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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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Answer:

h=18.05 cm

Explanation:

Given that

m= 25 kg

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x=26.4 cm

θ= 19.5 ∘

When the block just leave the spring then the speed of block = v m/s

From energy conservation

\dfrac{1}{2}Kx^2=\dfrac{1}{2}mv^2

Kx^2=mv^2

v=\sqrt{\dfrac{kx^2}{m}}

By putting the values

v=\sqrt{\dfrac{kx^2}{m}}

v=\sqrt{\dfrac{1300\times 0.264^2}{25}}

v=1.9 m/s

When block reach at the maximum height(h) position then the final speed of the block will be zero.

We know that

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By putting the values

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4 0
3 years ago
You are comparing two diffraction gratings using two different lasers: a green laser and a red laser. You do these two experimen
Nikolay [14]

Answer:

a. (a) grating A has more lines/mm; (b) the first maximum less than 1 meter away from the center

Explanation:

Let  n₁ and n₂ be no of lines per unit length  of grating A and B respectively.

λ₁ and λ₂ be wave lengths of green and red respectively , D be distance of screen and d₁ and d₂ be distance between two slits of grating A and B ,

Distance of first maxima for green light

= λ₁ D/ d₁

Distance of first maxima for red light

= λ₂ D/ d₂

Given that

λ₁ D/ d₁ = λ₂ D/ d₂

λ₁ / d₁ = λ₂ / d₂

λ₁ / λ₂  = d₁ / d₂

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λ₁  <  λ₂

d₁ < d₂

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no of lines per unit length  ∝ 1 / d

If grating B is illuminated with green light first maxima will be at distance

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λ₁ D/ d₂ < λ₂ D/ d₂

λ₁ D/ d₂ < 1 m

In this case position of first maxima will be less than 1 meter.

Option a is correct .

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3 years ago
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Answer:

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