What are factor of safety for brittle and ductile material

Answer:

Point to point control motion system:

  In point to point control motion system tool perform specific task at a particular location.Point to point control motion system is also called positioning system.It perform intermittent operation.

Ex:  Drilling operation is a point to point motion control system.

Continuous path control system:

 Continuous path control system is continuous operation that perform by tool.The program used in  continuous path control system is more complex than point to point motion control system.

Ex :Milling operation is a  continuous path control system.

To avoid cavitation for water at 160 °F flowing through a sharp bend, the minimum absolute pressure should be slightly above the vapor pressure of water at this temperature, which is approximately 0.363 psi.

When water flows through a sharp bend, cavitation can occur if the local pressure falls to or below the fluid's vapor pressure. To estimate the minimum absolute pressure without causing cavitation for water at 160 °F, we must consider water's vapor pressure at this temperature. At 160 °F (about 71 °C), the vapor pressure of water is approximately 0.363 psi. Since fluids cannot have a negative absolute pressure, and to avoid cavitation, the absolute pressure must stay above this vapor pressure. Therefore, considering atmospheric pressure to be approximately 14.7 psi, to avoid cavitation, the minimum absolute pressure in the system should be slightly above 0.363 psi to ensure no cavitation occurs.

Answer:

461 C

862 F

Explanation:

The specific gas constant for CO2 is

R = 189 J/(kg*K)

Using the gas state equation:

p * v = R * T

T = p * v / R

v = 1/δ

T = p  / (R * δ)

T = 9.3*10^6  / (189 * 67) = 734 K

734 - 273 = 461 C

461 C = 862 F

Answer:

1) Acceleration of player is 7.8125 [tex]m/s^{2}[/tex].

2) Constant speed of player is 12.5 m/s.

Explanation:

Let the acceleration of the player by 'a' and let he complete the initial 10 meters in [tex]t_1[/tex] time

The initial 10 meters are case of uniformly accelerated motion and hence we can relate the above quantities using second equation of kinematics as

[tex]s=ut+\frac{1}{2}at^{2}\\\\[/tex]

now since the player starts from rest hence u = 0 thus the equation can be written as

[tex]10=\frac{1}{2}at_1^2\\\\at_{1}^{2}=20...........(i)[/tex]

The speed the player reaches after [tex]t_{1}[/tex] time is obtained using first equation of kinematics as

[tex]v=u+at\\v=at_{1}[/tex]

Since the total distance traveled by the player is 40 meters hence the total time of trip is 4 seconds hence we infer that he covers 30 meters of distance at a constant speed in time of [tex]4-t_{1}[/tex] seconds

Hence we can write

[tex](4-t_{1})\cdot at_{1}=30.......(ii)\\\\[/tex]

Solving equation i and ii we get

from equation 'i' we obtain [tex]a=\frac{10}{t_{1}^{2}}[/tex]

Using this in equation 'ii' we get

[tex](4-t_{1})\cdot \frac{20}{t_{1}^{2}}\cdot t_{1}=30\\\\30t_{1}=80-20t_{1}\\\\\therefore t_{1}=\frac{80}{50}=1.6seconds\\\\\therefore a=\frac{20}{1.6^2}=7.8125m/s^{2}[/tex]

Thus constant speed equals [tex]v=7.8125\times 1.6=12.5m/s[/tex]

Answer:

a) 152000 slugs

b) 2220000 kg or 2220 metric tons

Explanation:

A body with a weight of 4.9*10^6 lbf has a mass of

4.9*10^6 lbm * 1 lbf/lbm = 4.9*10^6 lbm

This mass value can then be converted to other mass values.

1 slug is 32.17 lbm

Therefore:

4.9*10^6 lbm * 1 slug / (32.17 lbm) = 152000 slugs

1 lb is 0.453 kg

Therefore:

4.9*10^6 lbm / (1/0.453) * kg/lbm = 2220000 kg

Answer:

Please, see the attachment.

Explanation:

First, we have to create two input boxes that allows the user to write the current year in one of them and his/her birth year in the another one. Also, we have to create a label that will show the result of the desired variable. We can write a message "Your age is:" and it will be attached to the result.

For the algorithm, let's call the variables as follows:

CY = Current Year

BY = Birth Year

X = Age of user

When the user inserts the current year and his/her birth year, the program will do the following operation:

X = CY - BY; this operation will give us the age of the user

After this the user will see something like "Your age is:" X.

Answer:

Q=1312.5 W

Explanation:

Given that

Cooling load or power input = 750 W

COP=1.75

We know that COP can be given as

COP is the ratio of cooling effect to the input power .

Lets take cooling effect is Q.So now by using COP formula

COP=Q/750

1.75=Q/750

Q=1312.5 W

So the cooling effect produce by air conditioning will be 1312.5 W.

The net effect on laboratory,the laboratory temperature will reduce.

Answer:d

Explanation:

Given

Temperature[tex]=200^{\circ}\approc 473 K[/tex]

Also [tex]\gamma for air=1.4[/tex]

R=287 J/kg

Flow will be In-compressible when Mach no.<0.32

Mach no.[tex]=\frac{V}{\sqrt{\gamma RT}}[/tex]

(a)[tex]1000 km/h\approx 277.78 m/s[/tex]

Mach no.[tex]=\frac{277.78}{\sqrt{1.4\times 287\times 473}}[/tex]

Mach no.=0.63

(b)[tex]500 km/h\approx 138.89 m/s[/tex]

Mach no.[tex]=\frac{138.89}{\sqrt{1.4\times 287\times 473}}[/tex]

Mach no.=0.31

(c)[tex]2000 km/h\approx 555.55 m/s[/tex]

Mach no.[tex]=\frac{555.55}{\sqrt{1.4\times 287\times 473}}[/tex]

Mach no.=1.27

(d)[tex]200 km/h\approx 55.55 m/s[/tex]

Mach no.[tex]=\frac{55.55}{\sqrt{1.4\times 287\times 473}}[/tex]

Mach no.=0.127

From above results it is clear that for Flow at velocity 200 km/h ,it will be incompressible.

Answer:

31.1 slug,  453.4 Kg

Explanation:

given,

mass of the beam is 1000 lb

to convert mass of beam into slugs and kilograms.

1 lb is equal to 0.0311 slug

1000 lb = 1000 × 0.0311

             = 31.1 slug

now, conversion of lb into kg

1 lb is equal to 0.4534 kg

so,

1000 lb = 1000 × 0.4534

             = 453.4 Kg

hence, 1000 lb of beam in slugs is equal to 31.1 slugs and in kilo gram is 453.4 Kg.

Answer:43.70 MPa

Explanation:

Given

mass of engine [tex] 700 lb \approx 317.515 kg[/tex]

diameter of cable [tex]0.375 in.\approx 9.525 mm[/tex]

[tex]A=\frac{\pi d^2}{4}=71.26 mm^2[/tex]

we know stress([tex]\sigma [/tex])[tex]=\frac{load\ applied}{area\ of\ cross-section}[/tex]

[tex]\sigma =\frac{317.515\times 9.81}{71.26\times 10^{-6}}=43.70 MPa[/tex]

Answer with Explanation:

By the equation or Torque we have

[tex]\frac{T}{I_{p}}=\frac{\tau }{r}=\frac{G\theta }{L}[/tex]

where

T is the torque applied on the shaft

[tex]I_{p}[/tex] is the polar moment of inertia of the shaft

[tex]\tau [/tex] is the shear stress developed at a distance 'r' from the center of the shaft

[tex]\theta [/tex] is the angle of twist of the shaft

'G' is the modulus of rigidity of the shaft

We know that for solid shaft [tex]I_{p}=\frac{\pi R^4}{2}[/tex]

For a hollow shaft [tex]I_{p}=\frac{\pi (R_o^4-R_i^4)}{2}[/tex]

Since the two shafts are subjected to same torque from the relation of Torque we have

