A disk with radius of 0.4 m is rotating about a centrally located axis with an angular acceleration of 0.3 times the angular position theta. The disk starts with an angular velocity of 1 rad/s when theta = 0. Determine the magnitudes of the normal and tangential components of acceleration of a point P on the rim of the disk when theta has completed one revolution.

Answer:

[tex]V = 0.208 m^3[/tex]

saturated pressure  = 0.91 bar 0r 91 kPa

Explanation:

Given data:

Mass of liquid water 200 kg

Steel tank temperature is 97 degree celcius = 97°C

At T = 97 degree celcius,  saturated pressure  = 0.91 bar 0r 91 kPa

And also for saturated liquid

Specific volume [tex]\nu[/tex] is 0.00104 m^3/kg

Volume of tank is given as

[tex]V = \nu * m[/tex]

[tex]V = 0.00104 * 200[/tex]

[tex]V = 0.208 m^3[/tex]

Answer and Explanation:

SPECIFIC HEAT :

Answer:

The mass of car in slug =108.5 slug.

The mass of car in kilogram =1575 Kg.

Explanation:

Given that

Mass of car = 3500 lbs

We know that

1 lbs =0.031 slug

So

3500 lbs = 0.031 x 3500 slug

So the mass of car in slug =108.5 slug.

We know that

1 lbs =0.45 kilograms

so

3500 lbs = 3500 x 0.45 Kg

So the mass of car in kilogram =1575 Kg.

Answer:

Open system

Explanation:

In refrigerator there is a interaction between refrigerator and environment.

In refrigerator heat moves from the system to the outer environment means there is transfer of heat from one system to environment.

We know that whenever there is transfer of energy or mass from system then it is known as pen system

So refrigerator is a open system

Answer:

The power required to cool the water is 1.11Kw.

Hence the correct option is (b).

Explanation:

Power needed to cool down is equal to heat extract from the water.

Given:

Volume of water is 1 liter.

Initial temperature is 100C.

Final temperature is 20C.

Time is 5 minutes.

Take density of water as 100 kg/m3.

Specific heat of water is 4.186 kj/kgK.

Calculation:

Step1

Mass of the water is calculated as follows:

[tex]\rho=\frac{m}{V}[/tex]

[tex]1000=\frac{m}{(1l)(\frac{1m^{3}}{1000l})}[/tex]

m=1kg

Step2

Amount of heat extraction is calculated as follows:

[tex]Q=mc\bigtriangleup T[/tex]

[tex]Q=1(4.186kj/kgk)(\frac{1000 j/kgk}{1 kj/kgk})\times(100-20)[/tex]

Q=334880 j.

Step3

Power to cool the water is calculated as follows:

[tex]P=\frac{Q}{t}[/tex]

[tex]P=\frac{334880}{(5min)(\frac{60s}{1min})}[/tex]

P=1116.26W

or

[tex]P=(1116.26W)(\frac{1Kw}{1000 W})[/tex]

P=1.11 Kw.

Thus, the power required to cool the water is 1.11Kw.

Hence the correct option is (b).

Answer:

9%

Explanation:

An ideal gas is one that has its molecules widely dispersed and does not interact with each other, studies have shown that air behaves like an ideal gas, so the state change equation for ideal gases can be applied.

P1V1T2 = P2V2T1

where 1 corresponds to state 1 = 320kPa

and 2 is state 2 = 349kPa.

Given that the volume remains constant the equation is:

P1T2=P2T1

SOLVING for T2/T1

[tex]\frac{T2}{T1} =\frac{P2}{P1} =\frac{349}{320} =1.09\\\\[/tex]

The equation to calculate the percentage increase is as follows

%ΔT=[tex]\frac{(T2-T1)100}{T1} =(\frac{T2}{T1} -1)100=(1.09-1)100=(0.09)100=9%[/tex]

Explanation:

A ductile material can convert into brittle material due to following reasons

1.At very low temperature

2.Due to presence of notch

In titanic ,the base of ship strike to the large ice cube and lower part of titanic ship material was made of steel .We know that steel is ductile material and when steel came with very low temperature of ice due to this ductile material converted in to brittle material and titanic ship failed.Brittle material does not show any indication before failure.

Answer: 139 Km.

Explanation:

The question tells us that a planet A has an orbit period of 10 years and its circular orbit has a radius of 106 Km, whilst a planet B has an orbit period of 15 years (also assuming a circular orbit), both orbiting a nearby star.

This information allow us to use the Kepler's 3rd law, for the special case in which the orbit is circular.

