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maxonik [38]
3 years ago
5

Learning Goal: How do 2 ordinary waves build up a "standing" wave? A very generic formula for a traveling wave is: y1(x,t)=Asin(

kx−ωt). This general mathematical form can represent the displacement of a string, or the strength of an electric field, or the height of the surface of water, or a large number of other physical waves!
Part C Find ye(x) and yt(t). Remember that yt(t) must be a trig function of unit amplitude. Express your answers in terms of A, k, x, ω, and t. Separate the two functions with a comma. Use parentheses around the argument of any trig functions.

Part E At the position x=0, what is the displacement of the string (assuming that the standing wave ys(x,t) is present)? Part G From

Part F we know that the string is perfectly straight at time t=π2ω. Which of the following statements does the string's being straight imply about the energy stored in the string?

a.There is no energy stored in the string: The string will remain straight for all subsequent times.

b.Energy will flow into the string, causing the standing wave to form at a later time.

c.Although the string is straight at time t=π2ω, parts of the string have nonzero velocity. Therefore, there is energy stored in the string.

d.The total mechanical energy in the string oscillates but is constant if averaged over a complete cycle.
Physics
1 answer:
zheka24 [161]3 years ago
3 0

Answer:

Explanation:

=Asin(kx−ωt). This general mathematical form can represent the displacement of a string, or the strength of an electric field, or the height of the surface of water, or a large number of other physical waves!

Part C Find ye(x) and yt(t). Remember that yt(t) must be a trig function of unit amplitude. Express your answers in terms of A, k, x, ω, and t. Separate the two functions with a comma. Use parentheses around the argument of any trig functions.

Part E At the position x=0, what is the displacement of the string (assuming that the standing wave ys(x,t) is present)? Part G From

Part F we know that the string is perfectly straight at time t=π2ω. Which of the following statements does the string's being straight imply about the energy stored in the strJHJMNMMUJJHTGGHing?

a.There is no energy stored in the string: The string will remain straight for all subsequent times.

b.Energy will flow into the string, causing the standing wave to form at a later time.

c.Although the string is straight at time t=π2ω, parts of the string have nonzero velocity. Therefore, there is energy stored in the string.

d.The total mechanical energy in the string oscillates but is constant if averaged over a complete cycle.

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James is planning a science fair project on sound waves. He places an alarm inside a jar which he can remove the
ss7ja [257]

Answer:

D

Explanation:

this is simple, because when air is removed, means that there is no particles in the jar so vacuum is achieved, and when you can't hear a sound means that the sound couldn't travel through the vacuum. which means that sound cannot travel through vacuum,

as a result, sound requires a medium ( air) travel from one point to another.

hope it helps, if not please report it so that someone else gets to try it

7 0
3 years ago
A 2.0 kg wood block is launched up a wooden ramp that is inclined at a 30° angle. The block’s initial speed is 10 m/s. What vert
Romashka [77]

Answer:

Vertical Height = 0.784 meter, Speed back at starting point = 10 m/s

Explanation:

Given Data:

V is the overall velocity vector, Vi and Ui are its initial vertical and horizontal components

R = 10 m/s\\ Projection Angle (theta) = 30 degrees\\Vi   = 10*sin(30) = 5 m/s\\Ui  = 10*cos(30) = 8.66 m/s

To find:

Max Height h achieved

Calculation:

1) Using the 3^{rd} equation of motion, we know

2*a*s = Vf^{2}  - Vi^{2}

2) In terms of gravity g height h and  the vertical component of Velocity Vf , Vi.

3) As Vf = 0 as at maximum height the vertical component of velocity is zero maximum height achieved

2*g*h = Vf^{2}  -Vi^{2}

putting values

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3 0
4 years ago
Help with these three
Elenna [48]
The first: alright, first: you draw the person in the elevator, then draw a red arrow, pointing downwards, beginning from his center of mass. This arrow is representing the gravitational force, Fg.
You can always calculate this right away, if you know his mass, by multiplying his weight in kg by the gravitational constant
g = 9.81 \frac{m}{s {}^{2} }
let's do it for this case:
f_{g}  = m \times g \\ f _{g}  = 65kg \times 9.81 \frac{m}{s {}^{2} }  = 637.65
the unit of your fg will be in Newton [N]
so, first step solved, Fg is 637.65N
Fg is a field force by the way, and at the same time, the elevator is pushing up on him with 637.65N, so you draw another arrow pointing upwards, ending at the tip of the downwards arrow.
now let's calculate the force of the elevator
f = m \times a \\ f = 65 \times 5 \frac{m}{s {}^{2} }  \\ f = 325n
so you draw another arrow which is pointing downwards on him, because the elevator is accelating him upwards, making him heavier
the elevator force in this case is a contact force, because it only comes to existence while the two are touching, while Fg is the same everywhere
8 0
3 years ago
Which statements does the Second Law of Thermodynamics support?
Mazyrski [523]
According to the second law of thermodynamics, 
the answer is 
<span>4. The entropy of the universe is increasing. </span>
4 0
3 years ago
An airplane touches down on the runway with a speed of 70 m/s2. Determine the airplane after each second of its deceleration.
ivann1987 [24]
<span>vf^2 = vi^2 + 2*a*d
---
vf = velocity final
vi = velocity initial
a = acceleration
d = distance
---
since the airplane is decelerating to zero, vf = 0
---
0 = 55*55 + 2*(-2.5)*d
d = (-55*55)/(2*(-2.5))
d = 605 meters


</span>
5 0
3 years ago
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