Answer:
Option D
160 kHz
Explanation:
Since we must use at least one synchronization bit, total message signal is 15+1=16
The minimum sampling frequency, fs=2fm=2(5)=10 kHz
Bandwith, BW required is given by
BW=Nfs=16(10)=160 kHz
The piston makes four strokes in the crankshaft makes two revolutions between combustion firings. The diameter of the piston, and the inside diameter of the cylinder, is called a bore, so the area of the head of the piston is pi times the diameter squared divided by four.
Answer:
do it
Explanation:
they are giving you step by step instructions to do. i woul hate to have to work with you out anywhere in the world if you have to ask this. i am 15, and i would at least read the d a m n thing before i post it.
Answer:
Explanation:
Force on Q₃ due to charge Q₁
= 9 x 10⁹x 4 x 10⁻⁹ x1 x 10⁻⁹ / 5²
= 1.44 x 10⁻⁹ N
Force due to Q₂ will also be 1.44 x 10⁻⁹ N
component of these forces along x axis
-= 2 x 1.44 x 10⁻⁹ cosθ
= 2.88 x 10⁻⁹ x 4/5
= 2.30x10⁻⁹ N along x axis.
The y-component will calcel out.
b ) In this case , Q₁ will repel and Q₂ will attract.
In this case Q₁ will repel and Q₂ will attract. Component along y - axis will be same as earlier one or 2.30x10⁻⁹ N . Component along x axis will cancel out.
c ) Electric field in case 1 and case 2 will be
= 2.30x10⁻⁹ / 1 x 10⁻⁹
= 2.3 N / C , because field is force per unit charge. The sane field will be in case 2 .