Two streams of air enter a control volume: stream 1 enters at a rate of 0.05 kg / s at 300 kPa and 380 K, while stream 2 enters at 400 kPa and 300 K. Stream 3 leaves the control volume at 150 kPa and 270 K. The control volume does 3 kW of work on the surroundings while losing 5 kW of heat. Find the mass flow rate of stream 2. Neglect changes in kinetic and potential energy.

Answer:

The weight of the combined system is 2001.24 Newtons

Explanation:

From the basic relation between mass, density and volume we know that

[tex]density=\frac{Mass}{Volume}[/tex]

In our context we are given that the density of water is 1000 kg per cubic meters

Thus we can find the mass of 0.2 cubic meters of water using the above relation as

[tex]1000kg/m^{3}=\frac{Mass}{0.2m^{3}}\\\\\therefore Mass=1000kg/m^{3}\times 0.2m^{3}=200kg[/tex]

Hence the mass of water in the tank is 200 kilograms.

The total mass of water and the plastic tank thus becomes

[tex]200+4=204kg[/tex]

Now we know that weight of any given mass is calculated as

[tex]Weight=mass\g[/tex]

where,

'g' is the acceleration due to gravity with value = [tex]9.81m/s^{2}[/tex]

Applying the values in the above equation we get

[tex]Weight=204\times 9.81=2001.24Newtons[/tex]

Answer:

The temperature in degree Celsius will be -238.7055°C

Explanation:

We have given the substance temperature = 62°R

We have to convert degree Rankine to degree Celsius

For conversion from Rankine to Celsius we use formula

[tex]T_C=(T_R-491.67)\times\frac{5}{9}[/tex]

So [tex]T_C=(62-491.67)\times\frac{5}{9}[/tex]

[tex]T_C=-238.7055^{\circ}C[/tex]

So temperature in degree Celsius will be -238.7055°C  

After calculation i got -238.7055°C but in option this is not given

The temperature of the substance will be "94.4°C". To understand the calculation, check below.

According to the question,

Substance temperature, T°R = 62

or,

T°C = (T°R - 491.67) × [tex]\frac{5}{9}[/tex]

By substituting the values,

      = -238.706

If we take the value,

T°C = (662 - 491.67) × [tex]\frac{5}{9}[/tex]

      = 94.62°C or,

      = 94.4°C

Thus the above response "Option D" is correct.

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Answer:

11.125°

Explanation:

Given:

Radius of bend, R = 100 m

Speed around the bend = 50 Km/hr = [tex]\frac{5}{18}\times50[/tex] = 13.89 m/s

Now,

We have the relation

[tex]\tan\theta=\frac{v^2}{gR}[/tex]

where,

θ = angle of banking

g is the acceleration due to gravity

on substituting the respective values, we get

[tex]\tan\theta=\frac{13.89^2}{9.81\times100}[/tex]

or

[tex]\tan\theta=0.1966[/tex]

or

θ = 11.125°

To determine the instantaneous velocity and acceleration of a particle described by the position function s(t), one needs to calculate the first and second derivatives of the position function and evaluate them at the given time, t = 8 s. The average velocity and speed require the change in position over a specific time interval, which is not provided in the question.

The position of a particle along a straight-line path is given by the equation s = (t3 - 6t2 - 15t + 7) feet, where t is time in seconds. To find the particle's instantaneous velocity and instantaneous acceleration at t = 8 s, we need to take the first and second derivatives of the position function with respect to time, respectively.

The first derivative of s with respect to t gives us the velocity v(t), and the second derivative gives us the acceleration a(t). At t = 8 s, the velocities and accelerations can be calculated by plugging in the value of t into those derivatives.

To determine the average velocity and average speed, we take the change in position over the change in time interval for the specific time range provided.

Note: The question lacks sufficient specific information to calculate these values as the time interval for the average velocity and average speed is not provided. However, the general process of calculation has been explained.

Acceleration, instantaneous velocity, and particle position at different times are essential concepts in physics and particle motion analysis.

Acceleration is the rate of change of velocity with respect to time. In the given scenarios, acceleration is provided in terms of a function of time or as a constant value.

Instantaneous velocity is the velocity of the particle at a specific moment. It can be calculated by taking the derivative of the position function with respect to time.

Position of the particle at different times can be found by substituting the respective time values into the position function.

Answer:

The correct answer is 'velocity'of liquid flowing out of an orifice is proportional to the square root of the 'height'  of liquid above the center of the orifice.

Explanation:

Torricelli's theorem states that

[tex]v_{exit}=\sqrt{2gh}[/tex]

where

[tex]v_{exit}[/tex] is the velocity with which the fluid leaves orifice

[tex]h[/tex] is the head under which the flow occurs.

Thus we can compare the given options to arrive at the correct answer

Velocity is proportional to square root of head under which the flow occurs.

Explanation:

Step1

Factor of safety is the number that is taken for the safe design of any component. It is the ratio of failure stress to the maximum allowable stress for the material.

Step2

It is an important parameter for design of any component. This factor of safety is taken according to the environment condition, type of material, strength, type of component etc.

Step3

Different material has different failure stress. So, ductile material fails under shear force. Ductile material’s FOS is based on yield stress as failure stress as after yield point ductile material tends to yield. Brittle material’s FOS is based on ultimate stress as failure stress.

The expression for factor of safety for ductile material is given as follows:

[tex]FOS=\frac{\sigma_{yp}}{\sigma_{a}}[/tex]

Here,[tex]\sigma_{f}[/tex] is yield stress and [tex]\sigma_{a}[/tex] is allowable stress.

The expression for factor of safety for brittle material is given as follows:

[tex]FOS=\frac{\sigma_{ut}}{\sigma_{a}}[/tex]

Here,[tex]\sigma_{ut}[/tex] is ultimate stress and [tex]\sigma_{a}[/tex] is allowable stress.

Answer:

The correct answer is option 'b': 9.9 m/s

Explanation:

We know that for an ideal fluid the velocity of exit from a tank with the height of water 'h' is given by Torricelli's Law as

[tex]v=\sqrt{2gh}[/tex]

where,

'g' is acceleration due to gravity

'h' is the level of water

Applying the given values we obtain velocity as

[tex]v=\sqrt{2\times 9.81\times 5}=9.90m/s[/tex]

Answer:

It will be equivalent to 338.95 N-m

Explanation:

We have to convert 250 lb-ft to N-m

We know that 1 lb = 4.45 N

So foe converting from lb to N we have to multiply with 4.45

So 250 lb = 250×4.45 =125 N

And we know that 1 feet = 0.3048 meter

Now we have to convert 250 lb-ft to N-m

So [tex]250lb-ft=250\times 4.45N\times 0.348M=338.95N-m[/tex]

So 250 lb-ft = 338.95 N-m

Answer:

The mass in:

1) Grams =[tex]377765.5\times 10^{3}grams[/tex]

2) Kilograms =[tex]377765.5kilograms[/tex]

3)Tonnes =[tex]377.7655tonnes[/tex]

4) Megatonnes =[tex]0.378megatonnes[/tex]

Explanation: The given mass of the aircraft in pounds is 833,000 pounds.

Part 1)

Since we know that 1 pound equals 453.5 grams thus by ratio we have

833,000 pounds =[tex]833000lb\times 453.5\frac{g}{lb}=377765.5\times 10^{3}grams[/tex]

Part 2)

Since we know that 1000 grams equals 1 kilogram

thus the above mass in kilograms equals[tex]\frac{377765.5\times 10^{3}}{1000}=377765.5kg[/tex]

Part 3)

Since there are 1000 kilograms in 1 tonne

thus the given mass is converted into tonnes as

[tex]mass_{tonnes}=\frac{377765.5kg}{1000}=377.765tonnes[/tex]

Part 4)

Since 1 Mega tonne(Mton) equals 1000 tonnes thus the given mass is converted into mega tonnes as

[tex]mass_{M\cdot tonn}=\frac{377.7655tonnes}{1000}=0.378Megatonnes[/tex]

Explanation:

In shaping work piece will be stationary and tool will reciprocates,but on the other hand in planning work piece will reciprocates and tool will be stationary.Shaping is used for small work piece and planning is used for large work piece.Both shaping and planning are not continuous cutting process,cutting action take place in forward stroke and return stroke is idle stroke.The velocity of return stroke is much more than the forward stroke.

Answer:

The charpy test is used to determine amount of energy a material absorbs at fracture.

Explanation:

Charpy Impact test was developed by a French scientist to determine the amount of energy a material absorbs at fracture hence giving the toughness of the material. It is widely used in industrial applications since it is easy to perform and does not requires sophisticated equipment to perform.

The test is performed when a swinging pendulum of known weight  is dropped from a known height and is made to strike the metal specimen which is notched.The notch in the sample affects the results of the test hence the notch should be standardized while performing the test. The qualitative results obtained can also be used to compare ductility of different materials.

Answer:

Explanation:

Heat will flow from system B to system A. This is because system B has a higher temperature than system A. Temperature is a measurement od thermodynamic equilibrium. A difference of temperature between two systems is a thermal unbalance, which if they are in contact is compensated by a flow of heat to change the temperatures and reach an equilibrium.

Answer:

Modulus of resilience will be [tex]3216942.308j/m^3[/tex]

Explanation:

We have given yield strength [tex]\sigma _y=818MPa[/tex]

Elastic modulus E = 104 GPa

We have to find the modulus

Modulus of resilience is given by

Modulus of resilience [tex]=\frac{\sigma _y^2}{2E}[/tex], here [tex]\sigma _y[/tex] is yield strength and E is elastic modulus

Modulus of resilience [tex]=\frac{(818\times 10^6)^2}{2\times 104\times 10^9}=3216942.308j/m^3[/tex]  

Answer with Explanation:

i) Boundary layer thickness: It is the thickness of the boundary layer formed around an object that is placed in the path of a flowing viscous fluid.The boundary layer thickness is the thickness up to which the effect of the object on the flow can be felt. When a viscous flowing fluid encounters an object in it's path of flow, the flowing fluid forms a thin layer of fluid over the object and this layer of fluid is known as boundary layer. This is a phenomenon only observed in the viscous fluids. As shown in the below figure a uniform flow of a viscous fluid encounters a plate, as we can see the thickness of the boundary layer goes on increasing as we move away from the leading edge of the plate the thickness of the boundary layer at any position is termed as boundary layer thickness.

ii) Boundary layer transition: It is the transition of the flow from a laminar nature to fully developed turbulent flow as it moves over an object. It occurs due to change in the Reynolds number of the flow as the effect of boundary layer increases as we move away from the leading edge of the object.  

Answer:

Pipes, pressure vessels, tanks, reactors, tubes and nozzles

Explanation:

Thin walled cylinders are typically defines as having wall thickness of 1/10 of the radius (doesn't matter much if inner or outer, they should be similar). Also this is used mostly for things that will be subject to some radial load, as opposed to axles and shafts.

As such, some structures that can be modeled as thin walled cylinders are pipes, pressure vessels, tanks, reactors, tubes and nozzles.

Answer:

1) Dimensions of shear rate is [tex][T^{-1}][/tex] .

2)Dimensions of shear stress are [tex][ML^{-1}T^{-2}][/tex]

Explanation:

Since the dimensions of velocity 'v' are [tex][LT^{-1}][/tex] and the dimensions of distance 'y'  are [tex][L][/tex] , thus the dimensions of [tex]\frac{dv}{dy}[/tex] become

[tex]\frac{[LT^{-1}]}{[L]}=[T^{-1}][/tex] and hence the units become [tex]s^{-1}[/tex].

Now we know that the dimensions of coefficient of dynamic viscosity [tex]\mu [/tex] are [tex][ML^{-1}T^{-1}][/tex] thus the dimensions of shear stress can be obtained from the given formula as

[tex][\tau ]=[ML^{-1}T^{-1}]\times [T^{-1}]\\\\[\tau ]=[ML^{-1}T^{-2}][/tex]

Now we know that dimensions of momentum are [tex][MLT^{-1}][/tex]

The dimensions of [tex]Area\times time[/tex] are [tex][L^{2}T][/tex]

Thus the dimensions of [tex]\frac{Moumentum}{Area\times time}=\frac{MLT^{-1}}{L^{2}T}=[MLT^{-2}][/tex]

Which is same as that of shear stress. Hence proved.

Answer:

Acceleration in circular path will be 9

Explanation:

We have given speed of the particle in circular path = 6 m/sec

Radius of the circular path = 4 m

We have to find the centripetal acceleration [tex]a_c[/tex]

We know that centripetal acceleration is given by [tex]a_c=\frac{v^2}{r}=\frac{6^2}{4}=9[/tex]

As in question it is given that don't include the unit

So acceleration will be 9

Answer:

The heat from the sun melted it

Explanation:

If the street runs east to west, houses on the south (across the street) will project shadows on their sidewalk, while the northern sidewalk will be illuminated. This is for the northern hemisphere, on the southern hemisphere it would be the other way around.

Answer:

eccentrcity of orbit is 0.22

Explanation:

GIVEN DATA:

Initial velocity of satellite = 8333.3 m/s

distance from the sun is 600 km

radius of earth is 6378 km

as satellite is acting parallel to the earth therefore[tex] \theta angle = 0[/tex]

and radial component of given velocity is zero

we have[tex] h = r_o v_r_o = 6378+600 =6.97*10^6 m[/tex]

h = 6.97*10^6 *8333.3 = 58.08*10^9 m^2/s

we know that

[tex]\frac{1}{r} =\frac{GM}{h^2} \times ( i + \epsilon cos\theta)[/tex]

[tex]GM = gr^2 = 9.81*(6.37*10^6)^2 = 398*10^{12} m^3/s[/tex]

so

[tex]\frac{1}{6.97*10^6} =\frac{398*10^{12}}{(58.08*10^9)^2} \times ( i + \epsilon cos0)[/tex]

solvingt for [tex] \epsilon)[/tex]

[tex]\epsilon = 0.22)[/tex]

therefore eccentrcity of orbit is 0.22

Answer:

b)False

Explanation:

When one or more than one monomer  join to other monomer then they form a chain and this joining of monomers is called degree of polymerization .

Degree of polymerization :

[tex]Degree\ of\ polymerization=\dfrac{Mass\ of\ polymer}{Mass\ of\ monomer}[/tex]

So from above we can say that when chain will become longer then the weight of polymer will increase.

Answer:

6.8 kJ

Explanation:

p1 = 95 kPa

T1 = 300 K

T3 = 2100 K

m = 12 g

Ideal gas equation:

p * v = R * T

v = R * T / p

R for air is 0.287 kJ/(kg K)

v1 = 0.287 * 300 / 95 = 0.9 m^3/kg

v2 = v1 / cr

v2 = 0.9 / 18 = 0.05 m^3/kg

Assuming an adiabatic compression

p*v^k = constant

k is 1.4 for air

p1 * v1 ^ k = p2 * v2 ^ k

p2 = p1 * (v1 / v2) ^k

p2 = p1 * cr^k

p2 = 95 * 18^1.4 = 5.43 MPa

p1*v1/T1 = p2*v2/T2

T2 = p2*v2*T1/(p1*v1)

T2 = 5430 * 0.05 * 300 / (95 * 0.9) = 952 K

The first principle of thermodynamics

Q = W + ΔU

Since this is an adiabatic process Q = 0

W = -ΔU

W1-2 = -m * Cv * (T2 - T1)

The Cv of air is 0.72 kJ/kg

W1-2 = -0.012 * 0.72 * (952 - 300) = -5.63 kJ

Next the combustion happens and temperature increases suddenly.

v3 = v2 = 0.05 m^3/kg

T2 * p2^((1-k)/k) = T3 * p3^((1-k)/k)

p3 = p2 * (T2/T3)^(k/(1-k)

p3 = 5430 * (952/2100)^(1.4/(1-1.4) = 86.5 MPa

The work is zero because the piston doesn't move.

Next it expands adiabatically:

v4 = v1 = 0.9 m^3/kg

T * v^(k-1) = constant (adiabatic process)

T3 * v3^(k-1) = T4 * v4^(k-1)

T4 = T3 * (v3 / v4)^(k-1)

T4 = 2100 * (0.05 / 0.9)^(1.4-1) = 661 K

p3*v3/T3 = p4*v4/T4

p4 = p3*v3*T4/(v4*T3)

p4 = 86500*0.05*661/(0.9*2100) = 1512 kPa

L3-4 = -m * Cv * (T4 - T3)

L3-4 = -0.012 * 0.72 * (661 - 2100) = 12.43 kJ

Net work:

L1-2 + L3-4 = -5.63 + 12.43 = 6.8 kJ

Answer:

The correct answer is option 'b': 0.2%

Explanation:

The yield strength of a material is defined as the stress in the material when the material begin's to undergo plastic deformation which is also known as  yielding.

For materials with well defined yield point such as steel of grade Fe-250 the yield strength can be properly identified from the stress strain curve. But for high strength steel such as Fe-415, Fe-500 the yield point is not properly identified hence the yield strength is taken at 0.2% of proof strain.

Answer:

a)31%

b)34MW

Explanation:

A rankine cycle is a generation cycle using water as a working fluid, when heat enters the boiler the water undergoes a series of changes in state and energy until generating power through the turbine.

This cycle is composed of four main components, the boiler, the pump, the turbine and the condenser as shown in the attached image

To solve any problem regarding the rankine cycle, enthalpies in all states must be calculated using the thermodynamic tables and taking into account the following.

• The pressure of state 1 and 4 are equal

• The pressure of state 2 and 3 are equal

• State 1 is superheated steam

• State 2 is in saturation state

• State 3 is saturated liquid at the lowest pressure

• State 4 is equal to state 3 because the work of the pump is negligible.

Once all enthalpies are found, the following equations are used using the first law of thermodynamics

Wout = m (h1-h2)

Qin = m (h1-h4)

Win = m (h4-h3)

Qout = m (h2-h1)

The efficiency is calculated as the power obtained on the heat that enters

Efficiency = Wout / Qin

Efficiency = (h1-h2) / (h1-h4)

For this problem, we will first find the enthalpies in all states

h1=3231kJ/Kg

h2=2310kJ/Kg

h3=h4=272kJ/Kg

A) using the eficiency ecuation

Efficiency = (h1-h2) / (h1-h4)

Efficiency =(3231-2310)/(3231-272)=0.31=31%

b)using ecuation for Wout

Wout = m (h1-h2)

Wout=37(3231-2310)=34077KW=34.077MW

Answer:

a)Patm=135.95Kpa

b)Pabs=175.91Kpa

Explanation:

the absolute pressure is the sum of the water pressure plus the atmospheric pressure, which means that for point a we have the following equation

Pabs=Pw+Patm(1)

Where

Pabs=absolute pressure

Pw=Water pressure

Patm= atmospheric pressure

Water pressure is calculated with the following equation

Pw=γ.h(2)

where

γ=especific weight of water=9.81KN/M^3

H=depht

A)

Solving using ecuations 1 y 2

Patm=Pabs-Pw

Patm=185-9.81*5=135.95Kpa

B)

Solving using ecuations 1 y 2, and atmospheric pressure

Pabs=0.8x5x9.81+135.95=175.91Kpa

Answer:

1) Angle with x-axis = 42.03 degrees

2) Angle with y-axis =68.2 degrees

3) Angle with z-axis =   56.14 degrees

Explanation:

given any vector [tex]\overrightarrow{r}=x\widehat{i}+y\widehat{j}+z\widehat{k}[/tex]

and any x axis the angle between them is given by

[tex]\theta_x =cos^{-1}(\frac{\overrightarrow{r}\cdot\widehat{i}}{\sqrt{x^2+y^2+z^2}} )\\\\\theta_x=cos^{-1}(\frac{x\cdot i}{\sqrt{x^2+y^2+z^2}} )[/tex]

Applying values we get

[tex]\theta_x=cos^{-1}(\frac{160}{\sqrt{160^2+80^2+120^2}} )=42.03^{o}[/tex]

Angle between the vector and y axis is given by

[tex]\theta_y =cos^{-1}(\frac{\overrightarrow{r}\cdot\widehat{j}}{\sqrt{x^2+y^2+z^2}} )\\\\\theta_y=cos^{-1}(\frac{y\cdot j}{\sqrt{x^2+y^2+z^2}} )[/tex]

Applying values we get

[tex]\theta_x=cos^{-1}(\frac{80}{\sqrt{160^2+80^2+120^2}} )=68.2^{o}[/tex]

Similarly angle between z axis and the vector is given by

[tex]\theta_z =cos^{-1}(\frac{\overrightarrow{r}\cdot\widehat{k}}{\sqrt{x^2+y^2+z^2}} )\\\\\theta_x=cos^{-1}(\frac{z\cdot k}{\sqrt{x^2+y^2+z^2}} )[/tex]

Applying values we get

[tex]\theta_z=cos^{-1}(\frac{120}{\sqrt{160^2+80^2+120^2}} )=56.145^{o}[/tex]

The angles made by F will be "42.03°", "68.2°" and "56.14°".

According to the question,

Force, F = 160 i + 80 j + 120 kN

Let any vector,

[tex]\vec r = x \hat i + y \hat j+ z \hat k[/tex]

The angle between x-axis be:

[tex]\Theta_x[/tex] = Cos⁻¹ ([tex]\frac{\vec r. \hat i}{\sqrt{x^2 +y^2 + z^2} }[/tex])

    = Cos⁻¹ ([tex]\frac{x.i}{\sqrt{x^2+y^2+z^2} }[/tex])

By substituting the values,

    = Cos⁻¹ ([tex]\frac{160}{\sqrt{160^2+80^2+120^2} }[/tex])

    = 42.03°

and,

The angle between y-axis be:

[tex]\Theta_y[/tex] = Cos⁻¹ ([tex]\frac{\vec r. \hat j}{\sqrt{x^2 +y^2 + z^2} }[/tex])

    = Cos⁻¹ ([tex]\frac{y.i}{\sqrt{x^2+y^2+z^2} }[/tex])

By substituting the values,

    = Cos⁻¹ ([tex]\frac{80}{\sqrt{160^2+80^2+120^2} }[/tex])

    = 68.2°

and,

The angle between z-axis be:

[tex]\Theta_z[/tex] = Cos⁻¹ ([tex]\frac{\vec r. \hat k}{\sqrt{x^2 +y^2 + z^2} }[/tex])

    = Cos⁻¹ ([tex]\frac{z.k}{\sqrt{x^2+y^2+z^2} }[/tex])

By substituting the values,

    = Cos⁻¹ ([tex]\frac{120}{\sqrt{160^2+80^2+120^2} }[/tex])

    = 56.145°

Thus the above approach is correct.

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Answer:

a) 19094 N

b) 110.055 kPa

c) 1222 J

Explanation:

The force on the gas is the weight plus the atmospheric pressure multiplied by the piston area

F = P + p * A

F = m * g + p * π/4 * d^2

F = 150 * 9.813 + 101570 * π/4 * 0.47^2 = 19094 N

The pressure is the force divided by the area of the piston

p1 = F / A

p1 = F / (π/4 * d^2)

p1 = 19094 / (π/4 * 0.47^2) = 110055 Pa = 110.055 kPa

variation of gravitational potential energy is defined as

ΔEp = m * g * Δh

ΔEp = 150 * 9.813 * 0.83 = 1222 J

In this exercise we have to use the knowledge of force to calculate the required energies, so we have to:

a) 19094 N

b) 110.055 kPa

c) 1222 J

In the field of physics, force is a physical action that causes deformation or that changes the state of rest or movement of a given object.

a) Knowing that the force formula is defined by:

[tex]F = P + p * A\\F = m * g + p *\pi /4 * d^2\\F = 150 * 9.813 + 101570 * \pi /4 * 0.47^2 = 19094 N[/tex]

b) Knowing that the force exerted by an area is equal to the pressure in that area, we have:

[tex]p_1 = F / A\\p_1 = F / (\pi /4 * d^2)\\p_1 = 19094 / (\pi /4 * 0.47^2) = 110055 Pa = 110.055 kPa[/tex]

c)So calculating the potential energy we have:

[tex]\Delta E_p = m * g * \Delta h\\\Delta E_p = 150 * 9.813 * 0.83 = 1222 J[/tex]

See more about force at brainly.com/question/26115859

The total kinetic energy of the system is approximately 17128.26 J.T

What is the energy?

Calculate the moment of inertia (I) of the pulley:

= 0.5 * mass * (outer radius² + inner radius²)

= [tex]0.5 * 12.0 kg * ((0.250 m)² + (0.200 m)²)[/tex]

= 1.925 kg * m²

Use the parallel-axis theorem to find I:

I = [tex]1.925 kg * m² + 12.0 kg * (0.231 m)²[/tex]

I = 2.5683 kg * m²

Calculate the kinetic energy of the pulley:

= 0.5 * I * omega²

= [tex]0.5 * 2.5683 kg * m² * (69.0 rad/s)[/tex]

= 6555.63 J

Calculate the linear velocity of cylinder A:

v = outer radius * omega

v =[tex]0.250 m * 69.0 rad/s[/tex]

v = 17.25 m/s

Calculate the kinetic energy of cylinder A:

= 0.5 * mass * v²

= [tex]0.5 * 71.0 kg * (17.25 m/s)²[/tex]

= 10572.63 J

KEtotal = KEpulley + KEcylinder

=  [tex]6555.63 J + 10572.63 J[/tex]

= 17128.26 J

Answer:

f=1.59 Hz

Explanation:

Given that

Simple undamped system means ,system does not consists any damper.If system consists damper then it is damped spring mass system.

Velocity = 100 mm/s

Maximum amplitude = 10 mm

We know that for a simple undamped system spring mass system

[tex]V_{max}=\omega A[/tex]

now by putting the values

[tex]V_{max}=\omega A[/tex]

100 = ω x 10

ω = 10 rad/s

We also know that

ω=2π f

10 = 2 x π x f

f=1.59 Hz

So the natural frequency will be f=1.59 Hz.

Answer:

You can't divide by zero

Explanation:

The error appears when you divide each side by (a - b). If a = b, then (a - b) = 0 and you can't divide each side by 0. Moreover, the equation before division, that is, (a + b)(a − b) = b(a − b) is true after replacing (a - b) = 0  because it gives 0 = 0.

The error in the proof lies in the step where we divided both sides by (a - b) . Since [tex]\( a = b \), \( (a - b) \)[/tex] becomes 0, making the division by 0 undefined.

The error occurs when dividing both sides by (a - b) . Since ( a = b ), ( (a - b)  becomes 0. Division by 0 is undefined, leading to the invalid conclusion.

To elaborate, dividing both sides by (a - b)  in the equation (a + b)(a - b) = b(a - b) results in:

[tex]\[ \frac{(a + b)(a - b)}{(a - b)} = \frac{b(a - b)}{(a - b)} \]\[ \frac{(a + b)\cancel{(a - b)}}{\cancel{(a - b)}} = \frac{b\cancel{(a - b)}}{\cancel{(a - b)}} \]\[ a + b = b \][/tex]

This step assumes  (a - b)  is not equal to zero, leading to the false conclusion that ( a + b = b ), which is only true when  a  and ( b ) are equal.

Therefore, the proof incorrectly concludes that ( 2 = 1 ) due to the division by zero, highlighting the importance of avoiding such mathematical errors.

Answer:

Dimension of specific heat will be [tex]=L^2T^{-2}\Theta ^{-1}[/tex]

Explanation:

We know that heat [tex]Q=mc\Delta T[/tex], Q is heat generated, m is mass, c is specific heat and [tex]\Delta T[/tex] is temperature difference

From formula we can write [tex]c=\frac{Q}{m\times \Delta T}[/tex]

Now unit of Q is joule or N-m

Newton can be written as [tex]kgm/sec^2[/tex]

So unit of Q is [tex]kgm^2/sec^2[/tex]

For dimension we use M for kg, L for meter(m) ,T for sec and [tex]\Theta[/tex] for temperature

So dimension of Q is [tex]ML^2T^{-2}[/tex]

So dimension of specific heat will be [tex]\frac{ML^2T^{-2}}{M\Theta }=L^2T^{-2}\Theta ^{-1}[/tex]