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2 years ago

11

A local surf report provides the height of the wave from the trough to the crest of the wave. How does this relate to the wave’s

amplitude?

1 answer:

gulaghasi [49]2 years ago

7 0

Answer:

The relationship between the provided height of the wave from the trough to the crest of the wave and the wave's amplitude is;

It is about twice the wave amplitude

Explanation:

The amplitude of a wave is the measured distance of a particle on the wave that moves from the rest position to the point of maximum displacement. Hence it is the distance between the crest or through and the rest position

Therefore, as the surf report provides the height of the wave from the through to the crest of the wave, it is twice about twice the wave amplitude.

You might be interested in

A double-pane insulated window consists of two 1 cm thick pieces of glass separated by a 1.8 cm layer of air. The window measure

Elanso [62]

Answer:

(b). T = 22.55 ⁰C

(c). q = 557.8 W

Explanation:

we take follow a step by step process to solving this problem.

from the question, we have that

The two glass pieces is separated by a 1.8 cm distance layer of air.

the thickness of glass piece is 1 cm

width = 4 m

the height = 3 m

(a). the sketch of the thermal circuit is uploaded in the picture below.

(b). the thermal resistance due to the conduction in the first glass plane is given thus;

R₁ = Lg / Kg A ................(1)

given that Kg rep. the thermal conductivity of the glass plane

A = conduction surface area

Lg = Thickness of glass plane4

taking the thermal conductivity of glass plane as Kg = 0.78 w/mk

inputting values into equation (1) we have,

R₁ = [1 (cm) ˣ 1 (m)/100 (cm)] / [(0.78 w/mk)(4m ˣ 3m)]

R₁ = 1.068 ˣ 10 ⁻³ k/w

Being that we have same thermal resistance in the first and second plane,

therefore R₁ = R₃ = 1.068 ˣ 10 ⁻³ k/w

⇒ Also the thermal resistance between air and glass as a result of the conduction by the layer is given thus

R₂ = La/KaA .....................(2)

given Ka = thermal conductivity of air

A = surface area

La = thickness of air

substituting values into the equation we have

R₂ = [1.8 (cm) ˣ 1 (m)/100 (cm)] / [(0.0262 w/mk)(4m ˣ 3m)]

R₂ = 5.73 ˣ 10⁻² k/w

Given the thermal resistance on the outer surface due to convection, we have

R₄ = 1/hA

inputting value gives R₄ = 1 / (12 w/m² ˣ 12m) = 6.94 ˣ 10⁻³k/w

R₄ = 6.94 ˣ 10⁻³k/w

Finally the sum total of thermal resistance = R₁ + R₂ + R₃ + R₄

R-total = 0.0663 kw

From this we can calculate the rate of heat loss

using q = Ti - To / R-total ..............(3)

given Ti and To is the inside and outside temperature i.e. 27⁰C and -10⁰C

from equation (3),

q = 27- (-10) / 0.0063 = 557.8 W

q = 557.8 W

⇒ Applying the heat transfer formula for inside surface glass temperature gives;

q = Ti - T₂ / R₃ + R₄

T₂ = Ti - q (R₃ + R₄)

T₂ = 27 - 557.8 (1.068ˣ10⁻³ + 6.94ˣ10⁻³ ) = 22.55°C

T₂ = 22.55°C

cheers i hope this helps

8 0

2 years ago

Q1: The first option should always be to get out safely (RUN)

nekit [7.7K]

Answer:

Q1 true

Q2 true

And other I am confuse

6 0

2 years ago

The hot and cold inlet temperatures to a concentric tube heat exchanger are Th,i = 200°C, Tc,i = 100°C, respectively. The outlet

alexgriva [62]

Answer:Counter,

0.799,

1.921

Explanation:

Given data

$T_{h_i}=200^{\circ}C$

$T_{h_o}=120^{\circ}C$

$T_{c_i}=100^{\circ}C$

$T_{c_o}=125^{\circ}C$

Since outlet temperature of cold liquid is greater than hot fluid outlet temperature therefore it is counter flow heat exchanger

Equating Heat exchange

$m_hc_{ph}\left [ T_{h_i}-T_{h_o}\right ]=m_cc_{pc}\left [ T_{c_o}-T_{c_i}\right ]$

