Power radiated , P = σεAT⁴ where σ = Stefan-Boltzmann constant = 5.6704 × 10⁻⁸ W/m²K⁴, ε = emissivity of bulb filament = 0.8, A = surface area of bulb = 30 mm² = 30 × 10⁻⁶ m² and T = operating temperature of filament.

So, T = ⁴√(P/σεA)

Since P = 60 W, we substitute the vales of the variables into T. So,

T = ⁴√(P/σεA)

= ⁴√(60 W/(5.6704 × 10⁻⁸ W/m²K⁴ × 0.8 × 30 × 10⁻⁶ m²)

= ⁴√(60 W/(136.0896 × 10⁻¹⁴ W/K⁴)

= ⁴√(60 W/(13608.96 × 10⁻¹⁶ W/K⁴)

= ⁴√(0.00441 × 10¹⁶K⁴)

= 0.2577 × 10⁴ K

= 2577 K

6 0

2 years ago

A tensile test specimen has a gage length = 50 mm and its cross-sectional area = 100 mm2. The specimen yields at 48,000 N, and t

Shalnov [3]

Answer:

a) yield strength

$\sigma_y = \dfrac{F_y}{A} = =\dfrac{48000}{100} = 480 MPa$

b) modulus of elasticity

strain calculation

$\varepsilon_0=\dfrac{L-L_0}{L_0}=\dfrac{50.23-50}{50} = 0.0046$

strain for offset yield point

$\varepsilon_{new} = \varepsilon_0 -0.002$

=0.0046-0.002 = 0.0026

now, modulus of elasticity

$E = \dfrac{\sigma_y}{\varepsilon_{new}}=\dfrac{480}{0.0026}$

= 184615.28 MPa = 184.615 GPa

c) tensile strength

$\sigma_u =\dfrac{F_{max}}{A}=\dfrac{87000}{100}=870MPa$

d) percentage elongation

$\% Elongation = \dfrac{L-L_0}{L_0}\times 100 = \dfrac{67.3-50}{50}\times 100 = 34.6\%$

e) percentage of area reduction

$\% Area\ reduction = \dfrac{A-A_f}{A}\times 100=\dfrac{100-53}{100}= 47 \%$

7 0

3 years ago

Jonathan is studying solutions in his chemistry class. The teacher instructs the students to measure 2 tablespoons of salt into

bulgar [2K]

8 tablespoons of water.
2 x 4 = 8

I hope this helps :)

3 0

2 years ago