Kepler's 3rd law, tells that there exist a direct proportionality between the square of the orbit period, and the cube of the orbit radius (in the more general case, with the cube of the semi-major axis of the elipse), for celestial bodies orbiting a same star.

(like Earth and Mars orbiting Sun).

So, for planet A and planet B (orbiting a same star), we can write the following:

(TA)²/ (TB)² = (rA)³ / (rB)³

Replacing by TA= 10 years, TB= 15 years, rA= 106 Km, and solving for Rb, we get  RB= 139 Km.

Answer:

The acceleration is [tex]2220.00m/s^{2}[/tex]

Explanation:

We know that the acceleration is given by

[tex]a=v\frac{dv}{ds}........................(i)[/tex]

The velocity as a function of position is given by [tex]v=14\cdot s^{7/6}[/tex]

Thus the acceleration as obtained from equation 'i' becomes

[tex]a=14\cdot s^{7/6}\times \frac{d}{ds}(14\cdot s^{7/6})\\\\a=14\cdot s^{7/6}\times \frac{7}{6}\times 14\cdot s^{1/6}\\\\a=\frac{686}{3}\cdot s^{8/6}[/tex]

Hence acceleration at s = 5.5 equals

[tex]a(5.5)=\frac{686}{3}\times (5.5)^{8/6}\\\\\therefore a=2220.00mm/s^{2}[/tex]

Answer:

304.42 cc

Explanation:

When the aluminum expands the markings will be further apart. If the 300 cc mark was at a distance l0 of the origin at 20 C, at 100 C it will be

l = l0 * (1 + a * (t - t0))

l = l0 * (1 + 23*10^-6 * (100 - 20))

l = l0 * 1.0018

The volume of water would have expanded by

V = V0 (1 + a * (t - t0))

V = 300 (1 + 2.07*10^-4 * (100 - 20))

V = 304.968 cc

Since the markings expanded they would measure

304.968/1.0018 = 304.42 cc

Answer:

Network is a collection of computers

Explanation:

A network is a gathering of at least two gadgets that can convey. Practically speaking, a system is included various distinctive PC frameworks associated by physical or potentially remote associations.  

The scale can run from a solitary PC sharing out fundamental peripherals to huge server farms situated far and wide, to the Internet itself. Despite extension, all systems permit PCs or potentially people to share data and assets.

Answer:

Explanation:

Convection needs a fluid to transport the heat, while radiation doesn't.

Generally convection transports a lot more heat than radiation.

Since fluids tend to expand, when in a gravitational field such as that of Earth convection tends to move heat upwards while radiation is indifferent to gravity.

Answer:

Specific change in internal energy is - 2.025 kj/kg.

Explanation:

The process is constant pressure expansion. Apply first law of thermodynamic to calculate the change in internal energy.

Given:

Mass of gas is 4 kg.

Initial volume is 0.005 m³.

Final volume is 0.006 m³.

Pressure is 12 Mpa.

Heat is transfer to the gas. So it must be positive 3.9 kj.

Calculation:

Step1

Work of expansion is calculated as follows:

[tex]W=P(V_{f}-V_{i})[/tex]

[tex]W=12\times10^{6}(0.006-0.005)[/tex]

W=12000 j.

Or,

W=12 Kj.

Step2

Apply first Law of thermodynamic as follows:

Q=W+dU

3.9=12+dU

dU = - 8.1 kj.

Step3

Specific change in internal energy is calculated as follows:

[tex]u=\frac{U}{m}[/tex]

[tex]u=\frac{-8.1}{4}[/tex]

u= - 2.025 kj/kg.

Thus, the specific change in internal energy is - 2.025 kj/kg.

The elongation of the steel bar under the given tensile force is approximately 2.10 millimeters.

The cross-sectional area (A) of the steel bar can be calculated using the formula for the area of a circle:

A = π * r²

Now, let's calculate A:

A = π * (0.01 m)²

≈ π * (0.0001 m²)

≈ 0.000314 m²

Now, we can calculate the elongation (ΔL) using Hooke's Law:

ΔL = (F * L) / (A * E)

Given:

F = 33,000 kg

L = 0.40 m

E = 2 * 10¹⁰ kg/m²

Now, plug in the values:

ΔL = (33,000 kg * 0.40 m) / (0.000314 m² * 2 * 10₈⁰ kg/m²)

Now, perform the calculation

ΔL ≈ (13,200 kg*m) / (6.28 * 10 kg/m²)

ΔL ≈ 2.10 * 10⁻₀ meters

Answer:

a) Power =50 KJ/min

b)Rate of heat transfer = 110 KJ/min

Explanation:

Given that

COP = 1.2

Heat removed from space Q = 60 KJ/min

As we know that COP of refrigerator is the ratio of heat removed to work input.

