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3 years ago

A wave with a frequency of 14 Hz has a wavelength of 3 meters. At what speed will this wave travel?

Engineering

2 answers:

kobusy [5.1K]3 years ago

6 0

Answer:

42m/s

Explanation:

Frequency (f) =14Hz and wavelength (λ )=3m

Frequency and wavelength can be related to speed:

speeds(V)=f*λ=14*3=42m/s.

vodka [1.7K]3 years ago

3 0

Answer:

42 meter Hertz

Explanation:

As the Frequency and wavelength are known

So the speed of the wave can be find by the product of its frequency and wavelength.

Frequency= 14 Hz and wavelength= 3 meters

So, Speed of the wave = 14 *3 = 42 meter Hertz

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An aluminum part will be subjected to cyclic loading where the maximum stress will be 300 MPa and the minimum stress will be-100

Dominik [7]

Answer:

a) The mean stress experimented by the aluminium part is 100 megapascals, b) The stress amplitude of the aluminium part is 400 megapascals, c) The stress ratio of the aluminium part is 4.

Explanation:

a) The mean stress is determined by this expression:

$\sigma_{m} = \frac{\sigma_{min}+\sigma_{max}}{2}$

Where:

$\sigma_{m}$ - Mean stress, measured in megapascals.

$\sigma_{min}$ - Minimum stress, measured in megapascals.

$\sigma_{max}$ - Maximum stress, measured in megapascals.

If we know that $\sigma_{min} = -100\,MPa$ and $\sigma_{max} = 300\,MPa$ , the mean stress is:

$\sigma_{m} = \frac{-100\,MPa+300\,MPa}{2}$

$\sigma_{m} = 100\,MPa$

The mean stress experimented by the aluminium part is 100 megapascals.

b) The stress amplitude is given by the following difference:

$\sigma_{a} = |\sigma_{max}-\sigma_{min}|$

Where $\sigma_{a}$ is the stress amplitude, measured in megapascals.

If we know that $\sigma_{min} = -100\,MPa$ and $\sigma_{max} = 300\,MPa$ , the stress amplitude is:

$\sigma_{a} = |300\,MPa-(-100\,MPa)|$

$\sigma_{a} = 400\,MPa$

The stress amplitude of the aluminium part is 400 megapascals.

c) The stress ratio ( $R$ ) is the ratio of the stress amplitude to mean stress. That is:

$R = \frac{\sigma_{a}}{\sigma_{m}}$

If we know that $\sigma_{m} = 100\,MPa$ and $\sigma_{a} = 400\,MPa$ , the stress ratio is:

$R = \frac{400\,MPa}{100\,MPa}$

$R = 4$

The stress ratio of the aluminium part is 4.

3 0

3 years ago

Air is compressed slowly in a piston-cylinder assembly from an initial state where P1 = 1.4 bar, V1= 4.25 m^3, to a final state

lord [1]

Answer:

W=-940.36 KJ

Explanation:

Given that

$P_1=1\ bar,V_1=4.25 {m^3}$

$P_2=6.8\ bar$

Process follows pv=constant

So this is the isothermal process and work in isothermal process given as

$W=P_1V_1\ln \dfrac{P_1}{P_2}$

Now by putting the values (1.4 bar =140 KPa)

$W=P_1V_1\ln \dfrac{P_1}{P_2}$

$W=140\times 4.25 \ln \dfrac{1.4}{6.8}$

W=-940.36 KJ

Negative sign indicates that this is a compression process and work will given to the system.

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4 years ago

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Diffusion of Ammonia in an Aqueous Solution Ammonia (A)-water (B) solution ta 278 K and 4 mm thick is in contact with an organic

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Answer:

Explanation: