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Bumek [7]
3 years ago
11

A lifeguard on a beach observes that waves have a speed of 2.60 m/s and a distance of 2.50 m between wave crests. What is the pe

riod of the wave motion? Please show all work. (ref: p.382-387)
Physics
1 answer:
Alja [10]3 years ago
4 0

Answer:

T = 0.96 seconds

Explanation:

Given that,

The speed of wave, v = 2.6 m/s

The distance between wave crests, \lambda=2.5\ m

We need to find the period of the wave motion. Let T be the period. So,

v=\dfrac{\lambda}{T}\\\\T=\dfrac{\lambda}{v}\\\\T=\dfrac{2.5}{2.6}\\\\T=0.96\ s

So, the period of the wave motion is equal to 0.96 seconds.

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Answer:

A. It changes all the time, allowing us to see if the experimental group stays the same.

control group mean the thing the stay the same during the Experiment.

I hope this help

I hope this is correct

4 0
4 years ago
The two vectors and in fig. 3-28 have equal magnitudes of 10.0 m and the angles are 30° and 105°. find the (a) x and (b) y compo
mylen [45]

You can just use basic trigonometry to solve for the x & y components.

<span>vector a = 10cos(30) i + 10sin(30) j = <5sqrt(3), 5></span>

vector b is only slightly harder because the angle is relative to vector a, and not the positive x-axis. Anyway, this just makes vector b with an angle of 135deg to the positive x-axis.

<span>vector b = 10cos(135) i + 10sin(135) j = <-5sqrt(2), 5sqrt(2)></span>

So now we can do the questions:

r = a + b

r = <5sqrt(3)-5sqrt(2), 5+5sqrt(2)>

(a) 5sqrt(3)-5sqrt(2)

(b) 5+5sqrt(2)

(c)

|r| = sqrt( (5sqrt(3)-5sqrt(2))2 + (5+5sqrt(2))2 )

= 12.175

(d)

θ = tan-1 ( (5+5sqrt(2)) / (5sqrt(3)-5sqrt(2)) )

θ = 82.5deg

<span> </span>

6 0
3 years ago
Approximately how many kelvins are equal to 60°f? <br> a. 333 <br> b. 323 <br> c. 413 <br> d. 289
Afina-wow [57]
D. 289
Take the formula:
K=5/9(Fahrenheit-32)+273
Plug in Fahrenheit
K=5/9 (60-32)+273
From here it is simple math and you can plug it into your calculator getting 288.5555556 and round to 289
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Tamiku [17]

Answer:

A

Explanation:

7 0
3 years ago
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