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rusak2 [61]
2 years ago
5

Is the alien theory a scientific claim? Why or why not?

Physics
2 answers:
spin [16.1K]2 years ago
8 0
I'm not sure what the alien Theory says or if there really is such a theory. If the theory says that aliens definitely exist and that they have visited Earth in the past then the theory is totally and completely without any kind of support. It's not scientific in any way because there is no evidence for such a claim. It may be thought to be probable but no solid evidence has ever been presented.
mina [271]2 years ago
7 0

Answer:

Sample Response: The alien theory is not a scientific claim because it is not based on valid data, there is no evidence to support it, and it does not come from a reliable source.

Explanation:

Dont worry i got yall. Mark me.

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A mass moves back and forth in simple harmonic motion with amplitude A and period T.
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a. 0.5 T

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- The period T is the time the system takes to complete one oscillation

During a full time period T, the mass on the spring oscillates back and forth, returning to its original position. This means that the total distance covered by the mass during a period T is 4 times the amplitude (4A), because the amplitude is just half the distance between the maximum and the minimum position, and during a time period the mass goes from the maximum to the minimum, and then back to the maximum.

So, the time t that the mass takes to move through a distance of 2 A can be found by using the proportion

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b. 1.25T

Now we want to know the time t that the mass takes to move through a total distance of 5 A. SInce we know that

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3 years ago
A cannon with a muzzle speed of 1 000 m/s is used to start an avalanche on a mountain slope. The target is 2 000 m from the cann
Nataliya [291]

Answer:

∅ = 89.44°

Explanation:

In situations like this air resistance are usually been neglected thereby making g= 9.81 m/s^{2}

Bring out the given parameters from the question:

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Target height (h) =  800 m

Projection angle ∅ = ?

Horizontal distance = V_{1x}tcos ∅     .......................... Equation 1

where V_{1x} = velocity in the X - direction

           t = Time taken

Vertical Distance = y = V_{1y} t - \frac{1}{2}gt^{2}        ................... Equation 2

Where   V_{1y} = Velocity in the Y- direction

              t  = Time taken

V_{1y} = V_{1}sin∅

Making time (t) subject of the formula in Equation 1

                    t = d/(V_{1x}cos ∅)

                      t = \frac{2000}{1000coso} = \frac{2}{cos0}  =    \frac{d}{cos o}             ...................Equation 3

substituting equation 3 into equation 2

Vertical Distance = d = V_{1y} \frac{d}{cos o} - \frac{1}{2}g\frac{2}{cos0}   ^{2}

                                  Vertical Distance = h = sin∅ \frac{d}{cos o} - \frac{1}{2}g\frac{2}{cos0}   ^{2}

  Vertical Distance = h = dtan∅   - \frac{1}{2}g\frac{2}{cos0}   ^{2}

  Applying geometry

                              \frac{1}{cos o} = tan^{2} o + 1

  Vertical Distance = h = d tan∅   - 2 g (tan^{2} o + 1)

               substituting the given parameters

               800 = 2000 tan ∅ - 2 (9.81)( tan^{2} o + 1)

              800 = 2000 tan ∅ - 19.6( tan^{2} o + 1)  Equation 4

Replacing tan ∅ = Q     .....................Equation 5

In order to get a quadratic equation that can be easily solve.

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Rearranging 19.6Q^{2} - 2000 Q + 780.4 = 0

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Taking the Tan inverse of each value of Q

                  ∅ = 89.44°     ∅ = 22.37°

             

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