1) For solid shaft

[tex]\frac{2T}{\pi R^4}\times r=\tau _{solid} [/tex]

2) For hollow shaft we have

[tex]\tau _{hollow}=\frac{2T}{\pi (R^4-0.7R^4)}\times r=\frac{2T}{\pi 0.76R^4} [/tex]

Comparing the above 2 relations we see

[tex]\frac{\tau _{solid}}{\tau _{hollow}}=0.76[/tex]

Similarly for angle of twist we can see

[tex]\frac{\theta _{solid}}{\theta _{hollow}}=\frac{\frac{LT}{I_{solid}}}{\frac{LT}{I_{hollow}}}=\frac{I_{hollow}}{I_{solid}}=1.316[/tex]

Part b)

Strength of solid shaft = [tex]\tau _{max}=\frac{T\times R}{I_{solid}}[/tex]

Weight of solid shaft =[tex]\rho \times \pi R^2\times L[/tex]

Strength per unit weight of solid shaft = [tex]\frac{\tau _{max}}{W}=\frac{T\times R}{I_{solid}}\times \frac{1}{\rho \times \pi R^2\times L}=\frac{2T}{\rho \pi ^2R^5L}[/tex]

Strength of hollow shaft = [tex]\tau '_{max}=\frac{T\times R}{I_{hollow}}[/tex]

Weight of hollow shaft =[tex]\rho \times \pi (R^2-0.7R^2)\times L[/tex]

Strength per unit weight of hollow shaft = [tex]\frac{\tau _{max}}{W}=\frac{T\times R}{I_{hollow}}\times \frac{1}{\rho \times \pi (R^2-0.7^2)\times L}=\frac{5.16T}{\rho \pi ^2R^5L}[/tex]

Thus [tex]\frac{Strength/Weight _{hollow}}{Strength/Weight _{Solid}}=5.16[/tex]

Answer:

36 kg

Explanation:

For water to boil at 105 C it needs a pressure of 121 kPa (this is the vapor pressure of water at 105 C).

In the lid there will be a difference of pressure from one side to the other, this will be compensated by the weight of the lid.

Δp = pwater - patm

Δp = 121 - 101 = 20 kPa

The pressure caused by the weigh of the lid is:

Δp = w / A

Δp = m * g / A

Rearranging

m = Δp * A / g

m = Δp * π/4 * d^2 / g

m = 20000 * π/4 * 0.15^2 / 9.81 = 36 kg

Answer:

Plastic deformation, irreversible or permanent. Deformation mode in which the material does not return to its original shape after removing the applied load. This happens because, in plastic deformation, the material undergoes irreversible thermodynamic changes by acquiring greater elastic potential energy.

Elastic deformation, reversible or non-permanent. the body regains its original shape by removing the force that causes the deformation. In this type of deformation, the solid, by varying its tension state and increasing its internal energy in the form of elastic potential energy, only goes through reversible thermodynamic changes.

Answer:

origin

Explanation:

The point of intersection of axis is called origin.

In 2D origin is the intersection point of x-axis and y-axis if we go right to the origin then it is positive x axis, if we go left side of origin then it is negative x- axis

Similarly when we go above the origin then it positive y axis , and if we go bellow the origin then it is negative x axis

In 3D origin is the intersection of x-axis, y-axis and z-axis

NOTE- For defining i take here x axis as horizontal axis and y-axis as vertical axis

Answer:

wr = 6.32 rad/s

Explanation:

m = 750 kg

k = 30 kN/m

This system has no dampening, therefore the resonance frequency will simply be the natural frequency of the system.

[tex]wr = w0 = \sqrt{\frac{k}{m}}[/tex]

[tex]wr = \sqrt{\frac{30000}{750}} = 6.32 rad/s[/tex]

In this case the force applied doesn't matter. because we are calculating the resonance frequency.