Lets take power consume by refrigerator is W

So

COP= Q/W

1.2=60/W

W=50 KJ/min

So the power consume is 50 KJ/min.

From first law of thermodynamic

Heat removed from the kitchen = 50 + 60 KJ/min

Heat removed from the kitchen =110 KJ/min

Answer:

52.2538 psia

Explanation:

The absolute pressure at depth of 27 inches can be calculated by:

Pressure = Local pressure + Gauge pressure

Also,

[tex]P_{gauge}=\rho\times g\times h[/tex]

Where,

[tex]\rho[/tex] is the density of glycerin ([tex]\rho=74.9\ lbm/ft^3[/tex])

g is the gravitational acceleration = 32.1741 ft/s²

h = 27 in

Also, 1 in = 1/12 ft

So,

h = 27 / 12 ft = 2.25 ft

So,

[tex]P_{gauge}=74.9\times 32.1741\times 2.25\ lbf/ft^2=5422.1402\ lbf/ft^2[/tex]

Also,

1 ft = 12 inch

1 ft² = 144 in²

So,

[tex]P_{gauge}=5422.1402\ lbf/ft^2=\frac {5422.1402\ lbf}{144\ in^2}=37.6538\ lbf/in^2=37.6538\ psia[/tex]

Local pressure = 14.6 psia

So,

Absolute pressure = 14.6 psia + 37.6538 psia=52.2538 psia

Answer:

Part B will be 37.2 m to the east of the launch point at an altitude of 80.8 m

Explanation:

At t0 the rocket is at 60 m and has a speed of 28 m/s.

Then it explodes when it reaches max altitude.

Since the rocket will have its engine off after t0 it will be in free fall.

It will be affected only by the acceleration of gravity, so it moves with constant acceleration, we can use this equation.

Y(t) = Y0 + V0*t + 1/2*a*t^2

Y0 = 60 m

V0 - 28 m/s

a = g = -9.81

We also have the equation for speed:

V(t) = V0 + a*t

And we know that when it reaches its highest point it ill have a speed of zero.

0 = V0 + a * t

a * t = -V0

t = -V0 / a

t = -28 / -9.81 = 2.85 s

This is the time after t0 when the engine ran of of fuel.

Using this value on the position equation:

Y(2.85) = 60 + 28*2.85 + 1/2*(-9.81)*2.85^2 = 100 m

At an altitude of 100 m it explodes in two parts. An explosion is like a plastic collision in reverse, momentum is conserved.

Since the speed of the rocket is zero at that point, the total momentum will be zero too.

Part A, with a mass of 1 kg fell 74.4 m west of the launch point. It fell in a free fall with a certain initial speed given to it by the explosion. This initial speed had vertical and horizontal components.

First the horizontal. In the horizontal there is no acceleration (ignoring aerodynamic drag).

X(t) = X1 + Vx1 * (t - t1)

X1 = 0 because it was right aboce the launch point. t1 is the moment of the explosion.

t1 = 2.85 s

We know it fell to the ground at t = 5.85 s

Vx1 * (t2 - t1) = X(t2) - X1

Vx1 = (X(t2) - X1) / (t2 - t1)

Vx1 = (74.4 - 0) / (5.85 - 2.85) = 24.8 m/s

Now the vertical speed

Y(t2) = Y1 + Vy1*(t2 - t1) + 1/2*a*(t2 - t1)^2

Vy1*(t2 - t1) = Y(t2) - 1/2*a*(t2 - t1)^2 - Y1

Y(t2) = 0 because it is on the ground.

Vy1 = (-1/2*a*(t2 - t1)^2 - Y1) / (t2 - t1)

Vy1 = (-1/2*-9.81*(5.85 - 2.85)^2 - 100) / (5.85 - 2.85) = -18.6 m/s

If the part A had a speed of (24.8*i - 18.6*j) m/s, we calcultate the speed of part 2

Horizontal:

vxA * mA + vxB * mB = 0

-vxA * mA = vxB * mB

vxB = -vxA * mA / mB

vxB = -24.8 * 1 / 2 = -12.4 m/s

Vertical:

vyA * mA + vyB * mB = 0

-vyA * mA = vyB * mB

vyB = -vyA * mA / mB

vyB = -(-16.6) * 1 / 2 = 8.3 m/s

Now that we know its speed and position at t1 we can know ehre it will be at t2

X(t) = 0 - 12.4 * (t - t1)

