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olga nikolaevna [1]
3 years ago
14

Two Velocities in a Traveling Wave? Wave motion is characterized by two velocities: the velocity with which the wave moves in th

e medium (e.g., air or a string) and the velocity of the medium (the air or the string itself).
Consider a transverse wave traveling in a string. The mathematical form of the wave is
y(x,t) = A sin(kx-omega...
Physics
1 answer:
Lyrx [107]3 years ago
3 0

For the answer to the question above,
<span>There is nothing in the equations to suggest that the string moves in the x direction so D) v_x(x,t)=0. 
</span>
 y(x,t) = A sin(kx-omega t) 
d{y(x,t)}/d{x} = A k cos(kx - omega t)
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Answer:

encyclopedia most reliable I think

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The following are the Earth–Sun distance at the equinoxes and solstices: March equinox 149.0 million km June solstice 152.0 mill
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Answer:

During <u>winter (late December/early January)</u> the Earth is closest to the Sun and during <u>summer (late June/early July)</u> the Earth is farthest from the Sun.

Explanation:

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On the other hand, the earth is farthest from the sun during the time of summer season, mostly in late June or early July.

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5 0
3 years ago
A ball is thrown horizontally from the top of a 60.0-m building and lands 100.0 m from the base of thebuilding. Ignore air resis
Bumek [7]

Answer:

a)3.5s

b)28.57m/S

c)34.33m/S

d)44.66m/S

Explanation:

Hello!

we will solve this exercise numeral by numeral

a) to find the time the ball takes in the air we must consider that vertically the ball experiences a movement with constant acceleration whose value is gravity (9.81m / S ^ 2), that the initial vertical velocity is zero, we use the following equation for a body that moves with constant acceleration

Y= VoT+0.5gt^{2}

where

Vo = Initial speed =0

T = time

g=gravity=9.81m/s^2

y = height=60m

solving for time

Y=0.5gt^2\\t=\sqrt{\frac{Y}{0.5g} } \\t=\frac{60}{0.5(9.81)} \\

T=3.5s

b)The horizontal speed remains constant since there is no horizontal acceleration. with the value of the distance traveled (100m) and the time that lasts in the air (3.5s) we estimate the horizontal speed

V=\frac{x}{t} =\frac{100}{3.5}=28.57m/s

c)

to find the final vertical velocity we use the equations for motion with constant velocity as follows

Vf=Vo+g.t    

Vf=0+(9.81 )(3.5)=34.335m/S          

d)Finally, to find the resulting velocity, we add the horizontal and vertical velocities vectorially, this is achieved by finding the square root of the sum of its squares

V=\sqrt{Vx^2+Vy^2} =\sqrt{34.33^2+28.57^2} =44.67m/S

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I'm stuck on the same question, as well :(
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