Answer:

1113kN

Explanation:

The ouside diameter OD of the pipe is 61cm and the thickness T is 0.9cm, so the inside diameter ID will be:

Inside Diameter = Outside Diameter - Thickness

Inside Diameter = 61cm - 0.9cm = 60.1cm

Converting this diameter to meters, we have:

[tex]60.1cm*\frac{1m}{100cm}=0.601m[/tex]

This inside diameter is useful to calculate the volume V of water inside the pipe, that is the volume of a cylinder:

[tex]V_{water}=\pi  r^{2}h[/tex]

[tex]V_{water}=\pi (\frac{0.601m}{2})^{2}*120m[/tex]

[tex]V_{water}=113.28m^{3}[/tex]

The problem gives you the water density d as 1.0kg/L, but we need to convert it to proper units, so:

[tex]d_{water}=1.0\frac{Kg}{L}*\frac{1L}{1000cm^{3}}*(\frac{100cm}{1m})^{3}[/tex]

[tex]d_{water}=1000\frac{Kg}{m^{3}}[/tex]

Now, water density is given by the equation [tex]d=\frac{m}{V}[/tex], where m is the water mass and V is the water volume. Solving the equation for water mass and replacing the values we have:

[tex]m_{water}=d_{water}.V_{water}[/tex]

[tex]m_{water}=1000\frac{Kg}{mx^{3}}*113.28m^{3}[/tex]

[tex]m_{water}=113280Kg[/tex]

With the water mass we can find the weight of water:

[tex]w_{water}=m_{water} *g[/tex]

[tex]w_{water}=113280kg*9.8\frac{m}{s^{2}}[/tex]

[tex]w_{water}=1110144N[/tex]

Finally we find the total weight add up the weight of the water and the weight of the pipe,so:

[tex]w_{total}=w_{water}+w_{pipe}[/tex]

[tex]w_{total}=1110144N+2500N[/tex]

[tex]w_{total}=1112644N[/tex]

Converting this total weight to kN, we have:

[tex]1112644N*\frac{0.001kN}{1N}=1113kN[/tex]

Answer:

Infinite slow thermodynamic process is Quasi-equilibrium process. All the thermodynamic model or equation is based on Quasi-equilibrium process. So, Quasi-equilibrium process is very important process in engineering.

Explanation:

Step1

Quasi-equilibrium process is the thermodynamic process which is infinitely slow. All the thermodynamic variables or properties are taken as uniform. Pressure or temperature is uniform throughout the process. Quasi-equilibrium process is represented by complete joint line in thermodynamics not with the dash line. So, this process has infinite equilibrium points near to each other.  

Step2

All the thermodynamic analysis or equations are based on Quasi-equilibrium process. This gives estimation of heat, work, enthalpy, entropy etc. Quasi-equilibrium process gives maximum power output in power producing devices like turbine or engine. The entire thermodynamic engineering model is designed on the basis of Quasi-equilibrium process. Thus, this process is very important in terms of engineering.  

Answer:

6243.6 lbs

Explanation:

We have given weight = 2838 kg

We have to convert this weight into lbs

Both kg and lbs are unit of measuring the weight of the body so they are changeable

We know that 1 lbs = 2.20 kg

So for converting kg into lbs we have to multiply with 2.20

So 2838 kg = 2838×2.2 = 6243.6 pound

So weight of 2838 kg will be equivalent to 6243.6 lbs

Answer:

Hybrid topology is the connection of one or more than one topology.

Explanation:

Topology:

 Topology is the arrangement of network.These network connects by line and nodes.

Type of topology:

1.Bus topology

2.Star topology

3.Ring topology

4.Mesh topology

Along with given above topology one topology is also used is known as hybrid topology.Hybrid topology is the connection of one or more than two one above given topology.

Answer:

The diameter of the test cylinder should be 7.65 meters.

