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Mandarinka [93]
3 years ago
14

A woman is standing in the ocean, and she notices that after a wavecrest passes, five more crests pass in a time of 50.0 s. Thed

istance between two successive crests is 32 m. Determine, ifpossible, the wave’s (a) period, (b) frequency, (c) wavelength, (d) speed, and (e) amplitude. If it is not possible todetermine any of these quantities, then so state.I gotT=50/5 = 10sf=1/T = 0.1Hzv=f/lamda = 0.1/32 = 4.93m/s
Physics
1 answer:
Vadim26 [7]3 years ago
6 0

Answer:

(a) T = 10 s

(b) f = 0.1 Hz

(c) λ = 32 m

(d) v = 3.2 m/s

(e) Insufficient data

Explanation:

(a)

Time period is defined as the time interval required for one wave to pass. Therefore, the time period can be given as:

T = Period = Time Taken/No. of Waves

T = 50 s/5

<u>T = 10 s</u>

<u></u>

(b)

Frequency is the reciprocal of time period:

f = frequency = 1/T

f = 1/10 s

<u>f = 0.1 Hz</u>

<u></u>

(c)

Wavelength is the distance between two consecutive crests or troughs:

<u>λ = Wavelength = 32 m</u>

<u></u>

(d)

Speed of wave is given by the following formula:

Speed = v = fλ

v = (0.1 Hz)(32 m)

v = 3.2 m/s

(e)

Amplitude cannot be found with given data.

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If a runner has a speed of 8.66m/s and runs for 46.2s what distance is covered? tv = d
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Riders in an amusement park ride shaped like a Viking ship hung from a large pivot are rotated back and forth like a rigid pendu
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Answer:

a)  v = 16.57 m / s, b)  a = 19.6 m / s², d)    N = 1.76 10³ N,     N / W = 3

Explanation:

This exercise looks interesting, but I think you have some problem with the writing, the questions seem a bit disconnected from the initial text.

Let's answer the questions.

a) For this part we can use energy considerations.

Starting point. The upper part of the trajectory indicates that the arm is horizontally

          Em₀ = U = m g h

in this case h = r

Final point. For lower of the trajectory

          Em_f = K = ½ m v²

as they indicate that there is no friction

         Em₀ = em_f

         mgh = ½ m v²

         v = \sqrt{2gh}

let's calculate

        v = \sqrt{2 \ 9.8 \ 14.0}

         v = 16.57 m / s

b) the centripetal acceleration has the formula

           a = v² / r

           a = 16.57² / 14.0

           a = 19.6 m / s²

c) see attached where the diagram is

where N is the normal and w the weight

d) let's use Newton's second law

               N-W = m a

               N - mg = m ar

               N = m (g + a)

let's calculate

               N = 60.0 (9.8 + 19.6)

               N = 1.76 10³ N

the relationship with weight is

              N / W = 1.76 10³/( 60 9.8)

              N / W = 3

normal is three times greater than body weight

e) the answer is reasonable since by Newton's first law the body must continue in a straight line, therefore to change its trajectory a force must be applied to deflect it

6 0
2 years ago
a person pushing a stroller starts from rest, uniformly accelerating at a rate of 0.500m/s^2. what is the velocity of the stroll
grandymaker [24]
You're going to use the constant acceleration motion equation for velocity and displacement:
(V)final²=(V)initial²+2a(Δx)

Given:
a=0.500m/s²
Δx=4.75m
(V)intial=0m
(V)final= UNKNOWN

(V)final= 2.179m/s
5 0
3 years ago
Read 2 more answers
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