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
Answer:
183.75 cubic inches.
Explanation:
The volume of the wood board is determine by means of this expression:
By replacing variables:
Answer:
IA = 80/3 kgm^2
Explanation:
Given:-
- The mass of rod AB, m1 = 20 kg
- The length of rod AB, L1 = 2m
- The mass of rod CD, m2 = 10 kg
- The length of rod CD, L2 = 1m
Find:-
What is the moment of inertia about A for member AB?
Solution:-
- The moment of inertia About point "O" the center of rod AB is given as:
IG = 1/12*m1*L^2
- To shift the axis of moment of inertia for any object at a distance "d" from the center of mass of that particular object we apply the parallel axis theorem. The new moment of inertia about any arbitrary point, which in our case A end of rod AB is:
IA = IG + m1*d^2
- Where the distance "d" from center of rod AB to its ends is 1/2*L1 = 1 m.
So the moment of inertia for rod AB at point A would be:
IA = 1/12*m1*L^2 + m1*0.5*L1^2
IA = 1/3 * m1*L1^2
IA = 1/3*20*2^2
IA = 80/3 kgm^2
Answer:
False
Explanation:
I got it wrong picking true