Explanation:

The Hooke's law relation between stress and strain is mathematically represented as

[tex]Stress=E\times strain\\\\\sigma =e\times \epsilon[/tex]

Where 'E' is modulus of elasticity of the material

Now by definition of strain we have

[tex]\epsilon =\frac{\Delta L}{L_{o}}[/tex]

Applying values to obtain strain we get

[tex]\epsilon =\frac{0.5}{380}=0.001316[/tex]

Thus the stress developed in the material equals

[tex]\sigma = 110\times 10^{3}\times 0.001316=144.76N/m^{2}[/tex]

Now by definition of stress we have

[tex]\sigma =\frac{Force}{Area}\\\\\therefore Area=\frac{Force}{\sigma }\\\\\frac{\pi D^{2}}{4}=\frac{6660N}{144.76}=46m^{2}[/tex]

Solving for 'D' we get

[tex]D=\sqrt{\frac{4\times 46}{\pi }}=7.653meters[/tex]

Answer:

The gravitational force between the masses is [tex]1.0\times 10^{-8}Newtons[/tex]

Explanation:

For 2 masses 'm' and 'M' separated by a distance 'd' the gravitational force between them is given by Newton as

[tex]F=G\cdot \frac{mM}{d^{2}}[/tex]

where

'G' is universal gravitational constant whose value is [tex]6.67\times 10^{-11}m^3kg^{-1}s^{-2}[/tex]

Applying the values in the above relation we get

[tex]F=6.67\times 10^{-11}\times \frac{8\times 12}{(800\times 10^{-3})^{2}}=1.0\times 10^{-8}Newtons[/tex]

Weight of 8 kg mass =[tex]8\times 9.81=78.45Newtons[/tex]

Weight of 12 kg mass =[tex]12\times 9.81=117.72Newtons[/tex]

thus we see that gravitational force between the masses is completely negligible as compared to the weight of the masses.

Answer:

The force between the charges when the separation decreases to 0.7 meters equals 126.955 Newtons

Explanation:

We know that for two point charges of magnitude [tex]q_{1},q_{2}[/tex] the magnitude of force between them is given by

[tex]F=\frac{k_{e}q_{1}\cdot q_{2}}{r^{2}}[/tex]

where

[tex]k_{e}[/tex] is constant

[tex]r[/tex] is the separation between the charges

Initially when the charges are separated by 2.4 meters the force can be calculated as

[tex]F_{1}=\frac{k_{e}\cdot q_{1}q_{2}}{2.4^{2}}\\\\10.8=\frac{k_{e}\cdot q_{1}q_{2}}{2.4^{2}}\\\\\therefore k_{e}\cdot q_{1}q_{2}=10.8\times 2.4^{2}=62.208[/tex]

Now when the separation is reduced to 0.7 meters the force is similarly calculated as

[tex]F_{2}=\frac{k_{e}\cdot q_{1}q_{2}}{0.7^{2}}[/tex]

Applying value of the constant we get

[tex]F_{1}=\frac{62.208}{0.7^{2}}[/tex]

Thus [tex]F_{2}=126.955Newtons[/tex]

Answer:

 inside dimension  [tex]= 142 mm \times 42 mm[/tex]

cross section area [tex]= 7.5\times 10^{-3} m^2[/tex]

mass of 1.2 meter log steel  [tex] = 1.843\times 10^{-3} \rho[/tex]

Explanation:

given data:

Outside dimension of steel rectangular [tex]= 150 mm\times 50mm[/tex]

Thickness = 4 mm

Long = 1.22 m

inside dimension will be [tex]= (150- 8)mm \times ( 50-8)mm[/tex]

                                         [tex]= 142 mm \times 42 mm[/tex]

cross section area [tex]= 150\times 50 mm^2[/tex]

                               [tex]= 7500\times 10^{-6} m^2[/tex]

                               [tex]= 7.5\times 10^{-3} m^2[/tex]

let the density be assumed as \rho

mass of 1.2 meter log steel will be [tex]= 1.2  \times (7.5\times 10^{-3} -  0.142\times 0.048)\times \rho[/tex]

                                                       [tex] = 1.843\times 10^{-3} \rho[/tex]

Answer:

-273.16 °C

-459.677 °F

0 °K

0 °R

Explanation:

The lowest temperature is the absolute zero.

Absolute zero is at 0 degrees Kelvin, or 0 degrees Rankine, because these are absolute scales that have their zero precisely at the absolute zero.

