Answer:
Explanation:
Answer:
31.12. Accounts Receivable a/c Dr. 1000
To Service Revenue 1000
10 x 100 = 1000
Answer:
Alice is correct.
The loop are dependent.
Explanation:
for(i = 1; i <= N; i = (i*2)+17 )
for(k = i+1; k <= i+N; k = k+1) // notice i in i+1 and i+N
printf("B")
This is a nested for-loop.
After the first for-loop opening, there is no block of statement to be executed rather a for-loop is called again. And the second for-loop uses the value of i from the first for-loop. The value of N is both called from outside the loop.
So, the second for-loop depend on the first for loop to get the value of i. For clarity purpose, code indentation or use of curly brace is advised.
Answer:
w=2.25
Explanation:
It is necessary to determine the maximum w so that the normal stress in the AB and CD rods does not exceed the permitted normal stress.
The surface of the cross-section of the stapes was determined:
A_ab= 10 mm^2
A-cd= 15 mm^2
The maximum load is determined from the condition that the normal stresses is not higher than the permitted normal stress σ_allow.
σ_ab = F_ab/A_ab
σ_allow
σ_cd = F_cd/A_cd
σ_allow
In the next step we will determine the static size: Picture b).
We apply the conditions of equilibrium:
∑F_x=0
∑F_y=0
∑M=0
∑M_a=0 ==> -w*6*0.5*6*0.75*F_cd*6 =0
==> F_cd = 2*w*k*N
∑F_y=0 ==> F_cd+F_ab - 6*w*0.5 ==>2*w+F_ab -6*w*0.5 =0
==> F_ab = w*k*N
Now we determine the load w
<u>Sector AB: </u>
σ_ab = F_ab/A_ab
σ_allow=300 KPa
= w/10*10^-6
σ_allow=300 KPa
w_ab = 3*10^-3 kN/m
<u>Sector CD: </u>
σ_cd = F_cd/A_cd
σ_allow=300 KPa
= 2*w/15*10^-6
σ_allow=300 KPa
w_cd = 2.25*10^-3 kN/m
w=min{w_ab;w_cd} ==> w=min{3*10^-3;2.25*10^-3}
==> w=2.25 * 10^-3 kN/m
<u>The solution is: </u>
w=2.25 N/m
note:
find the attached graph
i believe the correct answer is c but i’m sorry if i’m not correct
Answer: its an Ignition coil