Answer:
The temperature attains equilibrium with the surroundings.
Explanation:
When the light bulb is lighted we know that it's temperature will go on increasing as the filament of the bulb has to constantly dissipates energy during the time in which it is on. Now this energy is dissipated as heat as we know it, this heat energy is absorbed by the material of the bulb which is usually made up of glass, increasing it's temperature. Now we know that any object with temperature above absolute zero has to dissipate energy in form of radiations.
Thus we conclude that the bulb absorbs as well as dissipates it's absorbed thermal energy. we know that this rate is dependent on the temperature of the bulb thus it the temperature of the bulb does not change we can infer that an equilibrium has been reached in the above 2 processes i.e the rate of energy absorption equals the rate of energy dissipation.
Steady state is the condition when the condition does not change with time no matter whatever the surrounding conditions are.
Answer:
Maximum Normal Stress σ = 8.16 Ksi
Maximum Shearing Stress τ = 4.08 Ksi
Explanation:
Outer diameter of spherical container D = 17 ft
Convert feet to inches D = 17 x 12 in = 204 inches
Wall thickness t = 0.375 in
Internal Pressure P = 60 Psi
Maximum Normal Stress σ = PD / 4t
σ = PD / 4t
σ = (60 psi x 204 in) / (4 x 0.375 in)
σ = 12,240 / 1.5
σ = 8,160 P/in
σ = 8.16 Ksi
Maximum Shearing Stress τ = PD / 8t
τ = PD / 8t
τ = (60 psi x 204 in) / (8 x 0.375 in)
τ = 12,240 / 3
τ = 4,080 P/in
τ = 4.08 Ksi
Answer:
does the question entail anything else?
Explanation:
Answer:
the percent increase in the velocity of air is 25.65%
Explanation:
Hello!
The first thing we must consider to solve this problem is the continuity equation that states that the amount of mass flow that enters a system is the same as what should come out.
m1=m2
Now remember that mass flow is given by the product of density, cross-sectional area and velocity
(α1)(V1)(A1)=(α2)(V2)(A2)
where
α=density
V=velocity
A=area
Now we can assume that the input and output areas are equal
(α1)(V1)=(α2)(V2)

Now we can use the equation that defines the percentage of increase, in this case for speed

Now we use the equation obtained in the previous step, and replace values

the percent increase in the velocity of air is 25.65%