All categories

2 years ago

PLEASE HELP WITH THIS ASAP! Thanks

Engineering

1 answer:

Tpy6a [65]2 years ago

8 0

Answer:

Architect

Explanation: You need to be unique and creative to be an architect.

You might be interested in

A cylindrical specimen of a titanium alloy having an elastic modulus of 107 GPa (15.5 × 106 psi) and an original diameter of 3.7

Keith_Richards [23]

Answer:

the maximum length of specimen before deformation is found to be 235.6 mm

Explanation:

First, we need to find the stress on the cylinder.

Stress = σ = P/A

where,

P = Load = 2000 N

A = Cross-sectional area = πd²/4 = π(0.0037 m)²/4

A = 1.0752 x 10^-5 m²

σ = 2000 N/1.0752 x 10^-5 m²

σ = 186 MPa

Now, we find the strain (∈):

Elastic Modulus = Stress / Strain

E = σ / ∈

∈ = σ / E

∈ = 186 x 10^6 Pa/107 x 10^9 Pa

∈ = 1.74 x 10^-3 mm/mm

Now, we find the original length.

∈ = Elongation/Original Length

Original Length = Elongation/∈

Original Length = 0.41 mm/1.74 x 10^-3

Original Length = 235.6 mm

5 0

2 years ago

Determine the dimensions for W if W = P L^3 / (M V^2) where P is a pressure, L is a length, M is a mass, and V is a velocity.

Eva8 [605]

Correct answer is option E. No dimensions

As we know formula Pressure (P) is $\frac{F}{A}$

also,

Dimensional formula of Pressure is $M^{1}L^{-1}T^{-2}$
Dimensional formula of length is L
Dimensional formula of mass is M
Dimensional formula of velocity is $L^{1} T^{-1}$

So, as given W= $\frac{P*L^{3} }{M*V^{2} }$

Dimensional formula of W = $\frac{M^{1}L^{-1}T^{-2} L^{3} }{M^{1} L^{2}T^{-2} }$

since all terms get cancelled

Work is dimensionless i.e no dimensions

Learn more about dimensions here brainly.com/question/20351712

#SPJ10

6 0

1 year ago

A PMMA plate with a 25 mm (width) x 6.5 mm (thickness) cross-section has a contained crack of length 2c = 0.5 mm in the center o

victus00 [196]

Answer:

LAOD = 6669.86 N

Explanation:

Given data:

width $= 25 mm = 25\times 10^{-3} m$

thickness $= 6.5 mm = 6.5\times 10^{-3} m$

crack length 2c = 0.5 mm at centre of specimen

$\sigma _{applied} = 1000 N/cross sectional area$

stress intensity factor = k will be

$\sigma_{applied} = \frac{1000}{25\times 10^{-3}\times 6.5\times 10^{-3}}$

$= 6.154\times 10^{6} Pa$

we know that

$k =\sigma_{applied} (\sqrt{\pi C})$

$=6.154\sqrt{\pi (2.5\times 10^{-04})}$ [c =0.5/2 = 2.5*10^{-4}]

K = 0.1724 Mpa m^{1/2} for 1000 load

if $K_C = 1.15 Mpa m^{1/2}$ then load will be

$Kc = \sigma _{frac}(\sqrt{\pi C})$

$1.15 MPa = \sigma _{frac}\times \sqrt{\pi (2.5\times 10^{-04})}$

$\sigma _{frac} = 41.04 MPa$

$load = \sigma _{frac}\times Area$

$load = 41.04 \times 10^6 \times 25\times 10^{-3}\times 6.5\times 10^{-3} N$

LAOD = 6669.86 N

3 0

2 years ago

A thick oak wall initially at 25°C is suddenly exposed to gases for which T =800°C and h =20 W/m2.K. Answer the following questi

Schach [20]

Answer:

a) What is the surface temperature, in °C, after 400 s?

T (0,400 sec) = 800°C

b) Yes, the surface temperature is greater than the ignition temperature of oak (400°C) after 400 s

c) What is the temperature, in °C, 1 mm from the surface after 400 s?

T (1 mm, 400 sec) = 798.35°C

Explanation:

oak initial Temperature = 25°C = 298 K

oak exposed to gas of temp = 800°C = 1073 K

h = 20 W/m².K

From the book, Oak properties are e=545kg/m³ k=0.19w/m.k Cp=2385J/kg.k

Assume: Volume = 1 m³, and from energy balance the heat transfer is an unsteady state.

From energy balance: $\frac{T - T_{\infty}}{T_i - T_{\infty}} = Exp (\frac{-hA}{evCp})t$

Initial temperature wall = $T_i$

Surface temperature = T

Gas exposed temperature = $T_{\infty}$

6 0

3 years ago

2. BCD uses 6 bits to represent a symbol. a) True b) False

Goshia [24]

Answer:

true because BCD used 6 bits to represent a symbol .

Explanation:

mark me brainlist

4 0

2 years ago