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
An infant galaxy that is studied through the use of radio waves is called QUASAR.
Quasar are astronomical object, which have very high luminosity and they are found in the centers of some galaxies. Scientists believed that quasar are centers of distant galaxies. Radio waves is used to study quasar because they emit a huge amount of radio waves.
Answer:
Part b)
h = 78.5 m
Part c)
v = 39.24 m/s
Explanation:
Part b)
If ball need t = 0 to t = 4 s then height of the tower is the total displacement of the ball in t = 4 s interval
here if ball start from rest
then its displacement is given as



Part c)
Speed of the bearing at the end of the motion of the ball



I need help on this too lol