Answer:
This filter can either be Band Pass or Band
Answer:
Speed=1.633 m/s
Force= 20 N
Explanation:
Ideally, hence where v is the speed of collar, m is the mass of collar, k is spring constant and s is the displacement.
In this case, s=100-0=100mm=0.1m since 1 m is equivalent to 1000mm
k is given as 200 N/m and mass is 0.75 Kg
Substituting the given values
Therefore, <u>the speed is 1.633 m/s</u>
The sum of vertical forces is given by mg where g is acceleration due to gravity and it's value taken as
Therefore,
The sum of forces in normal direction is given by therefore
Therefore, <u>normal force on the rod is 20 N</u>
Answer:
- 1.55
- 260 N.s
- 3370 m
- 1.6
- 43.75 kg/s
Explanation:
1) Thrust coefficient at sea level.
Cfsl = TSL / Pca
TSL = Mp * Vc + ( Pc - Pa )Ac
Mp = mass flux = 43.75 kg/s
∴ Cfsl = Mp Vc / Pca + ( Pc - Pa )/Pc * ( Ac / A* )
= 1.6 - 0.04923 = 1.55
<u>2) Specific impulse at sea </u>
Isp = Vc / g = 2549.75 / 9.81
= 260 N.s
3) Altitude at optimal expansion
H = 3370 m
<u>4) thrust coefficient at optimal expansion </u>
CF = 1.6
attached below is the detailed solution
<u>5) Mass flux through the throat </u>
Mass flux = P1 * At / Cc
= ( 7*10^6 * 0.01 ) / 1600
= 43.75 kg/s