true or false modeling is making predictions about future data points not associated with your equation? Explain why.
1 answer:
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Answer:
vertical load = 10 kN
Modulus of elasticity = 200GPa
Yield stress on the cable = 400 MPa
Safety factor = 2.0
Explanation:
Data
let L =
= 3.35 m
substituting 1.5 m for h and 3 m for the tern (a + b)
= tan⁻¹(
)
= 45⁰
substituting 1.5 for h and 3 m for (a+ b) yields:
₂ = tan⁻¹ (
)
=25.56⁰
checking all the forces, they add up to zero. This means that the system is balanced and there is no resultant force.
Answer:
<em>number of mole is 31342.36 moles</em>
<em>mass is 125.369 kg</em>
<em></em>
Explanation:
Diameter of the spherical balloon d = 9 m
radius r = d/2 = 9/2 = 4.5 m
The volume pf the sphere balloon ca be calculated from
V =
V = = 381.75 m^3
Temperature of the gas T = 20 °C = 20 + 273 = 293 K
Pressure of the helium gas = 200 kPa = 200 x 10^3 Pa
number of moles n = ?
Using
PV = nRT
where
P is the pressure of the gas
V is the volume of the gas
n is the mole number of the gas
R is the gas constant = 8.314 m^3⋅Pa⋅K^−1⋅mol^−1
T is the temperature of the gas (must be converted to kelvin K)
substituting values, we have
200 x 10^3 x 381.75 = n x 8.314 x 293
number of moles n = 76350000/2436 = <em>31342.36 moles</em>
We recall that n = m/MM
or m = n x MM
where
n is the number of moles
m is the mass of the gas
MM is the molar mass of the gas
For helium, the molar mass = 4 g/mol
substituting values, we have
m = 31342.36 x 4
m = 125369.44 g
m =<em> 125.369 kg</em>
Answer:
the minimum component thickness for which the condition of plane strain is valid is 0.005377 m or 5.38 mm
Explanation:
Given the data in the question;
yield strength σ = 690 Mpa
plane strain fracture toughness K = 32 MPa-
minimum component thickness for which the condition of plane strain is valid = ?
Now, for plane strain conditions, the minimum thickness required is expressed as;
t ≥ 2.5( K / σ
)²
so we substitute our values into the formula
t ≥ 2.5( 32 / 690 )²
t ≥ 2.5( 0.0463768 )²
t ≥ 2.5 × 0.0021508
t ≥ 0.005377 m or 5.38 mm
Therefore, the minimum component thickness for which the condition of plane strain is valid is 0.005377 m or 5.38 mm