Answer:
The program is as follows:
i = 1
while(i<11):
j = 1
while(j<=i):
print('*', end = '')
j += 1
i += 1
print()
Explanation:
Initialize i to 1
i = 1
The outer loop is repeated as long as i is less than 11
while(i<11):
Initialize j to 1
j = 1
The inner loop is repeated as long as j is less than or equal i
while(j<=i):
This prints a *
print('*', end = '')
This increments j and ends the inner loop
j += 1
This increments i
i += 1
This prints a blank and ends the inner loop
print()
Answer:
Q = 63,827.5 W
Explanation:
Given:-
- The dimensions of plate A = ( 10 mm x 1 m )
- The fluid comes at T_sat , 1 atm.
- The surface temperature, T_s = 75°C
Find:-
Determine the total condensation rate of water vapor onto the front surface of a vertical plate
Solution:-
- Assuming drop-wise condensation the heat transfer coefficient for water is given by Griffith's empirical relation for T_sat = 100°C.
h = 255,310 W /m^2.K
- The rate of condensation (Q) is given by Newton's cooling law:
Q = h*As*( T_sat - Ts )
Q = (255,310)*( 0.01*1)*( 100 - 75 )
Q = 63,827.5 W
Answer:
P = 4.745 kips
Explanation:
Given
ΔL = 0.01 in
E = 29000 KSI
D = 1/2 in
LAB = LAC = L = 12 in
We get the area as follows
A = π*D²/4 = π*(1/2 in)²/4 = (π/16) in²
Then we use the formula
ΔL = P*L/(A*E)
For AB:
ΔL(AB) = PAB*L/(A*E) = PAB*12 in/((π/16) in²*29*10⁶ PSI)
⇒ ΔL(AB) = (2.107*10⁻⁶ in/lbf)*PAB
For AC:
ΔL(AC) = PAC*L/(A*E) = PAC*12 in/((π/16) in²*29*10⁶ PSI)
⇒ ΔL(AC) = (2.107*10⁻⁶ in/lbf)*PAC
Now, we use the condition
ΔL = ΔL(AB)ₓ + ΔL(AC)ₓ = ΔL(AB)*Cos 30° + ΔL(AC)*Cos 30° = 0.01 in
⇒ ΔL = (2.107*10⁻⁶ in/lbf)*PAB*Cos 30°+(2.107*10⁻⁶ in/lbf)*PAC*Cos 30°= 0.01 in
Knowing that PAB*Cos 30°+PAC*Cos 30° = P
we have
(2.107*10⁻⁶ in/lbf)*P = 0.01 in
⇒ P = 4745.11 lb = 4.745 kips
The pic shown can help to understand the question.
Answer:
The solution code is written in Java.
System.out.println(numItems);
Explanation:
Java <em>println() </em>method can be used to display any string on the console terminal. We can use <em>println()</em> method to output the value held by variable <em>numItems.</em> The <em>numItems </em>is passed as the input parameter to <em>println()</em> and this will output the value of <em>numItems</em> to console terminal and at the same time the output with be ended with a newline automatically.
GIVEN:
Amplitude, A = 0.1mm
Force, F =1 N
mass of motor, m = 120 kg
operating speed, N = 720 rpm
= 
Formula Used:
Solution:
Let Stiffness be denoted by 'K' for each mounting, then for 4 mountings it is 4K
We know that:
so,
= 75.39 rad/s
Using the given formula:
Damping is negligible, so, 
will give the tranfer function
Therefore,
= 
=
Required stiffness coefficient, K = 173009 N/m = 173.01 N/mm
