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
It has been hypothesized, and some studies have supported the conjecture,
that certain species of insects and birds are able to sense the direction of external
magnetic fields.
I don't think there is any such notion where human beings are concerned.
Answer:
Explanation:
circumference of the tyre = 2πr = 2 x 3.14 x 0.26 = 1.6328m
76000km = 76000000m
no of revolutions required
= 76000000/1.6328 = 46546 revolutions.
Answer:
Explanation:
When objects collide, energy can be transferred from one object to another, thereby changing their motion. In such collisions, some energy is typically also transferred to the surrounding air; as a result, the air gets heated and sound is produced.
<span>Cobalt-60 is undergoing a radioactivity decay.
The formula of the decay is n=N(1/2)</span>∧(T/t).
<span>Where N </span>⇒ original mass of cobalt
<span> n </span>⇒ remaining mass of cobalt after 3 years
T ⇒ decaying period
t ⇒ half-life of cobalt.
So,
0.675 = 1 × 0.5∧(3/t)
log 0.675 = log 0.5∧(3/t)
3/t = log 0.675 ÷log 0.5
3/t= 0.567
t = 3÷0.567
= 5.290626524
the half-life of Cobalt-60 is 5.29 years.
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