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photoshop1234 [79]
3 years ago
9

Imagine the door to a theatre is opened and you are standing outside. How is it possible that you can hear the sound of the movi

e or play even if you cannot see the screen or stage ?
A) reflection only happens with sound waves not waves of light so the sound waves bounce out to your ears.

B) the sound waves refract as they move from one medium to another

C) the darkness in the theatre absorbed light but not sound so you can hear

D) the sound waves diffract much more than the light waves so try bend and spread as they move through the doorway.
Physics
1 answer:
meriva3 years ago
8 0
Imagine the door to a theatre is opened and you are standing outside. You can hear the sound of the movie or play even if you cannot see the screen or stage because <span>the sound waves diffract much more than the light waves so try bend and spread as they move through the doorway. The answer is letter D</span>
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Explanation:

Below is an attachment containing the solution.

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3 years ago
When an object oscillating in simple harmonic motion is at its maximum displacement from the equilibrium position, which of the
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Answer:

E. Zero Maximum

Explanation:

At the point of maximum displacement, the speed is zero while the restoring force is maximum. In fact:

- The restoring force is given by F=kx, where k is the spring constant and x is the displacement - at the point of maximum displacement, x is maximum, so F is maximum as well

- the total energy of the system is sum of kinetic energy and elastic potential energy:

E=K+U=\frac{1}{2}mv^2+\frac{1}{2}kx^2

where m is the mass of the system and v is the speed. Since E (the total energy) is constant due to the law of conservation of energy, we have that when K increases, U decreases, and viceversa. As a result, when x increases, v decreases, and viceversa. At the point of maximum displacement, x is maximum, so v will have its minimum value (which is zero, since the system is changing direction of motion).

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3 years ago
On a drive from one city to? another, victor averaged 39 mph. if he had been able to average 72 ?mph, he would have reached his
PilotLPTM [1.2K]
Let the unknown distance be xmiles
x/39-x/72=11hr
72x-39x/2808=11hr
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U can substitue back to check
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at speed of 39mph, he would need 936/39=24hr
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5 0
2 years ago
The velocity of a ball changes from ‹ 9, −6, 0 › m/s to ‹ 8.96, −6.12, 0 › m/s in 0.02 s, due to the gravitational attraction of
Alenkasestr [34]

Answer:

a) a=(-2,-6,0)m/s^2, with a magnitude of 6.3m/s^2

b) \frac{\Delta p}{\Delta t}=(-0.24,-0.72,0)Kgm/s^2, with a magnitude of 0.76Kgm/s^2

c) F=(-0.24,-0.72,0)N, with a magnitude of 0.76N

Explanation:

We have:

v_{ix}=9m/s, v_{iy}=-6m/s, v_{iz}=0m/s\\v_{fx}=8.96m/s, v_{fy}=-6.12m/s, v_{fz}=0m/s\\t=0.02s, m=0.12Kg

We can calculate each component of the acceleration using its definition a=\frac{\Delta v}{\Delta t}

a_x=\frac{v_{fx}-v_{ix}}{t} = \frac{(8.96m/s)-(9m/s)}{0.02s} =-2m/s^2\\a_y=\frac{v_{fy}-v_{iy}}{t} = \frac{(-6.12m/s)-(-6m/s)}{0.02s} =-6m/s^2\\a_y=\frac{v_{fz}-v_{iz}}{t} = \frac{(0m/s)-(0m/s)}{0.02s} =0m/s^2\\

The rate of change of momentum of the ball is \frac{\Delta p}{\Delta t} = \frac{\Delta mv}{\Delta t} = \frac{m\Delta v}{\Delta t} = ma

So for each coordinate:

\frac{\Delta p_x}{\Delta t}=-0.24Kgm/s^2\\\frac{\Delta p_y}{\Delta t}=-0.72Kgm/s^2\\\frac{\Delta p_z}{\Delta t}=0Kgm/s^2

And these are equal to the components of the net force since F=ma.

If magnitudes is what is asked:

a=\sqrt{a_x+a_y+a_z} =6.3m/s^2\\F=ma=\frac{\Delta p}{\Delta t}=0.76N

<em>(N and </em>Kgm/s^2<em> are the same unit).</em>

3 0
3 years ago
If s is the specific heat capacity of 4kg water, what is the specific heat capacity of 16kg
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Answer: d

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