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Y_Kistochka [10]
3 years ago
7

ME And what you know about love

Engineering
1 answer:
antoniya [11.8K]3 years ago
7 0

Answer:

Everything I got what you need

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A 225 MPa conducted in which the mean stress was 50 MPa and the stress amplitude was (a) Compute the maximum and (b) Compute the
tamaranim1 [39]

Answer:

Explanation:

Given data in question

mean stress  = 50 MPa

amplitude stress  = 225 MPa

to find out

maximum stress, stress ratio, magnitude of the stress range.

solution

we will find first  maximum stress  and minimum stress

and stress will be sum of (maximum +minimum stress) / 2

so for stress 50 MPa and 225 MPa

\sigma _{m} =  \sigma _{maximum} + \sigma _{minimum}  / 2

50 =  \sigma _{maximum} + \sigma _{minimum}  / 2    ...........1

and

225 =  \sigma _{maximum} + \sigma _{minimum}  / 2      ...........2

from eqution 1 and 2 we get maximum and minimum stress

\sigma _{maximum} = 275 MPa        ............3

and \sigma _{minimum} = -175 MPa     ............4

In 2nd part we stress ratio is will compute by ratio of equation 3 and 4

we get ratio =  \sigma _{minimum} / \sigma _{maximum}

ratio = -175 / 227

ratio = -0.64

now in 3rd part magnitude will calculate by subtracting maximum stress - minimum stress i.e.

magnitude = \sigma _{maximum} - \sigma _{minimum}  

magnitude = 275 - (-175) = 450 MPa

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3 years ago
) A certain polymer is used for evacuation systems for aircraft. It is important that the polymer be resistant to the aging proc
bonufazy [111]

Answer:

it will be a scattered plot

Explanation:

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Which of the eight diagnostic steps for locating an engine performance problem is performed first?
Kay [80]

Answer:

D. Perform a thorough visual inspection.

4 0
2 years ago
What are the de Broglie frequencies and wavelengths of (a) an electron accelerated to 50 eV (b) a proton accelerated to 100 eV
DaniilM [7]

Answer:

(a) De-Brogie wavelength is 0.173 nm and frequency is 2.42 x 10^16 Hz

(b) De-Brogie wavelength is 2.875 pm and frequency is 4.8 x 10^16 Hz

Explanation:

(a)

First, we need to find velocity of electron. Since, it is accelerated by electric potential. Therefore,

K.E of electron = (1/2)mv² = (50 eV)(1.6 x 10^-19 J/1 eV)

(1/2)mv² = 8 x 10^(-18) J

Mass of electron = m = 9.1 x 10^(-31) kg

Therefore,

v² = [8 x 10^(-18) J](2)/(9.1 x 10^(-31) kg)

v = √1.75 x 10^13

v = 4.2 x 10^6 m/s

Now, the de Broglie's wavelength is given as:

λ = h/mv

where,

h = Plank's Constant = 6.626 x 10^(-34) kg.m²/s

Therefore,

λ = (6.626 x 10^(-34) kg.m²/s)/(9.1 x 10^(-31) kg)(4.2 x 10^6 m/s)

<u>λ = 0.173 x 10^(-9) m = 0.173 nm</u>

The frequency is given as:

Frequency = f = v/λ

f = (4.2 x 10^6 m/s)/(0.173 x 10^(-9) m)

<u>f = 2.42 x 10^16 Hz</u>

(b)

First, we need to find velocity of proton. Since, it is accelerated by electric potential. Therefore,

K.E of proton = (1/2)mv² = (100 eV)(1.6 x 10^-19 J/1 eV)

(1/2)mv² = 1.6 x 10^(-17) J

Mass of proton = m = 1.67 x 10^(-27) kg

Therefore,

v² = [1.6 x 10^(-17) J](2)/(1.67 x 10^(-27) kg)

v = √1.916 x 10^10

v = 1.38 x 10^5 m/s

Now, the de Broglie's wavelength is given as:

λ = h/mv

where,

h = Plank's Constant = 6.626 x 10^(-34) kg.m²/s

Therefore,

λ = (6.626 x 10^(-34) kg.m²/s)/(1.67 x 10^(-27) kg)(1.38 x 10^5 m/s)

<u>λ = 2.875 x 10^(-12) m = 2.875 pm</u>

The frequency is given as:

Frequency = f = v/λ

f = (1.38 x 10^5 m/s)/(2.875 x 10^(-12) m)

<u>f = 4.8 x 10^16 Hz</u>

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