List three technologies that pose ethical dilemmas?

A 9-m length of 6-mm-diameter steel wire is to be used in a hanger. The wire stretches 18mm when a tensile force P is applied. I

worty [1.4K]

Force P is 11304 N and normal stress is 400 N/mm²

<u>Explanation:</u>

Given-

Length, l = 9 m = 9000 mm

Diameter, d = 6 mm

Radius, r = 3 mm

Stretched length, Δl= 18 mm

Modulus of elasticity, E = 200 GPa = 200 X 10³MPa

Force, P = ?

According to Hooke's law,

Stress is directly proportional to strain.

So,

σ ∝ ε

σ = E ε

Where, E is the modulus of elasticity

We know,

ε = Δl / l

So,

σ = E X Δl/l

σ =

$200 X 10^3 * \frac{18}{9000} \\\\ = 400N/mm^2$

We know,

σ = P/A

And A = π (r)²

σ = P / π (r)²

$400 N/mm^2 = \frac{P}{3.14 X (3)^2} \\\\400 = \frac{P}{28.26} \\\\P = 11304N$

Therefore, Force P is 11304 N and normal stress is 400 N/mm²

4 0

3 years ago

A welding rod with κ = 30 (Btu/hr)/(ft ⋅ °F) is 20 cm long and has a diameter of 4 mm. The two ends of the rod are held at 500 °

SOVA2 [1]

Answer:

In Btu:

Q=0.001390 Btu.

In Joule:

Q=1.467 J

Part B:

Temperature at midpoint=274.866 C

Explanation:

Thermal Conductivity=k=30 (Btu/hr)/(ft ⋅ °F)= $\frac{30}{3600} (Btu/s)/(ft.F)=8.33*10^{-3} (Btu/s)/(ft.F)$

Thermal Conductivity is SI units:

$k=30(Btu/hr)/(ft.F) * \frac{1055.06}{3600*0.3048*0.556} \\k=51.88 W/m.K$

Length=20 cm=0.2 m= (20*0.0328) ft=0.656 ft

Radius=4/2=2 mm =0.002 m=(0.002*3.28)ft=0.00656 ft

T_1=500 C=932 F

T_2=50 C= 122 F

Part A:

In Joules (J)

$A=\pi *r^2\\A=\pi *(0.002)^2\\A=0.00001256 m^2$

Heat Q is:

$Q=\frac{k*A*(T_1-T_2)}{L} \\Q=\frac{51.88*0.000012566*(500-50}{0.2}\\ Q=1.467 J$

In Btu:

$A=\pi *r^2\\A=\pi *(0.00656)^2\\A=0.00013519 m^2$

Heat Q is:

$Q=\frac{k*A*(T_1-T_2)}{L} \\Q=\frac{8.33*10^{-3}*0.00013519*(932-122}{0.656}\\ Q=0.001390 Btu$

PArt B:

At midpoint Length=L/2=0.1 m

$Q=\frac{k*A*(T_1-T_2)}{L}$

On rearranging:

$T_2=T_1-\frac{Q*L}{KA}$

$T_2=500-\frac{1.467*0.1}{51.88*0.00001256} \\T_2=274.866\ C$

4 0

3 years ago

1. A gas pressure difference is applied to the legs of a U-tube manometer filled with a liquid with S = 1.5. The manometer readi

julia-pushkina [17]

Answer:

1) The pressure difference is 4.207 kilopascals.

2) 2.5 pounds per square inch equals 5.093 inches of mercury and 5.768 feet of water.

Explanation:

1) We can calculate the gas pressure difference from the U-tube manometer by using the following hydrostatic formula:

$\Delta P = \frac{S\cdot \rho_{w}\cdot g \cdot \Delta h}{1000}$ (Eq. 1)

Where:

$S$ - Relative density, dimensionless.

$\rho_{w}$ - Density of water, measured in kilograms per cubic meter.

$g$ - Gravitational acceleration, measured in meters per square second.

$\Delta h$ - Height difference in the U-tube manometer, measured in meters.

$\Delta P$ - Gas pressure difference, measured in kilopascals.

If we know that $S = 1.5$ , $\rho_{w} = 1000\,\frac{kg}{m^{3}}$ , $g = 9.807\,\frac{m}{s^{2}}$ and $\Delta h = 0.286\,m$ , then the pressure difference is:

$\Delta P = \frac{1.5\cdot \left(1000\,\frac{kg}{m^{3}} \right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.286\,m)}{1000}$

$\Delta P = 4.207\,kPa$

The pressure difference is 4.207 kilopascals.

2) From Physics we remember that a pound per square unit equals 2.036 inches of mercury and 2.307 feet of water and we must multiply the given pressure by corresponding conversion unit: ( $p = 2.5\,psi$ )