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3 years ago

Explain the concept of degrees of freedom in terms of physics and mechanics.

Engineering

1 answer:

Sophie [7]3 years ago

6 0

Explanation:

Step1

Degree of freedom is the number of independent variable which is required to define motion of the system. Degree of freedom is the number of possible motion of the system. Motion can be sliding or rotate. So, there are maximum 6 possible motions of the system, 3 of rotation and 3 of translation. There are 6 maximum possible degrees of freedoms.

Step2

Take an example of slider that is allowed to slide only. So, the slider will have one degree of freedom. Take an example of cylinder which can rotate as well as slide. So, it will have 2 degree of freedom.

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olchik [2.2K]

ave you ever seen a Rube Goldberg machine in action? You probably have, even if you didn’t know what it was. A Rube Goldberg machine is a contraption that uses a chain reaction to carry out a simple task. It performs a very basic job in a complicated way.

7 0

3 years ago

One method that is used to grow nanowires (nanotubes with solid cores) is to initially deposit a small droplet of a liquid catal

7nadin3 [17]

Answer: maximum length of the nanowire is 510 nm

Explanation:

From the table of 'Thermo physical properties of selected nonmetallic solids at At T = 1500 K

Thermal conductivity of silicon carbide k = 30 W/m.K

Diameter of silicon carbide nanowire, D = 15 x 10⁻⁹ m

lets consider the equation for the value of m

m = ( (hP/kAc)^1/2 ) = ( (4h/kD)^1/2 )

m = ( ((4 × 10⁵)/(30×15×10⁻⁹ ))^1/2 ) = 942809.04

now lets find the value of h/mk

h/mk = 10⁵ / ( 942809.04 × 30) = 0.00353

lets consider the value θ/θb by using the equation

θ/θb = (T - T∞) / (T - T∞)

θ/θb = (3000 - 8000) / (2400 - 8000)

= 0.893

the temperature distribution at steady-state is expressed as;

θ/θb = [ cosh m(L - x) + ( h/mk) sinh m (L - x)] / [cosh mL+ (h/mk) sinh mL]

θ/θb = [ cosh m(L - L) + ( h/mk) sinh m (L - L)] / [cosh mL+ (h/mk) sinh mL]

θ/θb = [ 1 ] / [cosh mL+ (h/mk) sinh mL]

so we substitute

0.893 = [ 1 ] / [cosh (942809.04 × L) + (0.00353) sinh (942809.04 × L)]

L = 510 × 10⁻⁹m

L = 510 nm

therefore maximum length of the nanowire is 510 nm

4 0

3 years ago

How would you describe the stability of the atmosphere if you noted a dry adiabatic rate of 10ºC/1000 meters, a wet adiabatic ra

muminat

Answer:

conditional instability (Γd > Γe > Γw)

Explanation:

Given;

dry adiabatic rate, Γd = 10ºC/1000 meters

wet adiabatic rate, Γw= 6.5ºC/1000 meters

environmental lapse rate, Γe = 7.8ºC/1000 meters

Stability of the atmosphere can be described as Absolute stability, Absolute instability or conditional instability.

Conditions for Absolute stability:

Γd > Γw > Γe

Conditions for Absolute instability:

Γe > Γd > Γw

Conditions for conditional instability:

Γd > Γe > Γw

Thus, conditional instability satisfies the given values of the atmospheric condition: Γd (10) > Γe (7.8) > Γw (6.5)

3 0

3 years ago

Read 2 more answers

1. The equilibrium number of vacancies in Ni at 1123 K is 4.7x1022m-3. The atomic weight and density of Ni at 1123 K are 58.69 g

Ahat [919]

Answer:

vacancy formation energy of Ni is 1.400 eV

Explanation:

given data

number of vacancies in Ni = 4.7 x $10^{22}$ $m^{-3}$

atomic weight = 58.69 g/mol

density = 8.8 g/cm³

solution

we get here N that is

N = $\frac{N_A \times \rho}{A}$ ...........1

N = $\frac{6.023\times 10^{23} \times 8.8 \times 10^6}{58.69}$

N = $9.030 \times 10^{28}$

and here no of vacancy will be

Nv = $N \times e^{\frac{-Qv}{kT}}$ .................2

put here value

$4.7 \times 10^{22} = 9.030 \times 10^28 \times e^{\frac{-Qv}{8.62\times 10^{-5}\times 1123}}$

$10^{-7} \times 5.20487 = e^{\frac{-Qv}{0.0968}}$

take ln both side

$ln (10^{-7} \times 5.20487 ) = ln (e^{\frac{-Qv}{0.0968}})$

$-14.468 = \frac{-Qv}{0.0968}$

Qv = 1.400 eV

so vacancy formation energy of Ni is 1.400 eV

3 0

3 years ago

A 3.0-m wide rectangular asphalt channel discharges 33.84 m^3/s of water with a depth of 2.0 m. What is the head loss in 100 m l

BaLLatris [955]

Answer:

Head loss in 100 m length equals 1.00 m.

Explanation:

The head loss in an open channel is calculated using manning's equation as follows

$Q=\frac{1}{n}\frac{A^{5/3}}{P^{2/3}}S_{f}^{1/2}$

For a asphalt rectangular channel we have

Area of flow = $3.0\times 2.0 = 6m^{2}$

Wetted Perimeter = $3.0+2\times 2.0=7.0m$

manning's roughness coefficient = 0.016

Applying values in the above equation we get

$33.84=\frac{1}{0.016}\times \frac{6^{5/3}}{7^{2/3}}S_{f}^{1/2}\\\\\therefore S^{1/2}_{f}=0.1\\\\\therefore S_{f}=0.01$

Now we know that

$S_{f}=\frac{H_{L}}{L}\\\\\therefore H_{l}=0.01\times 100m=1.00m$

8 0

3 years ago