Celsius and Fahrenheit degrees are relative scales, these have their zeroes above the absolute zero.

Celsius scale has the same degree separation as the Kelvin scale, but the zero is separated by 273.16 degrees. Therefore the lowest temperature in the Celsius scale is -273.16 °C.

The Fahrenheit degrees have the same degree separation as the Rankine degrees, and the zero is 459.67 degrees. Therefore the lowest temperature in the Fahrenheit scale is -459.67 °F.

Answer:

9.7g / cm^3

Explanation:

To calculate a conbined density we must find the ratio  between the sum of the masses and the sum of the volumes remembering that the equation to find the density is α=m/v, taking into account the above the following equation is inferred.

αc=copper density

αs=silver density

Vs=volume of silver

Vc=volume of copper

α= density of alloy

[tex]\alpha =\frac{({\alpha c}{Vc} +{\alpha s }{Vs} )}{Vs+Vc} \\\alpha =\frac{(8.9)(10) +(10.5)(10) }{10+10} \\\\\alpha =9.7g / cm^3[/tex]

the density of the alloy is 9.7g / cm^3

Explanation:

Balancing:

  Generally balancing are of two types

1.Static balancing:In this only force balancing is done.

2.Dynamic balancing:in this force as well as moment balancing is done.

Balancing become compulsory for over hanging rotor because unbalance force produce lots of vibration and lots of sound due to this rotor or the whole system in which rotor is attached can be damage.

Answer:

The maximum length of the specimen is 0.2927 m or 292.7 mm

Solution:

Modulus of elasticity, E = 126 GPa = [tex]126\times 10^{9}[/tex]

Diameter of the cross-section, D = 4.0 mm = [tex]4.0\times 10^{- 3} m[/tex]

Force due to tension, F = 2380 N

Maximum elongation, [tex]\Delta L = 0.44 mm = 0.44\times 10^{- 3} m[/tex]

Now,

The maximum length of the specimen, [tex]L_{m}[/tex] can be calculated as follows:

The cross-sectional area, [tex]A_{c} = \frac{\pi D^{2}}{4} = \frac{\pi\times (4.0\times 10^{- 3})^{2}}{4} = 1.256\times 10^{- 5} m^{2}[/tex]

Now, the stress on the specimen, [tex]\sigma_{s} = \frac{F}{A_{c}} = \frac{2380}{1.256+\times 10^{- 5}}[/tex]

[tex]\sigma_{s} = 1.89\times 10^{8} N/m^{2}[/tex]

Now,

The strain on the specimen, [tex]\epsilon_{s}[/tex]:

[tex]\epsilon_{s} = \frac{\Delta L}{L_{m}}[/tex]

Also, from Hooke's law:

[tex]E = \frac{\sigma_{s}}{epsilon_{s}}[/tex]

⇒ [tex]E = \frac{1.89\times 10^{8}}{\frac{\Delta L}{L_{m}}}[/tex]

⇒ [tex]L_{m} = \frac{\Delta Ltimes E}{1.89\times 10^{8}}[/tex]

⇒ [tex]L_{m} = \frac{0.44\times 10^{- 3}\times 126\times 10^{9}}{1.89\times 10^{8}} = 0.2927 m[/tex]

The maximum length of the specimen before deformation is:           292.72 mm (0.2927 m).

For solving this question, it's necessary to know some concepts about the material's properties.

The tensile stress (σ) is determined from the ratio between load and original area before the load applied (σ=[tex]\frac{F}{Ao}[/tex]). Depending on the load applied, the material can have an elastic deformation (temporary deformation) and plastic deformation (permanent deformation). Both deformations can be calculated by the equation: ε=ΔL/Lo, where ΔL= deformation elongation and Lo= the original length before the load applied.

When the material is in the elastic portion, there is a linear relationship between stress and strain given by: σ=Eε. Due to this relationship, it is possible to find the elastic deformation (ε) when we know the stress (σ) and elastic modulus (E).

