Answer:
Explanation:
An information contains
25Hz and 75Hz sine wave
Sample frequency is 500Hz
The analogy signal are generally
y(t) = Asin(2πx/λ - wt), w=2πf
y1(t)=Asin(2πx/λ - wt)
y1(t)=Asin(2πx/λ - 2π•25t)
y1(t)=Asin(2πx/λ - 50πt)
Similarly
y2(t)=Asin(2πx/λ - 150πt)
Using Nyquist theorem
Nyquist Theorem states that in order to adequately reproduce a signal it should be periodically sampled at a rate that is 2 times the highest frequency you wish to record.
From sampling
f(nyquist)=f(sample)/2
f(nyquist)=500/2
f(nyquist)=250Hz
From signal
The highest frequency is 150Hz
F(nyquist) = 2×F(highest)
f(nyquist)= 2×150
f(nyquist)= 300Hz
Sample per frequency Ns is given as
Ns=F(sample)/F(highest signal)
Ns=500/150
Ns=3.33sample/period
This is above nyquist rate of 2sample/period
So signal below 300Hz reproduced without aliasing.
The highest resulting frequency is 300Hz
Mass of first car = Initial mass (Mi) = 2 kg
Initial velocity (Vi) = 2 m/s
Mass of both cars together = Final mass (Mf) = 2 + 3 kg = 5 kg
Final Velocity (Vf) = ?
Applying law of conservation of momentum,
Mi x Vi = Mf x Vf
2 x 2 = 5 x Vf
Vf = 4/5 = 0.8 m/s
The object will continue moving in a straight line at constant speed.
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Answer:
(a) 42 N
(b)36.7 N
Explanation:
Nomenclature
F= force test line (N)
W : fish weight (N)
Problem development
(a) Calculating of weight of the heaviest fish that can be pulled up vertically, when the line is reeled in at constant speed
We apply Newton's first law of equlibrio because the system moves at constant speed:
∑Fy =0
F-W= 0
42N -W =0
W = 42N
(b) Calculating of weight of the heaviest fish that can be pulled up vertically, when the line is reeled with an acceleration whose magnitude is 1.41 m/s²
We apply Newton's second law because the system moves at constant acceleration:
m= W/g , m= W/9.8 , m:fish mass , W: fish weight g:acceleration due to gravity
∑Fy =m*a
m= W/g , m= W/9.8 , m:fish mass , W: fish weight g:acceleration due to gravity
F-W= ( W/9.8 )*a
42-W= ( W/9.8 )*1.41
42= W+0.1439W
42=1.1439W
W= 42/1.1439
W= 36.7 N