Answer:
1.505
Explanation:
cylindrical part of diameter d is loaded by an axial force P. This causes a stress of P/A, where A = πd2/4. If the load is known with an uncertainty of ±11 percent, the diameter is known within ±4 percent (tolerances), and the stress that causes failure (strength) is known within ±20 percent, determine the minimum design factor that will guarantee that the part will not fail.
stress is force per unit area
stress=P/A
A = πd^2/4.
uncertainty of axial force P= +/-.11
s=+/-.20, strength
d=+/-.04 diameter
fail load/max allowed
minimum design=fail load/max allowed
minimum design =s/(P/A)
sA/P
A=(
.96d^2)/4, so Amin=
(because the diameter at minimum is (1-0.04=0.96)
minimum design=Pmax/(sminxAmin)
1.11/(.80*.96^2)=
1.505
Answer:
Bulk zoning.
Explanation:
Zoning is a method used by municipal authorities or government authorities to divide a land area into zones. It is a urban planning method. There are several types of zoning; bulk zoning being one of them.
Bulk zoning can be defined as a set of regulations used to control lot size, ratio of floor area, open space, yards, setbacks, bulding height, etc. These setbacks and restrictions, then, help authorities to determine maximun size of a building on a zoning area, and also the positioning of the building.
Therefore, bulk zoning is the correct answer.
Answer: 33.35 minutes
Explanation:
A(t) = A(o) *(.5)^[t/(t1/2)]....equ1
Where
A(t) = geiger count after time t = 100
A(o) = initial geiger count = 400
(t1/2) = the half life of decay
t = time between geiger count = 66.7 minutes
Sub into equ 1
100=400(.5)^[66.7/(t1/2)
Equ becomes
.25= (.5)^[66.7/(t1/2)]
Take log of both sides
Log 0.25 = [66.7/(t1/2)] * log 0.5
66.7/(t1/2) = 2
(t1/2) = (66.7/2 ) = 33.35 minutes
Answer:
clear, clc
prob3_5([1,2,3],[6,5,7],12,11,22,55,76)
function T=prob3_5(x,y,N,L,W,T1,T2)
w=zeros(1,length(x));
for n=1:2:N
for i=1:length(x)
w(i)=w(i)+(2/pi)*(2/n)*sin(n*pi*x(i)/L).*sinh(n*pi*y(i)/L)/sinh(n*pi*W/L);
end
end
T=(T2-T1)*w+T1;
end
Explanation:
Please input the commands into MATLAB
Answer:
Fluids
Explanation:
Fluids has special properties that allow forces and pressure to be distributed evenly within them.
- Fluids are gases and liquids whose intermolecular forces of attraction are generally weak or non-existence.
- Therefore, when pressure is applied to them, it permeates evenly on all parts.
- Their ability to tend to randomness makes liquids and gases very viable for distributing pressure.
