Answer:
Ano klassing tanong yn?
Explanation:
Ang taas namn yn? Paki linaw po para matulungan po kita.!!
Answer:
water rise = 22 mm
Explanation:
weight of packet IN AIR = 40 *9.81 =392.4 N
weight of packet IN WATER= 18 *9.81 =176.58 N
by Archimedi's principle
difference in weight = weight of displaced water
w_a - w_w = \rho_w v_d g
392.4 - 176.58 = 1000* v_d* 9.81
v_d = 0.022 m^3
v_d = A*H_rise
0.022 =1*H_rise
H_rise = 0.022 m = 22 mm
water rise = 22 mm
Answer:
0.71 lbf
Explanation:
Use ideal gas law:
PV = nRT
where P is absolute pressure,
V is volume,
n is number of moles,
R is universal gas constant,
and T is absolute temperature.
The absolute pressure is the sum of the atmospheric pressure and the gauge pressure.
P = 32 lbf/in² + 14.7 lbf/in²
P = 46.7 lbf/in²
Absolute temperature is in Kelvin or Rankine:
T = 75 + 459.67 R
T = 534.67 R
Given V = 3.0 ft³, and R = 10.731 ft³ psi / R / lb-mol:
PV = nRT
(46.7 lbf/in²) (3.0 ft³) = n (10.731 ft³ psi / R / lb-mol) (534.67 R)
n = 0.02442 lb-mol
The molar mass of air is 29 lbm/lb-mol, so the mass is:
m = (0.02442 lb-mol) (29 lbm/lb-mol)
m = 0.708 lbm
The weight of 1 lbm is lbf.
W = 0.708 lbf
Rounded to two significant figures, the weight of the air is 0.71 lbf.
Answer:
389.6 W/m²
Explanation:
The power radiated to the surroundings by the small hot surface, P = σεA(T₁⁴ - T₂⁴) where σ = Stefan-Boltzmann constant = 5.67 × 10⁻⁸ W/m²-K⁴, ε = emissivity = 0.8. T₁ = temperature of small hot surface = 430 K and T₂ = temperature of surroundings = 400 K
So, P = σεA(T₁⁴ - T₂⁴)
h = P/A = σε(T₁⁴ - T₂⁴)
Substituting the values of the variables into the equation, we have
h = 5.67 × 10⁻⁸ W/m²-K⁴ × 0.8 ((430 K )⁴ - (400 K)⁴)
h = 5.67 × 10⁻⁸ W/m²-K⁴ × 0.8 (34188010000 K⁴ - 25600000000 K⁴)
h = 5.67 × 10⁻⁸ W/m²-K⁴ × 0.8 × 8588010000K⁴
h = 38955213360 × 10⁻⁸ W/m²
h = 389.55213360 W/m²
h ≅ 389.6 W/m²