X(5.85) = -12.4 * (5.85 - 2.85) = -37.2 m

Y(t) = 100 + 8.3 * (t - t1) + 1/2 * (-9.8) * (t - t1)^2

Y(t) = 100 + 8.3 * (5.85 - 2.85) + 4.9 * (5.85 - 2.85)^2 = 80.8 m

Part B will be 37.2 m to the east of the launch point at an altitude of 80.8 m

Answer:

(a)27.12 N-m (b) 69.96 [tex]KN/m^3[/tex] (c) 1.27 mm/sec

Explanation:

We have to convert

(a) 20 lb-ft to N-m

We know that 1 lb = 4.45 N

So 20 lb = 20×4.45 = 89 N

1 feet = 0.3048 m

So [tex]20lb-ft=20\times 4.45N\times 0.3048N=27.12N-m[/tex]

(b) 450 [tex]lb/ft^3[/tex] to KN/[tex]m^3[/tex]

We know that 1 lb = 4.45 N = 0.0044 KN

[tex]1ft^3=0.0283m^3[/tex]

So [tex]450lb/ft^3=\frac{450\times 0.0044KN}{0.0283m^3}=69.96KN/m^3[/tex]

(c) 15 ft/hr to mm/sec

We know that 1 feet = 0.3048 m = 304.8 mm

And 1 hour = 60×60=3600 sec

So [tex]15ft/h=\frac{15\times 304.8mm}{3600sec}=1.27 mm/sec[/tex]

Answer:

a)20 lb.ft=27.2 N.m

b)[tex]450\dfrac{lb}{ft^3}=70.65\dfrac{KN}{m^3}[/tex]

c)15 ft/h = 1.26 mm/s

Explanation:

a)

As we know that

1 ft.lb= 1.36 N.m

So 20 lb.ft = 1.36 x 20 N.m

20 lb.ft=27.2 N.m

b)

We know that

[tex]1\dfrac{lb}{ft^3}=0.157\dfrac{KN}{m^3}[/tex]

So

[tex]450\dfrac{lb}{ft^3}=450\times 0.157\dfrac{KN}{m^3}[/tex]

[tex]450\dfrac{lb}{ft^3}=70.65\dfrac{KN}{m^3}[/tex]

c)

As we know that

1 ft/h=0.084 mm/s

So

15 ft/h = 0.084 x 15 mm/s

15 ft/h = 1.26 mm/s

Answer:

The nail exerts a force of 573.88 Pounds on the Hammer in positive j direction.

Explanation:

Since we know that the force is the rate at which the momentum of an object changes.

Mathematically [tex]\overrightarrow{F}=\frac{\Delta \overrightarrow{p}}{\Delta t}[/tex]

The momentum of any body is defines as [tex]\overrightarrow{p}=mass\times \overrightarrow{v}[/tex]

In the above problem we see that the moumentum of the hammer is reduced to zero in 0.023 seconds thus the force on the hammer is calculated using the above relations as

[tex]\overrightarrow{F}=\frac{m(\overrightarrow{v_{f}}-\overrightarrow{v_{i}})}{\Delta t}[/tex]

[tex]\overrightarrow{F}=\frac{m(0-(-73.33)}{0.23}=\frac{1.8\times 73.33}{0.23}=573.88Pounds[/tex]

Answer:

1000000 watts

Explanation:

1 kJ/sec= 1000 watts (or 1 watt= 0.001 kJ/sec)

Therefore,

1000 kJ/sec = 1000 x 1000 watts

                    = 1000000 watts

A kJ/sec (Kilojoule per second) is a unit used to measure power. It comes from the SI (Standard International) unit J/sec (Joule per second).

1 J/sec= 1 watt

Power is the energy transferred by a force per unit of time.

Energy is measured in Joules.

In this question, we are working out how much energy is transferred in watts.

Answer:

a)38.65%

b)1221KJ/kg

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)

first we calculate the enthalpies in all states

h1=3264kJ/Kg

h2=2043kJ/Kg

h2=h3=105.4kJ/Kg

a)we use the ecuation for efficiency

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

Efficiency = (3264-2043) / (3264-105.4)

=0.3865=38.65%

b)we use the ecuation for Wout

Wout = m (h1-h2)

for work ratio=

w = (h1-h2)

w=(3264-2043)=1221KJ/kg

Answer:

Fouling :

  When rust or undesired material deposit in the surface of heat exchanger,is called fouling of heat exchanger.

