C-14 is an isotope of the element carbon. How does it differ from the carbon atom seen here? A) C-14 has two more protons. B) C-
2 answers:
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Answer:
The location is at (3.535, 1.162) m
Solution:
As per the question:
Mass of the projectile, m = 9.6 kg
Initial velocity, v = 12.4 m/s
Angle,
Mass of one fragment, m = 6.5 kg
Time taken by the fragment, t = 1.42 s
Height of the fragment, y = 5.9 m
Horizontal distance, x = 13.6 m
Now,
To determine the location of the second fragment:
Horizontal Range,
Time of flight, t' =
Now, for the fragments:
Mass of the other fragment, m' = M - m = 9.6 - 6.5 = 3.1 kg
Distance traveled horizontally:
Distance traveled vertically:
Now,
x' = 3.535 m
Similarly,
The location of the other fragment is at (3.535, 1.162)
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