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3 years ago

7

How does refraction cause wave crests to move when the waves approach the shore? A. at right angles to the shore B. about 80 deg

rees in relation to the shore C. nearly parallel to the shoreline D. about 45 degrees in relation to the shore

2 answers:

Svet_ta [14]3 years ago

7 0

Nearly Parallel to the shore, is the answer I'm pretty sure.

goldenfox [79]3 years ago

4 0

The answer is C. I just took the test.:)

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When you are paddling a canoe, you push the water backwards with your paddle, which in turn will push you forward. Which law of

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Newton's third law of motion. Because you are pushing the water backwards which will push you forwards

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4 years ago

A 13 g rubber stopper is swung in a horizontal circle with a radius of 0.85 m with a period of 0.65 s. (a) How fast is the stopp

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Explanation:

(a) v = 2π r / t

v = 2π (0.85 m) / (0.65 s)

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4 years ago

Olaf is standing on a sheet of ice that covers the football stadium parking lot in Buffalo, New York; there is negligible fricti

Bas_tet [7]

Answer:

v = 0.059 m/s

Explanation:

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m: mass of the ball = 0.400kg

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v2i: initial velocity of Olaf = 0m/s

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You solve the equation (1) for v and replace the values of all variables:

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Hence, after Olaf catches the ball, the velocity of Olaf and the ball is 0.059m/s

3 0

3 years ago

A hailstone traveling with a velocity of 43 meters/second comes to a virtual stop 0.28 seconds after hitting water. What is the

lisabon 2012 [21]

Magnitude of acceleration = (change in speed) / (time for the change) .

Change in speed = (ending speed) - (starting speed)

= zero    - (43 m/s)

=    -43 m/s .

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5 0

3 years ago

Read 2 more answers

A street musician sounds the A string of his violin, producing a tone of 440 Hz, What frequency does a bicyclist hear as he a) a

DanielleElmas [232]

Answer:

a) $f_o=454.11Hz$

b) $f_o=425.89Hz$

Explanation:

Let´s use Doppler effect, in order to calculate the observed frequency by the byciclist. The Doppler effect equation for a general case is given by:

$f_o=\frac{v\pm v_o}{v\pm v_s} *f_s$

where:

$f_o=Observed\hspace{3}frequency$

$f_s=Actual\hspace{3}frequency$

$v=Speed\hspace{3}of\hspace{3}the\hspace{3}sound\hspace{3}waves$

$v_s=Velocity\hspace{3}of\hspace{3}the\hspace{3}source$

$v_o=Velocity\hspace{3}of\hspace{3}the\hspace{3}observer$

Now let's consider the next cases:

$+v_o\hspace{3}is\hspace{3}used\hspace{3}when\hspace{3}the\hspace{3}observer\hspace{3}moves\hspace{3}towards\hspace{3}the\hspace{3}source$

$-v_o\hspace{3}is\hspace{3}used\hspace{3}when\hspace{3}the\hspace{3}observer\hspace{3}moves\hspace{3}away\hspace{3}from\hspace{3}the\hspace{3}source$

$-v_s\hspace{3}is\hspace{3}used\hspace{3}when\hspace{3}the\hspace{3}source\hspace{3}moves\hspace{3}towards\hspace{3}the\hspace{3}observer$

$+v_s\hspace{3}is\hspace{3}used\hspace{3}when\hspace{3}the\hspace{3}source\hspace{3}moves\hspace{3}away\hspace{3}from\hspace{3}the\hspace{3}observer$

The data provided by the problem is:

$f_s=440Hz\\v_o=11m/s$

The problem don't give us aditional information about the medium, so let's assume the medium is the air, so the speed of sound in air is:

$v=343m/s$

Now, in the first case the observer alone is in motion towards to the source, hence:

$f_o=\frac{v+v_o}{v}*f_s=\frac{343+11}{343} *440=454.1107872Hz$

Finally, in the second case the observer alone is in motion away from the source, so:

$f_o=\frac{v-v_o}{v}*f_s=\frac{343-11}{343} *425.8892128Hz$

6 0

4 years ago