaces and create havoc. 
3 0
3 years ago
The height of the tide measured at a seaside community varies according to the number of hours t after midnight. If the height h
antoniya [11.8K]

Explanation:

Given that, the height of the tide measured at a seaside community varies according to the number of hours t after midnight. The height is given by the equation as :

h=-\dfrac{1}{2}t^2+6t-9

When the tide first be at 6 ft, put h = 6 ft in above equation as :

-\dfrac{1}{2}t^2+6t-9=6

-t^2+12t-18=0

On solving the above equation to find the value of t. It is equal to :

t = 3.551 seconds

or

t = 8.449 seconds

So, the tide of 6 ft is at  3.551 seconds and 8.449 seconds. Hence, this is the required solution.

6 0
3 years ago
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