Now you have the necessary information to solve your question.

The question gives:

E (elastic modulus) =126 GPA

d (original cross-sectional diameter)=4 mm

F (tensile load)=2380 N

ΔL (maximum allowable elongation) =0.44 mm

       1. Find the area of the cylindrical specimen.

       2. Find the tensile stress.

          σ= [tex]\frac{F}{Ao} =\frac{2380 N }{12.57 mm^2} =189.39 MPa[/tex]

      3. Calculate the maximum length of the specimen before deformation.

       Knowing that  ε=ΔL/Lo and σ=Eε, you can rewrite these equations as:

                                      σ= E * (ΔL/Lo)

                                      σ= (E * ΔL)/Lo

   The question asks the maximum length of the specimen before deformation, therefore you should find Lo. Thus,

                                     Lo= (E * ΔL)/σ

                  [tex]Lo=\frac{126*10^3 MPa*0.44 mm}{189.39 MPa} =292.72 mm= 0.2927 m[/tex]

Read more about the tensile stress here:

https://brainly.com/question/19756298

Answer:

22.90 × 10⁸ kg

Explanation:

Given:

Diameter, d = 0.02 m

ωₙ = 0.95 rad/sec

Time period, T = 0.35 sec

Now, we know

T= [tex]2\pi\sqrt{\frac{L}{g}}[/tex]

where, L is the length of the steel cable

g is the acceleration due to gravity

0.35= [tex]2\pi\sqrt{\frac{L}{9.81}}[/tex]

or

L = 0.0304 m

Now,

The stiffness, K is given as:

K = [tex]\frac{\textup{AE}}{\textup{L}}[/tex]

Where, A is the area

E is the elastic modulus of the steel = 2 × 10¹¹ N/m²

or

K = [tex]\frac{\frac{\pi}{4}d^2\times2\times10^11}{0.0304}[/tex]

or

K = 20.66 × 10⁸ N

Also,

Natural frequency, ωₙ = [tex]\sqrt{\frac{K}{m}}[/tex]

or

mass, m = [tex]\sqrt{\frac{K}{\omega_n^2}}[/tex]

or

mass, m = [tex]\sqrt{\frac{20.66\times10^8}{0.95^2}}[/tex]

mass, m = 22.90 × 10⁸ kg

Answer:

Wind

Hydro electric

Explanation:

Energy are of two types

1.Renewable energy

 These have unlimited source of energy.

Ex: Solar energy,wind energy,geothermal energy,hydro power ,biomass etc

2.Non renewable energy

These have limited source of energy.

Ex:  Petroleum,Coal

Wind is the fastest renewable source of energy.This energy is produce by using the natural velocity of air.

Hydro electric power plant is the mostly used renewable source of energy to produce electricity.

Answer with Explanation:

The growth of renewable sources of energy depend on various factor's and their grown is also a regional dependent process. Many different countries use different renewable sources of energy and the growth of the renewable sources of energy depend on the location of the place. As an example in India the solar energy is the most widely growing source of energy due to the location of the place. while as in European union Bio energy is the source of renewable energy that has shown the most growth.

Water is the renewable source which is  used to produce the most electricity in the world accounting for 16.3% of the total global electricity production.

Answer:

4.186 × 10⁷ J

Explanation:

Heat gain by water = Q

Thus,    

[tex]m_{water}\times C_{water}\times \Delta T=Q[/tex]

For water:  

Volume = 1 m³ = 1000 L ( as 1 m³ = 1000 L)

Density of water= 1 kg/L

So, mass of the water:  

[tex]Mass\ of\ water=Density \times {Volume\ of\ water}[/tex]  

[tex]Mass\ of\ water=1 kg/L \times {1000\ L}[/tex]  

Mass of water  = 1000 kg

Specific heat of water = 4.186 kJ/kg K

ΔT = 10 K

So,

[tex]1000\times 4.186\times 10=Q[/tex]  

Q = 41860 kJ

Also, 1 kJ = 1000 J

So, Q = 4.186 × 10⁷ J