Answer:
I think it is B: an arc
Explanation:
hope this helps mark as brainiest
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 = ![\frac{8000 X 10^-^3}{0.7}](https://tex.z-dn.net/?f=%5Cfrac%7B8000%20X%2010%5E-%5E3%7D%7B0.7%7D)
σ = 11.43ksi
To calculate the normal strain we use the formula,
ε = δ / L
By plugging in the values we get,
ε = ![\frac{0.002}{5}](https://tex.z-dn.net/?f=%5Cfrac%7B0.002%7D%7B5%7D)
ε = 0.0004 in/in
Therefore, modulus of elasticity would be:
![E = \frac{11.43}{0.004} \\\\E = 28.6 X 10^3 ksi](https://tex.z-dn.net/?f=E%20%3D%20%5Cfrac%7B11.43%7D%7B0.004%7D%20%5C%5C%5C%5CE%20%3D%2028.6%20X%2010%5E3%20ksi)
Thus, modulus of elasticity is 28.6 X 10³ ksi
Answer:
0.304 L of Freon is needed
Explanation:
Q = mCT
Q is quantity of energy that must be removed = 47 BTU = 47×1055.06 = 49587.82 J
C is specific heat of Freon = 74 J/mol.K = 74 J/mol.K × 1 mol/120 g = 0.617 J/g.K
T is temperature in the area of Mars = 189 K
m = Q/CT = 49587.82/(0.617×189) = 452.23 g = 452.24/1000 = 0.45223 kg
Density of Freon = specific gravity of Freon × density of water = 1.49 × 1000 kg/m^3 = 1490 kg/m^3
Volume of Freon = mass/density = 0.45223/1490 = 0.000304 m^3 = 0.000304×1000 = 0.304 L
Answer:
A degree in architecture with 60 credit hours.
Explanation:
The requirements need for a student to qualify for a two year master of architecture degree are;
- 60 credit hours in architecture
- Complete 60 credit hours in related area of profession such as; planning, landscape architecture ,public health and others.
- 45 credit hours in architecture course at the level of 500/600
The answer is 2nd Step because the first step is to define the problem and third is to define your goals