Answer:
591.3
Explanation:
99.19 + (1.85 × 266) = 591.29
rounded = 591.3
Answer:
Explanation:
Arbitrary means That no restrictions where placed on the number rather still each number is finite and has finite length. For the answer to the question--
Find(A,n,i)
for j =0 to 10000 do
frequency[j]=0
for j=1 to n do
frequency[A[j]]= frequency[A[j]]+1
for j =1 to n do
if i>=A[j] then
if (i-A[j])!=A[j] and frequency[i-A[j]]>0 then
return true
else if (i-A[j])==A[j] and frequency[j-A[j]]>1 then
return true
else
if (A[j]-i)!=A[j] and frequency[A[j]-i]>0 then
return true
else if (A[j]-i)==A[j] and frequency[A[j]-i]>1 then
return true
return false
Answer:
P = 18035.25 N
Explanation:
Given
D = 10.4 mm
ΔD = 3.2 ×10⁻³ mm
E = 207 GPa
ν = 0.30
If
σ = P/A
A = 0.25*π*D²
σ = E*εx
ν = - εz / εx
εz = ΔD / D
We can get εx as follows
εz = ΔD / D = 3.2 ×10⁻³ mm / 10.4 mm = 3.0769*10⁻⁴
Now we find εx
ν = - εz / εx ⇒ εx = - εz / ν = - 3.0769*10⁻⁴ / 0.30 = - 1.0256*10⁻³
then
σ = E*εz = (207 GPa)*(-1.0256*10⁻³) = - 2.123*10⁸ Pa
we have to obtain A:
A = 0.25*π*D² = 0.25*π*(10.4*10⁻3)² = 8.49*10⁻⁵ m²
Finally we apply the following equation in order o get P
σ = P/A ⇒ P = σ*A = (- 2.123*10⁸Pa)*(8.49*10⁻⁵ m²) = 18035.25 N
Answer:
(a) 6.91 mm (b) 160 MPa
Explanation:
Solution
Given that:
E = 200 GPa
The rod length = 48 mm
P =P¹ = 6 kN
Recall that,
1 kN = 10^3 N
1 m =10^3 mm
I GPa = 10^9 N/m²
Thus
The rod deformation is stated as follows:
δ = PL/AE-------(1)
σ = P/A----------(2)
Now,
(a) We substitute the values in equation and obtain the following:
48 * 10 ^⁻3 m = (6 * 10³ N) (60 m)/A[ 200 * 10^9 N/m^2]
Thus, we simplify
A = (6 * 10³) (60)/ ( 200 * 10^9) (48 * 10 ^⁻3)m²
A =0.0375 * 10 ^⁻3 m²
A =37.5 mm²
A = π/4 d²
Thus,
d² = 4A /π
After inserting the values we have,
d = √37.5 * 4/3.14 mm
= 6.9116 mm
or d = 6.91 mm
Therefore, the smallest that should be used is 6.91 mm
(B) To determine the corresponding normal stress that is caused by the tensile force, we input the values in equation (2)
Thus,
σ = P/A
σ= 6 * 10 ^ 3 N/ 37. 5 * 10 ^⁻6 m²
σ= 160 MPa
Note: I MPa = 10^6 N/m²
Hence the the corresponding normal stress is σ= 160 MPa
