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A metallic material with yield stress of 140 MPa and cross section of 300 mm x 100 mm, is subjected to a tensile force of 8.00 M

Readme [11.4K]

Answer:Yes,266.66 MPa

Explanation:

Given

Yield stress of material =140 MPa

Cross-section of $300\times 100 mm^2$

Force(F)=8 MN

Therefore stress due to this Force( $\sigma$ )

$\sigma =\frac{F}{A}=\frac{8\times 10^6}{300\times 100\times 10^{-6}}$

$\sigma =266.66 \times 10^{6} Pa$

$\sigma =266.66 MPa$

Since induced stress is greater than Yield stress therefore Plastic deformation occurs

8 0

3 years ago

1. Springs____________<br> energy when compressed<br> And _________energy when they rebound.

densk [106]

Answer:

Springs store energy when compressed and release energy when they rebound

Explanation:

6 0

3 years ago

A motor driven water pump operates with an inlet pressure of 96 kPa (absolute) and mass flow rate of 120 kg/min. The motor consu

NeX [460]

Answer:

The maximum water pressure at the discharge of the pump (exit) = 496 kPa

Explanation:

The equation expressing the relationship of the power input of a pump can be computed as:

$E _{pump,u} = \dfrac{m(P_2-P_1)}{\rho}$

where;

m = mass flow rate = 120 kg/min

the pressure at the inlet $P_1$ = 96 kPa

the pressure at the exit $P_2$ = ???

the pressure $\rho$ = 1000 kg/m³

∴

$0.8 \times 10^{3} \ W = \dfrac{(120 \ kg/min * 1min/60 s)(P_2-96000)}{1000}$

$0.8 \times 10^{3}\times 1000 = {(120 \ kg/min * 1min/60 s)(P_2-96000)}$

$800000 = {(120 \ kg/min * 1min/60 s)(P_2-96000)}$

$\dfrac{800000}{2} = P_2-96000$

400000 = P₂ - 96000

400000 + 96000 = P₂

P₂ = 496000 Pa

P₂ = 496 kPa

Thus, the maximum water pressure at the discharge of the pump (exit) = 496 kPa

8 0

3 years ago

What’s another name for a service overcurrent device?

alexandr402 [8]

Overcurrent protective devices, or OCPDs

4 0

3 years ago

A hypothetical A-B alloy of composition 57 wt% B-43 wt% A at some temperature is found to consist of mass fractions of 0.5 for b

Dennis_Churaev [7]

Answer:

composition of alpha phase is 27% B

Explanation:

given data

mass fractions = 0.5 for both

composition = 57 wt% B-43 wt% A

composition = 87 wt% B-13 wt% A

solution

as by total composition Co = 57 and by beta phase composition Cβ = 87

we use here lever rule that is

Wα = Wβ ...............1

Wα = Wβ = 0.5

now we take here left side of equation

we will get

$\frac{C_\beta - Co}{C_\beta - Ca}$ = 0.5

$\frac{87 - 57}{87 - Ca}$ = 0.5

solve it we get

Ca = 27

so composition of alpha phase is 27% B

8 0

3 years ago