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zaharov [31]
2 years ago
10

A wave has a frequency of 34 Hz and a wavelength of 2.0 m. What is the speed of the wave? Use . A. 17 m/s B. 36 m/s C. 0.059 m/s

D. 68 m/s
Physics
2 answers:
mel-nik [20]2 years ago
8 0
F= (speed)/(wavelength)

Therefore, speed = Frequency x wavelength
  V = 68m/s
saul85 [17]2 years ago
5 0
Speed = frequency * wavelength
           = 34 * 2
           = 68m/s
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In a rocket-propulsion problem the mass is variable. Another such problem is a raindrop falling through a cloud of small water d
Alexxandr [17]

Answer:

a) a = g / 3

b) x (3.0) = 14.7 m

c) m (3.0) = 29.4 g

Explanation:

Given:-

- The following differential equation for (x) the distance a rain drop has fallen has the form:

                             x*g = x * \frac{dv}{dt} + v^2

- Where,                v = Speed of the raindrop

- Proposed solution to given ODE:

                             v = a*t

Where,                  a = acceleration of raindrop

Find:-

(a) Using the proposed solution for v find the acceleration a.

(b) Find the distance the raindrop has fallen in t = 3.00 s.

(c) Given that k = 2.00 g/m, find the mass of the raindrop at t = 3.00 s.

Solution:-

- We know that acceleration (a) is the first derivative of velocity (v):

                             a = dv / dt   ... Eq 1

- Similarly, we know that velocity (v) is the first derivative of displacement (x):

                            v = dx / dt  , v = a*t ... proposed solution (Eq 2)

                             v .dt = dx = a*t . dt

- integrate both sides:

                             ∫a*t . dt = ∫dt

                             x = 0.5*a*t^2  ... Eq 3

- Substitute Eq1 , 2 , 3 into the given ODE:

                            0.5*a*t^2*g = 0.5*a^2 t^2 + a^2 t^2

                                                = 1.5 a^2 t^2

                            a = g / 3

- Using the acceleration of raindrop (a) and t = 3.00 second and plug into Eq 3:

                           x (t) = 0.5*a*t^2

                           x (t = 3.0) = 0.5*9.81*3^2 / 3

                           x (3.0) = 14.7 m  

- Using the relation of mass given, and k = 2.00 g/m, determine the mass of raindrop at time t = 3.0 s:

                           m (t) = k*x (t)

                           m (3.0) = 2.00*x(3.0)

                           m (3.0) = 2.00*14.7

                           m (3.0) = 29.4 g

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2 years ago
During a compaction test in the lab a cylindrical mold with a diameter of 4in and a height of 4.58in was filled. The compacted s
Ray Of Light [21]

Answer:

part a : <em>The dry unit weight is 0.0616  </em>lb/in^3<em />

part b : <em>The void ratio is 0.77</em>

part c :  <em>Degree of Saturation is 0.43</em>

part d : <em>Additional water (in lb) needed to achieve 100% saturation in the soil sample is 0.72 lb</em>

Explanation:

Part a

Dry Unit Weight

The dry unit weight is given as

\gamma_{d}=\frac{\gamma}{1+\frac{w}{100}}

Here

  • \gamma_d is the dry unit weight which is to be calculated
  • γ is the bulk unit weight given as

                                              \gamma =weight/Volume \\\gamma= 4 lb / \pi r^2 h\\\gamma= 4 lb / \pi (4/2)^2 \times 4.58\\\gamma= 4 lb / 57.55\\\gamma= 0.069 lb/in^3

  • w is the moisture content in percentage, given as 12%

Substituting values

                                              \gamma_{d}=\frac{\gamma}{1+\frac{w}{100}}\\\gamma_{d}=\frac{0.069}{1+\frac{12}{100}} \\\gamma_{d}=\frac{0.069}{1.12}\\\gamma_{d}=0.0616 lb/in^3

<em>The dry unit weight is 0.0616  </em>lb/in^3<em />

Part b

Void Ratio

The void ratio is given as

                                                e=\frac{G_s \gamma_w}{\gamma_d} -1

Here

  • e is the void ratio which is to be calculated
  • \gamma_d is the dry unit weight which is calculated in part a
  • \gamma_w is the water unit weight which is 62.4 lb/ft^3 or 0.04 lb/in^3
  • G is the specific gravity which is given as 2.72

Substituting values

                                              e=\frac{G_s \gamma_w}{\gamma_d} -1\\e=\frac{2.72 \times 0.04}{0.0616} -1\\e=1.766 -1\\e=0.766

<em>The void ratio is 0.77</em>

Part c

Degree of Saturation

Degree of Saturation is given as

S=\frac{G w}{e}

Here

  • e is the void ratio which is calculated in part b
  • G is the specific gravity which is given as 2.72
  • w is the moisture content in percentage, given as 12% or 0.12 in fraction

Substituting values

                                      S=\frac{G w}{e}\\S=\frac{2.72 \times .12}{0.766}\\S=0.4261

<em>Degree of Saturation is 0.43</em>

Part d

Additional Water needed

For this firstly the zero air unit weight with 100% Saturation is calculated and the value is further manipulated accordingly. Zero air unit weight is given as

\gamma_{zav}=\frac{\gamma_w}{w+\frac{1}{G}}

Here

  • \gamma_{zav} is  the zero air unit weight which is to be calculated
  • \gamma_w is the water unit weight which is 62.4 lb/ft^3 or 0.04 lb/in^3
  • G is the specific gravity which is given as 2.72
  • w is the moisture content in percentage, given as 12% or 0.12 in fraction

                                      \gamma_{zav}=\frac{\gamma_w}{w+\frac{1}{G}}\\\gamma_{zav}=\frac{0.04}{0.12+\frac{1}{2.72}}\\\gamma_{zav}=\frac{0.04}{0.4876}\\\gamma_{zav}=0.08202 lb/in^3\\

Now as the volume is known, the the overall weight is given as

weight=\gamma_{zav} \times V\\weight=0.08202 \times 57.55\\weight=4.72 lb

As weight of initial bulk is already given as 4 lb so additional water required is 0.72 lb.

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Answer:

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Explanation:

all correct on edge

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Explanation:

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