An object of mass 0.400.40 kg, hanging from a spring with a spring constant of 8.08.0 N/m, is set into an up-and-down simple har
1 answer:
Answer:
Acceleration will be equal to
Explanation:
We have given mass of the object m = 0.4 kg
Spring constant k = 8 N/m
Maximum displacement of the spring is given x = 0.1 m
From newton's law force is equal to .....eqn 1
By hook's law spring force is equal to .....eqn 2
From equation 1 and equation 2
So acceleration will be equal to
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Answer:
Explanation:
Using the equation
H = Q/t = k A ( T hot - T cold) / L
where H is the rate of heat loss = 51.4 W, T cold be temperature of the outer surface, A is the surface area of the fat layer which is a model of sphere ( surface area of a sphere ) = 4πr² where diameter = 1.60 m
radius = 1.60 m / 2 = 0.80 m
A = 4 × 3.142 × ( 0.8²) = 8.04352 m²
making T cold subject of the formula
T cold = T hot - = 30.9° C - ( 51.4 W × 3.9 × 10⁻² m) / ( 0.2 W/mK × 8.04352 m² ) = 30.9° C - 1.25 ° C = 29.65° C
b) The thickness of air layer for the bear to lose heat t a rate of 51.4 W
thermal conductivity of air is 0.024 W/mK and rearranging the earlier formula
L = = (0.024 W/ m K × 8.04352 m²) ( 29.65° C - 2.8°C) / 51.4 W = 0.101 m = 10.1 m