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docker41 [41]
4 years ago
6

Answer the amplitude part of the questions in 7, 8, & 9

Physics
1 answer:
erastova [34]4 years ago
5 0
I actually believe for the first question, it would be complete destructive interference as the amplitude and the approximate wavelength for each are the same and will completely or entirely cancel out, rather than simply decreasing or lowering the amplitude as in the bottom question.

The amplitude for the first will be 0, as the 2 waves will cancel each other out. The amplitude of the second, will be 3x, I believe, assuming the amplitude of the first is 2x and the second is 1x, in a constructive interference, I believe the amplitudes would add up.

Likewise for the bottom, I believe you would be subtracting the supposed amplitude of the first which is 2x from 1x which would be 1x.
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zaharov [31]

Take the missile's starting position to be the origin. Assuming the angles given are taken to be counterclockwise from the positive horizontal axis, the missile has position vector with components

x=v_0\cos20.0^\circ t+\dfrac12a_xt^2

y=v_0\sin20.0^\circ t+\dfrac12a_yt^2

The missile's final position after 9.20 s has to be a vector whose distance from the origin is 19,500 m and situated 32.0 deg relative the positive horizontal axis. This means the final position should have components

x_{9.20\,\mathrm s}=(19,500\,\mathrm m)\cos32.0^\circ

y_{9.20\,\mathrm s}=(19,500\,\mathrm m)\sin32.0^\circ

So we have enough information to solve for the components of the acceleration vector, a_x and a_y:

x_{9.20\,\mathrm s}=\left(1810\,\dfrac{\mathrm m}{\mathrm s}\right)\cos20.0^\circ(9.20\,\mathrm s)+\dfrac12a_x(9.20\,\mathrm s)^2\implies a_x=21.0\,\dfrac{\mathrm m}{\mathrm s^2}

y_{9.20\,\mathrm s}=\left(1810\,\dfrac{\mathrm m}{\mathrm s}\right)\sin20.0^\circ(9.20\,\mathrm s)+\dfrac12a_y(9.20\,\mathrm s)^2\implies a_y=110\,\dfrac{\mathrm m}{\mathrm s^2}

The acceleration vector then has direction \theta where

\tan\theta=\dfrac{a_y}{a_x}\implies\theta=79.2^\circ

5 0
3 years ago
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Answer:

9013 m/s

Explanation:

hope it helped!!!

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3 years ago
A student, starting from rest, slides down a water slide. On the way down, a kinetic frictional force (a nonconservative force)
salantis [7]

Answer:

<em>The velocity with which the student goes down the bottom of glide is 12.48m/s.</em>

Explanation:

The Non conservative force is defined as a force which do not store energy or get he energy dissipate the energy from the system as the system progress with the motion.

Given are

   <em>  mass of the student 73 kg</em>

<em>      height of water glide 11.8 m</em>

<em>      work done as -5.5*10³ J</em>

Have to find speed at which the student goes down the glide.

According to<em> Law of Conservation of energy</em>,

          K.E =P.E+Work Done

 mv²/2=mgh +W

Rearranging the above eqn for v

v = √2(gh+W/m)

Substituting values,

V =  12.48 m/s.

<em>The velocity with which the student goes down the bottom of glide is 12.48m/s.</em>

 

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3 years ago
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John needs to see a physical therapist because he cannot walk very well.
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