Answer:
P=361.91 KN
Explanation:
given data:
brackets and head of the screw are made of material with T_fail=120 Mpa
safety factor is F.S=2.5
maximum value of force P=??
<em>solution:</em>
to find the shear stress
T_allow=T_fail/F.S
=120 Mpa/2.5
=48 Mpa
we know that,
V=P
<u>Area for shear head:</u>
A(head)=π×d×t
=π×0.04×0.075
=0.003×πm^2
<u>Area for plate:</u>
A(plate)=π×d×t
=π×0.08×0.03
=0.0024×πm^2
now we have to find shear stress for both head and plate
<u>For head:</u>
T_allow=V/A(head)
48 Mpa=P/0.003×π ..(V=P)
P =48 Mpa×0.003×π
=452.16 KN
<u>For plate:</u>
T_allow=V/A(plate)
48 Mpa=P/0.0024×π ..(V=P)
P =48 Mpa×0.0024×π
=361.91 KN
the boundary load is obtained as the minimum value of force P for all three cases. so the solution is
P=361.91 KN
note:
find the attached pic
Answer:
15.64 MW
Explanation:
The computation of value of X that gives maximum profit is shown below:-
Profit = Revenue - Cost
= 15x - 0.2x 2 - 12 - 0.3x - 0.27x 2
= 14.7x - .47x^2 - 12
After solving the above equation we will get maximum differentiate for profit that is
14.7 - 0.94x = 0
So,
x = 15.64 MW
Therefore for computing the value of X that gives maximum profit we simply solve the above equation.
Answer:
88750 N
Explanation:
given data:
plastic deformation σy=266 MPa=266*10^6 N/m^2
cross-sectional area Ao=333 mm^2=333*10^-6 m^2
solution:
To determine the maximum load that can be applied without
plastic deformation (Fy).
Fy=σy*Ao
=88750 N
Answer:
The graph representing the linear inequalities is attached below.
Explanation:
The inequalities given are :
y>x-2 and y<x+1
For tables for values of x and y and get coordinates to plot for both equation.
In the first equation;
y>x-2
y=x-2
y-x = -2
The table will be :
x y
-2 -4
-1 -3
0 -2
1 -1
2 0
The coordinates to plot are : (-2,-4) , (-1,-3), (0,-2), (1,-1) ,(2,0)
Use a dotted line and shade the part right hand side of the line.
Do the same for the second inequality equation and plot then shade the part satisfying the inequality.
The graph attached shows results.
Answer:
See explaination
Explanation:
A Finite state machines can be synchronous or asynchronous. The operation of asynchronous state machines does not require a clock signal. An Asynchronous state machine is classified basically on their operating mode, such as the fundamental mode, pulse mode or burst mode. An asynchronous state machine can have stable and transient states.
Please kindly refer to attachment for a step by step solution.
