Explanation:
It is given that,
Mass,
Acceleration due to gravity,
Acceleration due to gravity is given by :
r = distance from head
r = 1.42 m
So, at the distance of 1.42 meters from the head its gravitational pull on you match that of Earth's. Hence, this is the required solution.
Answer:
775.48 W
Explanation:
given,
diameter of disk = 0.6 cm
length of the disk = 0.4 m
T₁ = 450 K T₂ = 450 K T₃ = 300 K
= 1.33
now,
the value of view factor (F₁₂)corresponding to 1.33
F₁₂ = 0.265
F₁₃ = 1 - 0.265 = 0.735
now,
net rate of radiation heat transfer from the disk to the environment:
= 2 F₁₃ A₁ σ (T₁⁴ - T₃⁴)
= 2 x 0.735 x π x (0.3)² x (5.67 x 10⁻⁸ W/m²) (450⁴ - 300⁴)
= 775.48 W
Net radiation heat transfer from the disks to the environment = 775.48 W