The force required to accelerate a 0.50kg ball at 50 m/s2 is (Remember F=ma) 25N 25K 50N 250N
1 answer:
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Answer:
The distribution is as depicted in the attached figure.
Explanation:
From the given data
- The plane wall is initially with constant properties is initially at a uniform temperature, To.
- Suddenly the surface x=L is exposed to convection process such that T∞>To.
- The other surface x=0 is maintained at To
- Uniform volumetric heating q' such that the steady state temperature exceeds T∞.
Assumptions which are valid are
- There is only conduction in 1-D.
- The system bears constant properties.
- The volumetric heat generation is uniform
From the given data, the condition are as follows
<u>Initial Condition</u>
At t≤0
This indicates that initially the temperature distribution was independent of x and is indicated as a straight line.
<u>Boundary Conditions</u>
<u>At x=0</u>
<u /><u />
This indicates that the temperature on the x=0 plane will be equal to To which will rise further due to the volumetric heat generation.
<u>At x=L</u>
<u /><u />
This indicates that at the time t, the rate of conduction and the rate of convection will be equal at x=L.
The temperature distribution along with the schematics are given in the attached figure.
Further the heat flux is inferred from the temperature distribution using the Fourier law and is also as in the attached figure.
It is important to note that as T(x,∞)>T∞ and T∞>To thus the heat on both the boundaries will flow away from the wall.
Answer:
Δ S = -50 J
Explanation:
given,
Temperature, T = 500 K
Heat dissipates, Q = 25.0 kJ
Q = 25000 J
change in entropy, ΔS = ?
using equation of entropy
Δ S = -50 J
negative sign is used because heat is lost in the surrounding.
Hence, Change in entropy is equal to -50 J