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elena-s [515]
3 years ago
9

Why were phonograph records placed on the sides of the two voyager spacecraft launched in the 1970s?

Physics
1 answer:
LenaWriter [7]3 years ago
7 0

Answer:

The reason for phonograph installation is to provide information about humans to interstellar aliens.

Explanation:

Phonographs are archaic forms of gramophone which are able to record and reproduce sound.Phonographs were used in early days in voyagers to help transmit information about humans to nearby alien species in space.

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During a plane showcase, a pilot makes circular "looping" with a speed equal to the sound speed (340 m/s). However, the pilot ca
Dmitriy789 [7]

Answer:

1472.98 m

Explanation:

Data provided:

Speed of circular looping, v = 340 m/s

Acceleration, a = 8g

here,

g is the acceleration due to the gravity = 9.81 m/s²

Now,

the centripetal acceleration is given as,

a=\frac{v^2}{r}

r is the radius of the loop

on substituting the respective values, we get

8\times9.81=\frac{340^2}{r}

or

r = 1472.98 m

5 0
3 years ago
A solid cylinder is released from the top of an inclined plane of height 0.81 m. From what height, in meters, on the incline sho
Jlenok [28]

Answer:

same 0.81m

Explanation:

in this problem if we assume there no resistance of any sort. and we apply the energy conservation

change in Potential energy = change in kinetic energy

mgh = 0.5mv^2

gh = 0.5v^2

the above relation suggests that the speed at the bottom is only depending on the height it is released from not on the shape, mass or radius.

so at the bottom

put h = 0.81m

9.81 * 0.81 * 2 = v^2

v=3.99 m/s

both CYLINDER and SPHERE will have same velocity at the bottom if released from the same height irrespective of shape and size

3 0
3 years ago
What are the disadvantages and advantages for electromagnetic spectrum​
Basile [38]

Answer:

Advantages: In small quantities they help your body produce vitamin D.

Disadvantages: They can cause sunburn or even skin cancer if used in large quantities.

Explanation:

3 0
3 years ago
A child has an ear canal that is 1.3 cm long. Assume the speed of sound is v = 344 m/s.
kap26 [50]

Answer:

The  frequencies are (f, f_1) =  (6615.4 \ Hz , 19846.2\ Hz)

Explanation:

From the question we are told that

  The  length of the ear canal is  l = 1.3 \ cm  =\frac{1.3}{100}  =  0.013 \ m

   The  speed of sound is assumed to be  v_s  =  344 \ m/s

Now  taking look at a typical  ear canal  we see that we assume it is  a  closed pipe

   Now the fundamental harmonics for the pipe(ear canal) is mathematically represented as

            f = \frac{v_s}{4 * l }

 substituting values  

          f = \frac{344}{4 * 0.013 }

         f = 6615.4 \ Hz

Also the the second harmonic for the pipe (ear canal) is mathematically represented as

        f_1 =  \frac{3v_s}{4 * l}

 substituting values  

       f_1 =  \frac{3 *  344}{4 * 0.013}

       f_1 =   19846.2 \ Hz

Given that sound would be loudest in the pipe at the frequency, it implies that the child  will have an increased audible sensitivity at this  frequencies

6 0
3 years ago
A certain kind of glass has an index of refraction of 1.640 for blue light of wavelength 440 nm and an index of 1.595 for red li
seraphim [82]

Answer: 18.27°

Explanation:

Given

Index of refraction of blue light, n(b) = 1.64

Wavelength of blue light, λ(b) = 440 nm

Index of refraction of red light, n(r) = 1.595

Wavelength of red light, λ(r) = 670 nm

Angle of incident, θ = 30°

Angle of refraction of red light is

θ(r) = sin^-1 [(n(a)* sin θ) / n(r)], where n(a) = index of refraction of air = 1

So that,

θ(r) = sin^-1 [(1 * sin 30) / 1.595]

θ(r) = sin^-1 (0.5 / 1.595)

θ(r) = sin^-1 0.3135

θ(r) = 18.27°

4 0
3 years ago
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