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
Answer:
The velocity will be v = 22.1[m/s]
Explanation:
We can solve this problem by using the principle of energy conservation, where potential energy is converted to kinetic energy. For this problem we will take the point with maximum potential energy when the body is 25 [m] high. By the time the height is zero, the potential energy will have been transformed into kinetic energy, and we can find the velocity of the body.
Now we know that the energy will be transformed.
Answer: 4,438.96m
Explanation:
(kindly find attachment below)
From the attachment below, it can be seen that the resultant displacement and the other 2 displacements form a right angle triangle, with A+B as the hypotenus, 3.2km as the opposite and the displacement B as the adjacent.
By using phythagoras theorem
H² = O² + A²
(5.38)² = (3.20)² + B²
28.944 = 10.24 + B²
B² = 28.944 - 10.24
B² = 18.7044
B = √18.7044
B = 4.439km to meter is 4.439 * 1000 = 4,438. 96m
