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4vir4ik [10]
3 years ago
10

All of the following are true for slow-motion sequences EXCEPT a. slow motion ritualizes movement. b. slow motion brings gracefu

lness to the commonplace. c. slow motion is achieved by filming at a slower rate than is normal. d. slow motion solemnizes movement.
Physics
2 answers:
LiRa [457]3 years ago
8 0

Answer:

c. slow motion is achieved by filming at a slower rate than is normal

Explanation:

Slow motion is achieved by filming at a faster frame rate and playing the video in a slower frame rate. When filming is done at a slower frame rate and played at a faster frame rate this is used to make a time lapse video.

Suppose you film something at 240000 frames per second and play it at 24 frames per second then the video will be (240000/24) 10000 times slower.

german3 years ago
3 0

Answer:

c. slow motion is achieved by filming at a slower rate than is normal.

Explanation:

Slow motion is the name given to the special film and video effect in which movements and actions in frame are seen over a longer duration than normal, giving the feeling that time itself is slowing down. Although the effect is noticeable only when projected, it can be prepared when shooting or processing images. To cause the slow motion effect, you need to capture frames at a faster rate than normal, and then play back frames at a slower than normal rate.

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As more and more resistors are added in parallel to a circuit, the equivalent resistance of the circuit ____________ (increases,
Anestetic [448]

Answer:

Decreases, Increases

Explanation:

Resistance is parallel can be calculated using

1/Req = 1/R1 + 1/R2 + 1/R3 +....

Then, as more resistor is added in parallel the equivalent resistance is reduced.  

Let use a simple sample

Let all the resistor have equal resistances

Let say R = R1 = R2 = R3 =...Rn

Then, 1/Req = 1/R1 + 1/R2 + 1/R3 +....

1/Req = 1/R + 1/R + 1/R +.... 1/Rn

Req = R/n

Check attachment on how I got that.

This implies that, the equivalent resistance will always be less than the original resistance, since n>1

So, as n increases (I.e. as the number of resistance increases), the equivalent resistance reduces.

B. Now, to know if the current reduces or increases

Using Ohms law

V = iR

Then, I = V/R

So, let assume the voltage is constant, then, the current is inversely proportional to the resistance, so as we know that the resistance is reducing, then the current will be increasing.

So current increase as we add more resistor in parallel to a circuit

3 0
3 years ago
How can I solve this?
inysia [295]
There is no equation here
3 0
3 years ago
Swinging a golf club or baseball bat are examples of ______________ stretching.
kari74 [83]

Answer:

Static stretching is the answer.

Explanation:

Static stretching is the most common form that greatly improves flexibility. However, static stretches does little to contract the muscles needed to generate powerful golf swings. Dynamic stretches help improve your range of motion while reducing muscle stiffness.

6 0
3 years ago
The main reason that most professional research telescopes are reflectors is that
GuDViN [60]
<h3><u>Answer;</u></h3>

Large mirrors are easier to build than large lenses.

<h3><u>Explanation;</u></h3>
  • <em><u>Reflector telescopes have a number of advantages as compared to refracting telescopes and other types of telescopes. </u></em>
  • <em><u>Reflector telescopes do not suffer from chromatic aberration because all wavelengths will reflect off the mirror in the same way. The support for the objective mirror is all along the back side so they can be made very large.</u></em>
  • Additionally, reflector telescopes are cheaper to make than refractors of the same size. Also since in reflector telescopes light is reflecting off the objective, rather than passing through it, only one side of the reflector telescope's objective needs to be perfect.

7 0
3 years ago
The normal eye, myopic eye and old age
yanalaym [24]

Answer:

1)    f’₀ / f = 1.10, the relationship between the focal length (f'₀) and the distance to the retina (image) is given by the constructor's equation

2) the two diameters have the same order of magnitude and are very close to each other

Explanation:

You have some problems in the writing of your exercise, we will try to answer.

1) The equation to be used in geometric optics is the constructor equation

          \frac{1}{f} = \frac{1}{p} + \frac{1}{q}

where p and q are the distance to the object and the image, respectively, f is the focal length

* For the normal eye and with presbyopia

the object is at infinity (p = inf) and the image is on the retina (q = 15 mm = 1.5 cm)

        \frac{1}{f'_o} = 1/ inf + \frac{1}{1.5}

        f'₀ = 1.5 cm

this is the focal length for this type of eye

* Eye with myopia

the distance to the object is p = 15 cm the distance to the image that is on the retina is q = 1.5 cm

           1 / f = 1/15 + 1 / 1.5

           1 / f = 0.733

            f = 1.36 cm

this is the focal length for the myopic eye.

In general, the two focal lengths are related

         f’₀ / f = 1.5 / 1.36

         f’₀ / f = 1.10

The question of the relationship between the focal length (f'₀) and the distance to the retina (image) is given by the constructor's equation

2) For this second part we have a diffraction problem, the point diameter corresponds to the first zero of the diffraction pattern that is given by the expression for a linear slit

          a sin θ= m λ

the first zero occurs for m = 1, as the angles are very small

          tan θ = y / f = sin θ / cos θ

for some very small the cosine is 1

          sin θ = y / f

where f is the distance of the lens (eye)

           y / f = lam / a

in the case of the eye we have a circular slit, therefore the system must be solved in polar coordinates, giving a numerical factor

           y / f = 1.22 λ / D

           y = 1.22 λ f / D

where D is the diameter of the eye

          D = 2R₀

          D = 2 0.1

          D = 0.2 cm

           

the eye has its highest sensitivity for lam = 550 10⁻⁹ m (green light), let's use this wavelength for the calculation

         

* normal eye

the focal length of the normal eye can be accommodated to give a focus on the immobile retian, so let's use the constructor equation

      \frac{1}{f} = \frac{1}{p} + \frac{1}{q}

sustitute

       \frac{1}{f} = \frac{1}{25} + \frac{1}{1.5}

       \frac{1}{f}= 0.7066

        f = 1.415 cm

therefore the diffraction is

        y = 1.22  550 10⁻⁹  1.415  / 0.2

        y = 4.75 10⁻⁶ m

this is the radius, the diffraction diameter is

       d = 2y

       d_normal = 9.49 10⁻⁶ m

* myopic eye

In the statement they indicate that the distance to the object is p = 15 cm, the retina is at the same distance, it does not move, q = 1.5 cm

       \frac{1}{f} = \frac{1}{15} + \frac{1}{ 1.5}

        \frac{1}{f}= 0.733

         f = 1.36 cm

diffraction is

        y = 1.22 550 10-9 1.36 10-2 / 0.2 10--2

        y = 4.56 10-6 m

the diffraction diameter is

        d_myope = 2y

         d_myope = 9.16 10-6 m

         \frac{d_{normal}}{d_{myope}} = 9.49 /9.16

        \frac{d_{normal}}{d_{myope}} =  1.04

we can see that the two diameters have the same order of magnitude and are very close to each other

8 0
3 years ago
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