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Irina18 [472]
4 years ago
5

A rigid bar pendulum is attached to a cart, which moves along the horizontal plane. The rigid bar has a center of mass at L/2. T

he pendulum is pinned to the cart, and a torsional spring is located between the cart and the pendulum. Assume the torsional spring does not cause a dynamic effect on the cart (but does affect the rigid bar). Recall for a pinned attachment we can assume reaction forces in the horizontal and vertical directions. There is a horizontal force F(t) that acts on the cart, which is also attached to a wall on the opposite side through a linear spring. Ignore friction.
a. Identify the number of degrees of freedom, and define a choice of generalized coordinates.
b. Find the equations of motion using Newton's second law.

Engineering
1 answer:
Vikentia [17]4 years ago
4 0

Answer:

See the attached picture for answer.

Explanation:

See the attached picture for explanation.

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vapor fraction = 0.4 and 0.08

Explanation:

At reasonably high temperatures, a mixture will exist in the form of a sub cooled liquid. Between these extremes, the mixture exists in a two phrase region where it is a vapor liquid equilibrium. From a vapor-liquid phase diagram, a mixture of 40% A, 39% B, and 21% C separates to give the vapor compositions of 0.4 and 0.08.

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What should you, the worker, be aware of with regard to evacuation procedures at your workplace
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Answer:

As a worker, it is important to follow the proper set of instructions or emergency plans during an emergent situation. Not carefully following the rules may result to a bigger problem such as further injury and damage to property.

Explanation:

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A metal alloy has been tested in a tensile test with the following results for the flow curve
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4 0
3 years ago
Air fl ows isentropically through a duct. At section 1, the pressure and temperature are 250 kPa and 1258C, and the velocity is
abruzzese [7]

The correct temperature is 125°C

Answer:

A) M_a1 = 0.5

B) T2 = 232.17 K

C) V2 = 611 m/s

D) m' = 187 kg/s

Explanation:

We are given;

Pressure; P1 = 250 kPa

Temperature; T1 = 125°C = 398 K

Speed; v1 = 200 m/s

Area; A2 = 0.25 m²

M_a2 = 2

A) Formula for M_a1 is given by;

M_a1 = v/a1

Where;

v is speed

a1 = √kRT

k is specific heat capacity ratio of air = 1.4

R is a gas constant with a value of R = 287 J/kg·K

T is temperature

Thus;

M_a1 = 200/√(1.4 × 287 × 398)

M_a1 = 200/399.895

M_a1 = 0.5

B) To find T2, let's first find the Stagnation pressure T0

Thus;

T0/T1 = 1 + ((k - 1)/2) × (M_a1)²

T0 = T1(1 + ((k - 1)/2) × (M_a1)²)

T0 = 398(1 + ((1.4 - 1)/2) × (0.5)²)

T0 = 398(1 + (0.2 × 0.5²))

T0 = 398 × 1.05

T0 = 417.9 K

Now,similarly;

T0/T2 = 1 + ((k - 1)/2) × (M_a2)²

T2 = T0/[(1 + ((k - 1)/2) × (M_a2)²)]

T2 = 417.9/(1 + (0.2 × 2²))

T2 = 417.9/1.8

T2 = 232.17 K

C) V2 is gotten from the formula;

T0 = T2 + (V2)²/(2C_p)

Cp of air = 1005 J/Kg.K

Thus;

V2 = √(2C_p)[T0 - T2]

V2 = √((2 × 1005) × (417.9 - 232.17))

V2 =√373317.3

V2 = 611 m/s

D) mass flow is given by the formula;

m' = ρA2•V2

Where;

ρ is Density of air with an average value of 1.225 kg/m³

m' = 1.225 × 0.25 × 611

m' = 187 kg/s

6 0
3 years ago
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