The modulus of elasticity is 28.6 X 10³ ksi
<u>Explanation:</u>
Given -
Length, l = 5in
Force, P = 8000lb
Area, A = 0.7in²
δ = 0.002in
Modulus of elasticity, E = ?
We know,
Modulus of elasticity, E = σ / ε
Where,
σ is normal stress
ε is normal strain
Normal stress can be calculated as:
σ = P/A
Where,
P is the force applied
A is the area of cross-section
By plugging in the values, we get
σ = 
σ = 11.43ksi
To calculate the normal strain we use the formula,
ε = δ / L
By plugging in the values we get,
ε = 
ε = 0.0004 in/in
Therefore, modulus of elasticity would be:
Thus, modulus of elasticity is 28.6 X 10³ ksi
Answer:
16-bit wide
Explanation:
In order to find the width of the address bus, we need first to know how many memory cells it is needed to address.
If the size memory is 64 KB, this means that the memory size, in bytes, is equal to the following quantity:
64 KB = 2⁶ * 2¹⁰ bytes = 2¹⁶ bytes.
In order to address this quantity of cell positions, the address bus must be able to address 2¹⁶ bytes, so it must have 16-bit wide.
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:
6.99 x 10⁻³ m³ / s
Explanation:
Th e pressure difference at the two ends of the delivery pipe due to atmospheric pressure and water column will cause flow of water.
h = difference in the height of water column at two ends of delivery pipe
6 - 1 = 5 m
Velocity of flow of water
v = √2gh
= √ (2 x 9.8 x 5)
= 9.9 m /s
Volume of water flowing per unit time
velocity x cross sectional area
= 9.9 x 3.14 x .015²
= 6.99 x 10⁻³ m³ / s
