Answer:
2.9*10^14 electrons
Explanation:
An Ampere is the flow of one Coulomb per second, so 35 μA = is 35*10^-6 C per second.
An electron has a charge of 1.6*10^-19 C.
35*10^-6 / 1.6*10^-19 = 2.9*10^14 electrons
So, with a current o 35 μA you have an aevrage of 2.9*10^14 electrons flowing past a fixed reference cross section perpendicular to the direction of flow.
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
Answer:
Answer for the question:
Given a 8-bit ripple carry adder and the following four input scenarios: (i) A4 + 1F, (ii) AB+55, (iii) CA+34, (iv) 6D+29. a) Under which input scenario can adder generate correct output with the minimal delay? b) Under which input scenario can adder generate correct output with the maximum delay?
Is given in the attachment.
Explanation:
Answer:
True :)
Explanation:
If this is a true or false question.