Answer and Explanation:
clear all; close all;
N=512;
t=(1:N)/N;
fs=1000;
f=(1:N)*fs/N;
x= sin(2*pi*200*t) + sin(2*pi*400*t);
y= sin(2*pi*200*t) + sin(2*pi*900*t);
for n = 1:20
a(n) = (2/N)*sum(x.*(cos(2*pi*n*t)))
b(n) = (2/N)*sum(x.*(sin(2*pi*n*t)))
c(n) = sqrt(a(n).^2+b(n).^2)
theta(n) =-(360/(2*pi))*atan(b(n)./a(n));
end
plot(f(1:20),c(1:20),'rd');
disp([a(1:4),b(1:4),c(1:4),theta(1:4)])
Answer:
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Answer:
8 to 10 times
Explanation:
For dry road
u= 15 mph ( 1 mph = 0.44 m/s)
u= 6.7 m/s
Let take coefficient of friction( μ) of dry road is 0.7
So the de acceleration a = μ g
a= 0.7 x 10 m/s ² ( g=10 m/s ²)
a= 7 m/s ²
We know that
v= u - a t
Final speed ,v=0
0 = 6.7 - 7 x t
t= 0.95 s
For snow road
μ = 0.4
de acceleration a = μ g
a = 0.4 x 10 = 4 m/s ²
u= 30 mph= 13.41 m/s
v= u - a t
Final speed ,v=0
0 = 30 - 4 x t'
t'=7.5 s
t'=7.8 t
We can say that it will take 8 to 10 times more time as compare to dry road for stopping the vehicle.
8 to 10 times
Answer:
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