Answer:
See the attachment
Verification shown in explanation
Explanation:
wc= 1/(RC)
The value of R and C shown in the attachment can be calculated as follows:
R=100 ohms
1000= 1/(100×C)
C= 10 microfarads
Gain= R/√(R²+Xc²)
at f=500, wc= 3142.59 rad/s
Xc= 1/(2πfC)
Xc=31.83
Gain= 100/√(100²+31.83²)
Gain= 0.953
at fc= 100, wc= 628.32 rad/s
Xc= 159.15
gain= 100/√(100²+ 159.15²)
gain= 0.532
As see form calculations that gain at low frequencies is smaller and gain at very high frequencies is near to 1. This means that at high frequencies Vin≈ Vout
Hence this filter allows high frequencies to pass but blocks low frequencies by lowering the gai
Answer:
2 yes ot will 2 would be a yes but i dont know how i would put that into a paragraph
Answer:

Explanation:
given,
mass = 2.2 kg
altitude(r₀) = (70 j) m
speed = 30 m/s
m_a = 0.77 kg
m_b =1.43 kg
part A strike ground (r_a)= (80 i) m
t = 6 s


r = 63.6 j m
by conservation of energy



Answer:
curly brackets are missing
Explanation:
The body of the main() function need to be enclosed in curly brackets. Try this:
int P = 3000;
int main( ) {
for (int t = 0; t < 10; t++) {
cout << P;
}
}