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Anna35 [415]
3 years ago
9

You are the Engineering Duty Officer getting your submarine, the USS GREENVILLE, ready to put to sea. When nuclear material in t

he reactor is fissioning at an increasing rate, this is known as a:__________
Physics
1 answer:
podryga [215]3 years ago
7 0

Answer:

Super-critical mass

Explanation:

This term refers to the mass, in which the amount of fission processes per unit of time increases to the point, where some intrinsic feedback mechanism causes the reactor to reach an equilibrium point at a high temperature or power, that is, It becomes critical again, or it is destroyed due to the amount of processes.

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Problem 1: Spherical mirrorConsider a spherical mirror of radius 2 m, and rays which go parallel to the optic axis. What is thep
SIZIF [17.4K]

Answer:

1) iii i= 1m, 2)  iii and iv, 3)  i = f₂ (L-f₁) / (L - (f₁ + f₂))

Explanation:

Problem 1

For this problem we use two equations the equations of the focal distance in mirrors

              f = r / 2

              f = 2/2

             f = 1 m

The builder's equation

           1 / f = 1 / o + 1 / i

Where f is the focal length, "o and i" are the distance to the object and the image respectively.

For a ray to arrive parallel to the surface it must come from infinity, whereby o = ∞ and 1 / o = 0

              1 / f = 0 + 1 / i

              i = f

              i = 1 m

The image is formed at the focal point

The correct answer is iii

Problem 2

For this problem we have two possibilities the lens is convergent or divergent, in both cases the back face (R₂) must be flat

Case 1 Flat lens - convex (convergent)

              R₂ = infinity

              R₁ > 0

Cas2 Flat-concave (divergent) lens

             R₂ = infinity

              R₁ <0

Why the correct answers are iii and iv

Problem 3

For a thick lens the rays parallel to the first surface fall in their focal length (f₁), this is the exit point for the second surface whereby the distance to the object is o = L –f₁, let's apply the constructor equation to this second surface

          1 / f₂ = 1 / (L-f₁) + 1 / i

          1 / i = 1 / f₂ - 1 / (L-f₁)

           1 / i = (L-f₁-f₂) / f₂ (L-f₁)

           i = f₂ (L-f₁) / (L - (f₁ + f₂))

This is the image of the rays that enter parallel to the first surface

6 0
3 years ago
The force exerted by the wind on the sails of a sailboat is Fsail = 330 N north. The water exerts a force of Fkeel = 210 N east.
Elena L [17]

Answer:

The magnitude of the acceleration is a_r = 1.50 \ m/s^2

The direction is  \theta =  32.5 6^o north of  east

Explanation:

From the question we are told that

   The force exerted by the wind is  F_{sail} =  (330 ) \ N \ north

   The force exerted by water is  F_{keel} =  (210  ) \ N \ east

      The mass of the boat(+ crew) is  m_b  =  260  \ kg

Now Force is mathematically represented as

      F =  ma

Now the acceleration towards the north is mathematically represented as

      a_n  =  \frac{F_{sail}}{m_b}

substituting values

       a_n  =  \frac{330 }{260}

      a_n  =  1.269 \ m/s^2

Now the acceleration towards the east is mathematically represented as

       a_e = \frac{F_{keel}}{m_b }

substituting values

      a_e = \frac{210}{260}

      a_e =0.808 \ m/s^2

The resultant acceleration is  

      a_r =  \sqrt{a_e^2 + a_n^2}

substituting values

     a_r =  \sqrt{(0.808)^2 + (1.269)^2}

      a_r = 1.50 \ m/s^2

The direction with reference from the north is evaluated as

Apply SOHCAHTOA

        tan \theta =  \frac{a_e}{a_n}

       \theta = tan ^{-1} [\frac{a_e}{a_n } ]

substituting values

     \theta = tan ^{-1} [\frac{0.808}{1.269 } ]

    \theta = tan ^{-1} [0.636 ]

   \theta =  32.5 6^o

     

   

       

5 0
3 years ago
What do halo stars do differently from disk stars?
VLD [36.1K]

Answer: They orbit the galactic center with many different inclinations, while disk stars all orbit in nearly the same plane. ... They have vertical motions out of the plane, making them appear to bob up and down, but they never get "too far" from the disk.

Explanation:

6 0
2 years ago
As the drawing illustrates, a siren can be made by blowing a jet of air through 20 equally spaced holes in a rotating disk. The
Aneli [31]

Answer:

ω = 630.2663 = 630[rad/s]

Explanation:

Solution:

- We can tackle this question by simple direct proportion relation between angular speed for the disk to rotate a cycle that constitutes 20 holes. We will use direct relation with number of holes per cycle to compute the revolution per seconds i.e frequency of speed f.

                                  1rev(20 hole) -> 20(cycle)/rev  

                                        2006.2(cycle) -> f ?  

                              f = 2006.2/20 = 100.31rev at second  

- The relation between angular frequency and angular speed is given by:

                                 ω = 2πf

                                 ω = 2*3.14*100.31

                                 ω = 630.2663 = 630[rad/s]

4 0
3 years ago
Please help
Snezhnost [94]

Answer:

Range = 22.61 m

Explanation:

We can use the formula for the Range in flat ground, given by:

Range=v_i^2\frac{sin(2\theta)}{g}

which for our case renders:

Range=15^2\frac{sin(80^o)}{9.8} \approx 22.61\,\,m

4 0
2 years ago
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