Answer:
A) Linear Equation -
Linear equation has only one independent variable and when the linear equation plotted on a graph it forms a straight line. It is made up of two expressions equal to each other in a equation. Linear equation graph fits the Y= mx+a ( m=slope).
B) Laplace's equation is linear as it is a second order partial differential equation. So if we put dependent variable in differential equation it always show result in linear.
To develop this problem it is necessary to apply the definitions of entropy change within the bodies
The change of entropy in copper would be defined as

Where,
Q= Heat exchange
T = Temperature
For an incompressible substance, the change in the heat exchange is defined as

Where,
m = Mass
c = Specific heat
Replacing in our equation we have that


Since
, then


In this way for the change of enthalpy and internal energy you have to

As
, then

Therefore the correct option is A. No change at All
Answer:
Answer for the question:
A hard real-time system has been developed to for a fly-by-wire aviation system. It has sensors for the following pilot interface systems: System Sampling Frequency (Hz) CPU time required (ms) Yoke 20 5 Rudder pedals 15 2 Throttle 10 1 Overhead of the hard real-time system on the architecture selected for the plane is .21. Can these tasks be scheduled? Show your work. Regardless of whether or not the tasks are schedulable , what should a designer do if they are not.
is given below.
Explanation:
Yea. Scheduling of these tasks can be done in below order:
- Throttle
- Rudder pedals.
- Yoke.
The temperature at which the diffusion coefficient have a value of 2.1×10-5 m 2 /s is -47078 K.
Using the relation;
logD = logDo - Ea/2.303RT
D = diffusion coefficient
Do = preexponential
Ea = activation energy
R = gas constant
T = temperature
Substituting values;
log(2.1×10-5)= log (1.1×10-5 ) - 253,300/2.303 × 8.314 × T
log(2.1×10-5) - log (1.1×10-5 ) = - 253,300/2.303 × 8.314 × T
log[2.1×10-5/1.1×10-5] = - 253,300/2.303 × 8.314 × T
0.281 × (2.303 × 8.314 × T) = - 253,300
T = - 253,300/2.303 × 0.281 × 8.314
T = -47078 K
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