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3 years ago

Any help is appreciated <3

Engineering

1 answer:

Len [333]3 years ago

8 0

Answer:

forwarder

Explanation:

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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

3 years ago

The basic barometer can be used to measure the height of a building. If the barometric readings at the top and the bottom of a b

Levart [38]

Answer:

230.51 m

Explanation:

Pb = 695 mmHg

Pt = 675 mmHg

Pb - Pt = 20 mmHg

Calculate dP:

dP = p * g * H = (13600)*(9.81)*(20/1000) = 2668.320 Pa

Calculate Height of building as dP is same for any medium of liquid

dP = p*g*H = 2668.320

H = 2668.32 / (1.18 * 9.81) = 230.51 m

8 0

3 years ago

What classification is an E7018?<br><br> will give brainliest

Hunter-Best [27]

Answer: E7018 is a low-hydrogen iron powder type electrode that produces high quality x-ray welds. It can be used in all positions on AC or DC reverse polarity welding current.

Explanation: Hope this helps

3 0

3 years ago

Hello, how are you?

Kisachek [45]

Answer:

Hello, I'm good. Thank you for asking

8 0

2 years ago

Read 2 more answers

: During a heavy rainstorm, water from a parking lot completely fills an 18-in.- diameter, smooth, concrete storm sewer. If the

Montano1993 [528]

Answer

diameter of parking lot = 18 in

flowrate = 10 ft³/s

pressure drop = 100 ft

using general equation

$\dfrac{P_1}{\gamma}+\dfrac{v_1^2}{2g}+Z_1 = \dfrac{P_2{\gamma} + \dfrac{v_2^2}{2g} + Z_2 +\dfrac{fLV^2}{2\rho D}$

$V = \dfrac{Q}{A} = \dfrac{10}{\dfrac{\pi}{4}(\dfrac{18}{12})^2} = 5.66\ ft/s$

$\Delta P = \gamma (Z_2-Z_1) +\dfrac{fLV^2}{2\rho D}$

taking f = 0.0185

at Z₁ = Z₂

$\Delta P = \dfrac{0.0185 \times 100\times 1.94\times 5.66^2}{2\dfrac{18}{12} (2)}$

ΔP = 0.266 psi

b) when flow is uphill z₂-z₁ = 2

$\Delta P =62.4\times 2 \times \dfrac{1}{144} +0.266$

$\Delta P= 1.13\ psi$

c) When flow is downhill z₂-z₁ = -2

$\Delta P =62.4\times 2 \times \dfrac{1}{144} +0.266$

$\Delta P=-0.601\ psi$

7 0

3 years ago