$\frac{m_hc_{ph}}{m_cc_{pc}}$ = $\frac{125-100}{200-120}=\frac{25}{80}=C\left ( capacity rate ratio\right )$

we can see that heat capacity of hot fluid is minimum

Also from energy balance

$Q=UA\Delta T_m=\left ( mc_p\right )_{h}\left ( T_{h_i}-T_{h_o}\right )$

$NTU=\frac{UA}{\left ( mc_p\right )_{h}}$ = $\frac{\left ( T_{h_i}-T_{h_o}\right )}{T_m}$

$T_m=\frac{\left ( 200-125\right )-\left ( 120-100\right )}{\ln \frac{75}{20}}$

$T_m=41.63^{\circ}C$

NTU=1.921

$And\ effectiveness \epsilon =\frac{1-exp\left ( -NTU\left ( 1-c\right )\right )}{1-c\left ( -NTU\left ( 1-c\right )\right )}$

$\epsilon =\frac{1-exp\left ( -1.921\left ( 1-0.3125\right )\right )}{1-0.3125exp\left ( -1.921\left ( 1-0.3125\right )\right )}$

$\epsilon =\frac{1-exp\left ( -1.32068\right )}{1-0.3125exp\left ( -1.32068\right )}$

$\epsilon =\frac{1-0.2669}{1-0.0834}$

$\epsilon =0.799$

5 0

3 years ago

BigorU [14]

Answer: preventive maintenance

Explanation:

3 0

2 years ago

Air at 400kPa, 970 K enters a turbine operating at steady state and exits at 100 kPa, 670 K. Heat transfer from the turbine occu

Sonja [21]

Answer:

a

The rate of work developed is $\frac{\r W}{\r m}= 300kJ/kg$

b

The rate of entropy produced within the turbine is $\frac{\sigma}{\r m}= 0.0861kJ/kg \cdot K$

Explanation:

From the question we are told

The rate at which heat is transferred is $\frac{\r Q}{\r m } = - 30KJ/kg$

the negative sign because the heat is transferred from the turbine

The specific heat capacity of air is $c_p = 1.1KJ/kg \cdot K$

The inlet temperature is $T_1 = 970K$

The outlet temperature is $T_2 = 670K$

The pressure at the inlet of the turbine is $p_1 = 400 kPa$

The pressure at the exist of the turbine is $p_2 = 100kPa$

The temperature at outer surface is $T_s = 315K$

The individual gas constant of air R with a constant value $R = 0.287kJ/kg \cdot K$

The general equation for the turbine operating at steady state is \

$\r Q - \r W + \r m (h_1 - h_2) = 0$

h is the enthalpy of the turbine and it is mathematically represented as

$h = c_p T$

The above equation becomes

$\r Q - \r W + \r m c_p(T_1 - T_2) = 0$

$\frac{\r W}{\r m} = \frac{\r Q}{\r m} + c_p (T_1 -T_2)$

Where $\r Q$ is the heat transfer from the turbine

$\r W$ is the work output from the turbine

$\r m$ is the mass flow rate of air

$\frac{\r W}{\r m}$ is the rate of work developed

Substituting values

$\frac{\r W}{\r m} = (-30)+1.1(970-670)$

$\frac{\r W}{\r m}= 300kJ/kg$

The general balance equation for an entropy rate is represented mathematically as

$\frac{\r Q}{T_s} + \r m (s_1 -s_2) + \sigma = 0$

=> $\frac{\sigma}{\r m} = - \frac{\r Q}{\r m T_s} + (s_1 -s_2)$

generally $(s_1 -s_2) = \Delta s = c_p\ ln[\frac{T_2}{T_1} ] + R \ ln[\frac{v_2}{v_1} ]$

substituting for $(s_1 -s_2)$

$\frac{\sigma}{\r m} = \frac{-\r Q}{\r m} * \frac{1}{T_s} + c_p\ ln[\frac{T_2}{T_1} ] - R \ ln[\frac{p_2}{p_1} ]$

Where $\frac{\sigma}{\r m}$ is the rate of entropy produced within the turbine

substituting values

$\frac{\sigma}{\r m} = - (-30) * \frac{1}{315} + 1.1 * ln\frac{670}{970} - 0.287 * ln [\frac{100kPa}{400kPa} ]$

$\frac{\sigma}{\r m}= 0.0861kJ/kg \cdot K$

5 0

3 years ago