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BartSMP [9]
3 years ago
6

A prototype boat is 30 meters long and is designed to cruise at 9 m/s. Its drag is to be simulated by a 0.5-meter-long model pul

led in a tow tank. For Froude scaling find a)-The tow speed b)-The ratio of model to prototype drag c)-The ratio of model to prototype power.
Engineering
1 answer:
Ghella [55]3 years ago
6 0

Answer:

a) 1.16 m/s

b)  1/216000

c)  (√15)/6480000

Explanation:

The parameters given are;

Length of boat prototype, lp = 30 m

Speed of boat prototype = 9 m/s

Length of boat model, lm= 0.5 m

a) lm/lp = 0.5/30 = 1/60 = ∝

(vm/vp) = ∝^(1/2) = √∝ = (1/60)^(1/2)

vm = 9 × (1/60)^(1/2) = 1.16 m/s

b) The ratio of the model to prototype drag, Fm/Fp, is given as follows;

Fm/Fp = (vm/vp)²×(lm/lp)² = ∝³

Fm/Fp = (1/60)³ = 1/216000

c) The ratio of the model to prototype power  pm/p_p = (Fm/Fp) × (vm/vp) = ∝³×√∝

The ratio of the model to prototype power  pm/p_p = √(1/60) × (1/60)³

pm/p_p = √(1/60) × (1/60)³ = (√15)/6480000

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Which one of the following faults cause the coffee in a brewer to keep boiling after the brewing cycle is finished?
MrRa [10]

Answer:

  C.  Welded contacts on the thermostat

Explanation:

Any fault that keeps the heating element heating when it should not is a fault that will cause the symptom described. The details <em>depend on the design of the brewer</em> (not given).

"A short at the terminals" depends on what terminals are being referenced. The device on-off switch terminals are normally connected together when the brewer is turned on, so a short there may not be observable.

"Welded contacts on the thermostat" will have the observed effect if the thermostat is the primary means of ending the brewing cycle. If the thermostat of interest is an overheat protective device not normally involved in ending the brewing cycle, then that fault may not cause the observed symptom.

__

If the heating element is open-circuit, no heating will occur. A gasket leak may cause a puddle, but may have nothing to do with the end of the brewing cycle. (Loss of water can be expected to end boiling, rather than prolong it.)

8 0
3 years ago
A fatigue test was conducted in which the mean stress was 46.2 MPa and the stress amplitude was 219 MPa.
sleet_krkn [62]

Answer:

a)σ₁ = 265.2 MPa

b)σ₂ = -172.8 MPa

c)Stress\ ratio =-0.65

d)Range = 438 MPa

Explanation:

Given that

Mean stress ,σm= 46.2 MPa

Stress amplitude ,σa= 219 MPa

Lets take

Maximum stress level = σ₁

Minimum stress level =σ₂

The mean stress given as

\sigma_m=\dfrac{\sigma_1+\sigma_2}{2}

2\sigma_m={\sigma_1+\sigma_2}

2 x 46.2 =  σ₁ +  σ₂

 σ₁ +  σ₂ = 92.4 MPa    --------1

The amplitude stress given as

\sigma_a=\dfrac{\sigma_1-\sigma_2}{2}

2\sigma_a={\sigma_1-\sigma_2}

2 x 219 =  σ₁ -  σ₂

 σ₁ -  σ₂ = 438 MPa    --------2

By adding the above equation

2  σ₁ = 530.4

σ₁ = 265.2 MPa

-σ₂ = 438 -265.2 MPa

σ₂ = -172.8 MPa

Stress ratio

Stress\ ratio =\dfrac{\sigma_{min}}{\sigma_{max}}

Stress\ ratio =\dfrac{-172.8}{265.2}

Stress\ ratio =-0.65

Range = 265.2 MPa - ( -172.8 MPa)

Range = 438 MPa

8 0
3 years ago
Lydia is the CEO for a large pharmaceutical manufacturer. Her company is in the final stages of FDA
weqwewe [10]
OSHA inspections are generally unannounced. In fact, except in four exceptional circumstances when advance notice may be given.


It is a criminal offense for any person to give unauthorized advance notice of an OSHA inspection.
5 0
3 years ago
An engineer is testing the shear strength of spot welds used on a construction site. The engineer's null hypothesis at a 5% leve
lilavasa [31]

Answer:

b) The null hypothesis should be rejected.

Explanation:

The null hypothesis is  that the mean shear strength of spot welds is at least

3.1 MPa

H0: u ≥3.1 MPa  against the claim Ha: u< 3.1 MPa

The alternate hypothesis is  that the mean shear strength of spot welds is less than 3.1 MPa.

This is one tailed test

The critical region Z(0.05) < ± 1.645

The Sample mean= x`= 3.07

The number of welds= n= 15

Standard Deviation= s= 0.069

Applying z test

z= x`-u/s/√n

z= 3.07-3.1/0.069/√15

z= -0.03/0.0178

z= -1.68

As the calculated z= -1.68  falls in the critical region Z(0.05) < ± 1.645 the null hypothesis is rejected and the alternate hypothesis is accepted that the mean shear strength of spot welds is less than 3.1 MPa

8 0
3 years ago
Las dos torres que sostienen un puente colgante tienen una separación de 240m y una altura de 110m a partir de la carretera, si
Sveta_85 [38]

La altura es de 169.4 metros.

Dado que las dos torres que sostienen un puente colgante tienen una separación de 240m y una altura de 110m a partir de la carretera, si el cable tensor más corto mide 10m, para determinar cuál es altura de un cable que se encuentra a 100m de distancia del centro se debe realizar los siguientes cálculos, aplicando la ecuación parabólica:  

  • (240)² = 4P x (110-10)
  • 57600 = 4P x 100
  • 57600 = 400P
  • 57600/400 = P
  • 144 = P
  • 200 x 200 = 4 x 144 x (Altura - 100)
  • 40000 = 576Altura - 57600
  • 40000 + 57600 / 576 = Altura
  • 169.4 metros = Altura

Por lo tanto, la altura es de 169.4 metros.

Aprende más en brainly.com/question/20333463

8 0
2 years ago
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