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Zielflug [23.3K]
3 years ago
12

The vibrations along a transverse wave move in a direction _________.

Physics
1 answer:
GrogVix [38]3 years ago
5 0

Answer: perpendicular to it oscillations.

Explanation: A transverse wave is a wave whose oscillations is perpendicular to the direction of the wave.

By perpendicular, we mean that the wave is oscillating on the vertical axis (y) of a Cartesian plane and the vibration is along the horizontal axis (x) of the plane.

Examples of transverse waves includes wave in a string, water wave and light.

Let us take a wave in a string for example, you tie one end of a string to a fixed point and the other end is free with you holding it.

If you move the rope vertically ( that's up and down) you will notice a kind of wave traveling away from you ( horizontally) to the fixed point.

Since the oscillations is perpendicular to the direction of wave, it is a transverse wave

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How to do this question
Yuki888 [10]

Answer:

64°

Explanation:

The triangle is an isosceles triangles (both legs are equal to the radius of the circle), so that means the base angles are the same.

Angles of a triangle add up to 180°, so:

128 + 2x = 180

2x = 52

x = 26

∠1 is complementary to the base angle, so:

∠1 = 90 − 26

∠1 = 64

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3 years ago
Gary is trying to think of an object that could model a sunspot. His teacher tells him that such an object might be found right
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the answer is B a freckle I just had this as a Study Island and got it right.

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3 years ago
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2. Two projectiles thrown from the same point at angles 0,=60° and 02=30° with the horizontal,
Liono4ka [1.6K]

Answer:

The answer is below

Explanation:

The maximum height (h) of a projectile with an initial velocity of u, acceleration due to gravity g and at an angle θ with the horizontal is given as:

h=\frac{u^2sin^2\theta}{2g}

Given that the two projectile has the same height.

For\ the\ first\ projectile\ with\ an\ angle\ \60^0 \ with\ the\ horizontal\ and initial\ velocity\ u_1:\\\\h=\frac{u_1^2sin^260}{2g} =\frac{0.75u_1^2}{2g} \\\\For\ the\ second\ projectile\ with\ an\ angle\ \30 \ with\ the\ horizontal\ and initial\ velocity\ u_2:\\\\h=\frac{u_2^2sin^230}{2g} =\frac{0.25u_2^2}{2g}\\\\\frac{0.25u_2^2}{2g}=\frac{0.75u_1^2}{2g}\\\\0.25u_2^2=0.75u_1^2\\\\\frac{u_1^2}{u_2^2} =\frac{0.25}{0.75} \\\\\frac{u_1^2}{u_2^2}=\frac{1}{3} \\\\\frac{u_1}{u_2}=\sqrt\frac{1}{3}

8 0
3 years ago
Why are hurricanes considered more damaging than tornadoes when tornadoes have stronger winds
chubhunter [2.5K]

-- A tornado follows a path that's a few miles wide, for a few hours.
   Then it's all over.

-- A hurricane follows a path that's several hundred miles wide,
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8 0
3 years ago
Read 2 more answers
A 1.25-kg ball begins rolling from rest with constant angular acceleration down a hill. If it takes 3.60 s for it to make the fi
miv72 [106K]

Answer:

The time taken is  \Delta t  = 1.5 \ s

Explanation:

From the question we are told that

   The mass of the ball is  m =  1.25 \ kg

    The time taken to make the first complete revolution is  t= 3.60 s

    The displacement of the first complete revolution is  \theta  =  1 rev  =  2 \pi \  radian

Generally the displacement for one  complete revolution is mathematically represented as

       \theta =  w_i t  +  \frac{1}{2} *  \alpha  * t^2

Now given that the stone started from rest w_i  = 0 \ rad / s

     2 \pi =0   +  0.5*  \alpha  *(3.60)^2

     \alpha   =  0.9698 \  s

Now the displacement for two  complete revolution is

         \theta_2  =  2 *  2\pi

         \theta_2  = 4\pi

Generally the displacement for two complete revolution is mathematically represented as  

     4 \pi =   0  +  0.5 * 0.9698 * t^2

=>   t^2  =  25.9187

=>   t=  5.1 \ s

So

 The  time taken to complete the next oscillation is mathematically evaluated as

     \Delta t  =  t_2  - t

substituting values

      \Delta t  = 5.1 -  3.60

     \Delta t  = 1.5 \ s

           

 

7 0
3 years ago
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