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MAVERICK [17]
3 years ago
11

A rubber wheel on a steel rim spins freely on a horizontal axle that is suspended by a fixed pivot at point P. When the wheel sp

ins at a rate of 4.00 rev/s, it precesses smoothly about point P in a horizontal plane with a period of 3.50 s. The wheel's outer radius is 15.0 cm, and its total mass is measured to be 1.12 kg, 60% of this being the spinning wheel, and 40% its axle assembly. The wheel and its axle are both symmetric about the middle of the wheel, and the pivot point P is 4.70 cm from the wheel's middle.
a. its axle considered as a body.
b. What is the direction of the net torque on the wheel about the pivot point P?
c. If the wheel is spinning counterclockwise (0) as seen from point B on the wheel's axle, what is the direction of its spin angular momentum referenced to the diagram?
d. As seen from point A above the pivot, which way is the wheel precessing (C or O)? Please give your reasoning.
e. Using this information, determine the moment of inertia I of the wheel about its axle. Please be sure your work is clear.
Engineering
1 answer:
Nimfa-mama [501]3 years ago
3 0

Answer:

stan lee

Explanation:

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When comparing solids to fluids, the following is true: for elastic solids, the stress must be normal. For Newtonian fluids, the
Nat2105 [25]

Answer: D

Find the answer in the explanation

Explanation:

Elastic solid will obey Hooke's law which state that the force applied is proportional to the extension provided the elastic limit is not exceeded.

Examples of Newtonian fluid are water, glycerol, honey and all organic solvents

When comparing solids to fluids, the below statement is true

We can therefore conclude that

For elastic solids, stress is linearly related to strain, and for Newtonian fluids, stress is linearly related to strain rate

4 0
4 years ago
An AC power generator produces 50 A (rms) at 3600 V. The voltage is stepped up to 100 000 V by an ideal transformer and the ener
RSB [31]

Given:

I_{rms} = 50 A

voltage, V = 3600V

step-up voltage, V' = 100000 V

Resistance of line, R = 100\ohm

Solution:

To calculate % heat loss in long distance power line:

Power produced by AC generator, P = 50\times 3600 W

P = 180000 W = 180 kW

At step-up voltage, V = 100000V or 100 kV

current, I = \frac{P}{V'}

I = \frac{1800000}{100000}

I = 1.8 A

Power line voltage drop is given by:

V_{drop} = I\times R

V_{drop} = 1.8\times 100

V_{drop} = 180 V

Power dissipated in long transmission line P_{dissipated} = V_{drop}\times I

Power dissipated in long transmission line P_{dissipated} = 180\times 1.8 = 324 W

% Heat loss in power line, P_{loss} = \frac{P_{dissipated}}{P}\times 100

% Heat loss in power line, P_{loss} = \frac{324}{180000}\times 100

P_{loss} = 0.18%

 

5 0
3 years ago
Water flows at a rate of 0.011 m3/s in a horizontal pipe whose diameter increases from 6 to 11 cm by an enlargement section. If
lakkis [162]

Answer:

Pressure change in pipe is 766.96 N/m²

Explanation:

The continuity equation is stated below,

AV = Q

A1V1 = A2V2

Where A is the cross-sectional area, V is the velocity and Q is the fluid flow rate.

To calculate the inlet velocity of the pipe,

V1 = Q/A1

V1 = Q/(π x d1²)

d1 is the inlet diameter of the pipe

Substituting values,

V1 = 0.011/(π x 1/4 x 0.06²)

V1 = 3.89 ms-¹

To determine the outlet velocity,

V2 = Q/A2

d2 is the outlet diameter of pipe

V2 = 0.011/(π x 1/4 x 0.11²)

V2 = 1.157 ms-²

Applying Bernoulli's equation for steady flow between the points,

P1/pg + a1(V1²/2g) + z1 + hp = P2/pg + a2(V2²/2g) + z2 + ht + hL

Collecting like terms,

The kinetic energy correction factor, a = a1 = a2

a((V1² - V2²)/2g) - hL = (P2 - P1)/pg

apg((V1² - V2²)/2g) - hLpg = P2 - P1

ap((V1² - V2²)/2) - hLpg = ∆P

p - density of water, g is the acceleration due to gravity and hL is the head loss due to friction in pipe.

Substituting values,

a = 1.05, p = 1000kg/m³, g = 9.81m/s², hL = 0.66m

∆P = (1000)(1.05)((3.89² - 1.157²)/2) - (0.66 x 1000 x 9.81)

∆P = 7241.56 - 6474.6

∆P = 766.96 N/m²

4 0
3 years ago
Name the main classes of polymer and define their characteristic properties
Svetlanka [38]

Answer:

Polymers are the naturally occurring or synthetic macromolecules that are composed of repeating subunits, called monomers.

The three main classes of polymers are: thermoplastic, thermosetting, and the elastomers.

Thermoplastic polymers have linear bonding. These polymers can be melted again and thus can recycled.

Thermosetting polymers have cross-linked bonding. These polymers decompose when heated and thus can not be remelted and recycled.

Elastomers have linear bonding with some cross-linking. These polymers extreme elastic extensibility and thus can revert back to its original shape after deformation, without causing any permanent damage.

8 0
4 years ago
Please read and answer each question carefully.
Klio2033 [76]

the answer is (c)

After the vehicle is involved in a car accident or fire

5 0
4 years ago
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