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aliina [53]
3 years ago
13

A woman is standing in the ocean, and she notices that after a wave crest passes by, five more crests pass in a time of 29.4 s.

The distance between two successive crests is 31.4 m. What is the wave's (a) period, (b) frequency, (c) wavelength, and (d) speed
Physics
1 answer:
baherus [9]3 years ago
3 0

Answer:

(a) 5.88 s

(b) 0.17 Hz

(c) 31.4 m

(d) 5.338 m/s

Explanation:

From the question,

(a) Period = time between successive crest = t/n............ Equation 1

where t = time, n = number of wave crest

Given: t = 29.4, n = 5

therefore,

Period (T) = 29.4/5 = 5.88 s.

(b) Frequency = 1/period

Frequency = 1/5.88

Frequency = 0.17 Hz.

(c) wavelength = distance between two successive crest = 31.4 m

(d)  using,

v = λf............... Equation 2

Where v = speed, f = frequency, λ = wavelength

given: f = 0.17 Hz, λ = 31.4 m

Substitute into equation 2

v = 0.17(31.4)

v = 5.338 m/s

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barxatty [35]

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v1 = 9*10^4 m/s

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K1 +U1 = K2 +U2

0.5mv1^2 - GMm/r1 = 0.5mv2^2 - GMm/r2

0.5v1^2 - GM/r1 = 0.5v2^2 - GM/r2

M is mass of sun = 1.98*10^30 kg

G = 6.67*10^-11 N.m^2/kg^2

0.5*(9*10^4)^2 - (6.67*10^-11*1.98*10^30/(5*10^10)) = 0.5v2^2 - (6.67*10^-11*1.98*10^30/(6*10^12))

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3 years ago
5/6 When switched on, the grinding machine accelerates from rest to its operating speed of 3450 rev/min in 6 seconds. When switc
ludmilkaskok [199]

Answer:

Δθ₁ =  172.5 rev

Δθ₁h =  43.1 rev

Δθ₂ =   920 rev

Δθ₂h = 690 rev

Explanation:

  • Assuming uniform angular acceleration, we can use the following kinematic equation in order to find the total angle rotated during the acceleration process, from rest to its operating speed:

       \Delta \theta = \frac{1}{2} *\alpha *(\Delta t)^{2}  (1)  

  • Now, we need first to find the value of  the angular acceleration, that we can get from the following expression:

       \omega_{f1}  = \omega_{o} + \alpha * \Delta t  (2)

  • Since the machine starts from rest, ω₀ = 0.
  • We know the value of ωf₁ (the operating speed) in rev/min.
  • Due to the time is expressed in seconds, it is suitable to convert rev/min to rev/sec, as follows:

       3450 \frac{rev}{min} * \frac{1 min}{60s} = 57.5 rev/sec (3)

  • Replacing by the givens in (2):

       57.5 rev/sec = 0 + \alpha * 6 s  (4)

  • Solving for α:

       \alpha = \frac{\omega_{f1}}{\Delta t} = \frac{57.5 rev/sec}{6 sec} = 9.6 rev/sec2 (5)

  • Replacing (5) and Δt in (1), we get:

       \Delta \theta_{1} = \frac{1}{2} *\alpha *(\Delta t)^{2} = \frac{1}{2} * 6.9 rev/sec2* 36 sec2 = 172.5 rev  (6)

  • in order to get the number of revolutions during the first half of this period, we need just to replace Δt in (6) by Δt/2, as follows:

       \Delta \theta_{1h} = \frac{1}{2} *\alpha *(\Delta t/2)^{2} = \frac{1}{2} * 6.9 rev/sec2* 9 sec2 = 43.2 rev  (7)

  • In order to get the number of revolutions rotated during the deceleration period, assuming constant deceleration, we can use the following kinematic equation:

       \Delta \theta = \omega_{o} * \Delta t + \frac{1}{2} *\alpha *(\Delta t)^{2}  (8)

  • First of all, we need to find the value of the angular acceleration during the second period.
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       0 = 57.5 rev/sec + \alpha * 32 s  (9)

  • Solving for α in (9), we get:

       \alpha_{2}  =- \frac{\omega_{f1}}{\Delta t} = \frac{-57.5 rev/sec}{32 sec} = -1.8 rev/sec2 (10)

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  • In order to get finally the number of revolutions rotated during the first half of the second period, we need just to replace 32 s by 16 s, as follows:
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U = \frac{1}{R}

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R is the R-value of the material

For the insulator in this problem,

R = 29

Substituting into the equation, we find the heat transfer coefficient:

U=\frac{1}{29}=0.034

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