The first modern atomic scientist was:
2 answers:
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Answer:
The rate of heat conduction through the layer of still air is 517.4 W
Explanation:
Given:
Thickness of the still air layer (L) = 1 mm
Area of the still air = 1 m
Temperature of the still air ( T) = 20°C
Thermal conductivity of still air (K) at 20°C = 25.87mW/mK
Rate of heat conduction (Q) = ?
To determine the rate of heat conduction through the still air, we apply the formula below.
Q = 517.4 W
Therefore, the rate of heat conduction through the layer of still air is 517.4 W
Answer:
α = 0.0135 rad/s²
Explanation:
given,
t = 133 min = 133 x 60 = 7980 s
angular speed varies from 570 rpm to 1600 rpm
now,
570 rpm =
= 59.69 rad/s
1600 rpm = =
= 167.6 rad/s
using equation of rotational motion
ωf = ωi + αt
167.6 = 59.7 + α x 7980
α x 7980 = 107.9
α = 0.0135 rad/s²