Homie I don’t know either♀️Drop out of schl ig ;-;
For part a)
Since the conical surface is not exposed to the radiation coming from the walls only from the circular plate and assuming steady state, the temperature of the conical surface is also equal to the temperature of the circular plate. T2 = 600 K
For part b)
To maintain the temperature of the circular plate, the power required would be calculated using:
Q = Aσ(T₁⁴ - Tw⁴)
Q = π(500x10^-3)²/4 (5.67x10^-8)(600⁴ - 300⁴)
Q = 5410.65 W
Answer:
9] V = D ÷ T
Take any distance value from the graph and its relevant time.
V = 4 ÷ 2
V = 2 m/s
[You will notice that any distance values with its time will give you 2 m/s as its speed. This means that speed is constant throughout.]
10] Take the distance value and its time for the highest peak of B.
V = 20 ÷ 2
V = 10 m/s
Answer:
a) 
b) 
Explanation:
Given:
- initial rotational speed of phonograph,
- final rotational speed of phonograph,
- time taken for the acceleration,
a)
Now angular acceleration:
b)
Using eq. of motion:
