Answer:
Negative feedback
Explanation:
In Biology, negative feedback refers to the counteraction of an effect by its own influence on the process producing it. For instance, the presence of a high level of a particular hormone in the blood may inhibit further secretion of that hormone.
In other words, in negative feedback, the result of a certain action may inhibit further performance of that action
Answer:
vec(a) = 16 i + 16 j
mag(a) = 22.63 ft/s^2
Explanation:
Given,
- The two components of velocity are given for fluid flow:
u = 4*y ft/s
v = 4*x ft/s
Find:
What is the time rate of change of the velocity vector V (i.e., the acceleration vector) for a fluid particle at x = 1 ft. and y = 1 ft. at time t = 1 second?
Solution:
- The rate of change of velocity is given to be acceleration. We will take derivative of each components of velocity with respect to time t:
a_x = du / dt
a_x = 4*dy/dt
a_y = dv/dt
a_y = 4*dx/dt
- The expressions dx/dt is the velocity component u and dy/dt is the velocity component v:
a_x = 4*(4*y) = 16y
a_y = 4*(4*x) = 16x
- The acceleration vector can be expressed by:
vec(a) = 16y i + 16x j
- Evaluate vector (a) at x = 1 and y = 1:
vec(a) = 16*1 i + 16*1 j = 16 i + 16 j
- The magnitude of acceleration is given by:
mag(a) = sqrt ( a^2_x + a^2_y )
mag(a) = sqrt ( 16^2 + 16^2 )
mag(a) = 22.63 ft/s^2
Answer: y' = - x'
<u>Explanation:</u>
Let f(x) = 2x + y
then f'(x) = 2 + y'
Let f(y) = x - 2y
then f'(y) = x' - 2
Given: f'(x) + f'(y) = 0
2 + y' + x' - 2 = 0
y' + x'= 0
y' = -x'
This can also be written as: 
Kirchhoff's voltage law states that the algebraic sum of the potential differences in any loop must be equal to zero as: ΣV = 0. Since the two resistors, R1 and R2 are wired together in a series connection, they are both part of the same loop so the same current must flow through each resistor.
Answer:
See explaination and attachment for theprogram code and output.
Explanation:
matlab program:
x1(1)=0;
y1(1)=0;
x2(1)=0;
y2(1)=0;
x3(1)=0;
y3(1)=0;
for i=1:10
x1(i+1)=0.5*x1(i);
y1(i+1)=0.5*y1(i);
x2(i+1)=0.5*x2(i)+0.25;
y2(i+1)=0.5*y2(i)+(sqrt(3)/4);
x3(i+1)=0.5*x3(i)+0.5;
y3(i+1)=0.5*y3(i);
end
figure
hold on
plot(x1,y1,'*')
plot(x2,y2,'*')
plot(x3,y3,'*')
hold off
Please kindly check attachment for output of 10, 100 and 1000 respectively.