Effect of heat exchanger are as follows"

1.It decreases the heat transfer.

2.It increases the thermal resistance.

3.It decreases the overall heat transfer coefficient.

4.It leads to increase in pressure drop.

5.It increases the possibility of corrosion.

Answer:

Cut off ratio=2.38

Explanation:

Given that

[tex]T_1=300K[/tex]

[tex]P_1=100KPa[/tex]

[tex]P_2=P_3=7200KPa[/tex]

[tex]T_3=2250K[/tex]

Lets take [tex]T_1[/tex] is the temperature at the end of compression process

For air γ=1.4

[tex]\dfrac{T_2}{T_1}=\left(\dfrac{P_2}{P_1}\right)^{\dfrac{\gamma -1}{\gamma}}[/tex]

[tex]\dfrac{T_2}{300}=\left(\dfrac{7200}{100}\right)^{\dfrac{1.4-1}{1.4}}[/tex]

[tex]T_2=1070K[/tex]

At constant pressure

[tex]\dfrac{T_3}{T_2}=\dfrac{V_3}{V_2}[/tex]

[tex]\dfrac{2550}{1070}=\dfrac{V_3}{V_2}[/tex]

[tex]\dfrac{V_3}{V_2}=2.83[/tex]

So cut off ratio

[tex]cut\ off\ ratio =\dfrac{V_3}{V_2}[/tex]

Cut off ratio=2.38

Answer:

[tex]F = \frac{m * v^2}{r}[/tex]

Explanation:

Once the centrifugal forces are equivalent as an object rotates, it is called "in equilibrium." If the point of impact of the moving object does not align with the center of the rotation, unequal centrifugal forces are produced and the rotating component is "out of equilibrium."  

Two conditions for balanced condition are: static and coupled. Static equilibrium relates to a one plane mass in which the unit is balanced by  weight inverse the unbalanced mass

formula used to calculate centrifugal mass is [tex]F = \frac{m * v^2}{r}[/tex]

where, m is mass, v is speed of body. r is radius

Answer:

convection because it make the surrounding air warm. hence make you feel warm without getting physically connect to it.

Explanation:

convection because it make the surrounding air warm. hence make you feel warm without getting physically connect to it.

Convection is the transition of heat between distinct temperature fields by moving fluid (liquid or gas). Dry air is less thick than moist air and in the presence of a temperature difference, convection currents can form.

Answer:

Explanation:

Given

Force=9 kN

Diameter reduces from 7 mm to 5 mm

As volume remains constant therefore

[tex]A_0L_0=A_fL_f[/tex]

[tex]7^2\times L_0=5^2\times L_f[/tex]

[tex]\frac{L_0}{L_f}=\frac{25}{49}[/tex]

Thus Engineering Strain [tex]\epsilon _E=\frac{L_f-L_0}{L_0}[/tex]

[tex]=\frac{49}{25}-1=\frac{49-25}{25}=0.96[/tex]

Engineering stress[tex]=\frac{Load}{original\ Cross-section}[/tex]

[tex]\sigma _E=\frac{9\times 10^3}{\frac{\pi 7^2}{4}}=233.83 MPa[/tex]

(b)True stress

[tex]\sigma _T=\sigma _E\left ( 1+\epsilon _E\right )[/tex]

[tex]\sigma _T=233.83\times (1+0.96)=458.30 MPa[/tex]

True strain

[tex]\epsilon _T=ln\left ( 1+\epsilon _E\right )[/tex]

[tex]\epsilon _T=ln\left ( 1.96\right )=0.672[/tex]

Answer:

Horizontal acceleration will be [tex]2.07m/sec^2[/tex]

Explanation:

We have given radius of track = 180 m

velocity = 30 m/sec

Total acceleration [tex]a_t=7.07m/sec^2[/tex]

The centripetal;l acceleration which is always normal to the velocity also known as normal acceleration is given by [tex]a_n=\frac{v^2}{r}=\frac{30^2}{180}=5m/sec^2[/tex]

So the tangent acceleration will be [tex]a_t=7.07-5=2.07m/sec^2[/tex]

Answer:

The mass flow rate of jet =69.44 kg/s

Explanation:

Given that

velocity of jet v= 60 m/s

Power  P=250 KW

As we know that force offered by water F

[tex]F=\rho\ A \ v^2[/tex]

Power P= F.v

So now power given as

[tex]P=\rho\ A \ v^3[/tex]

We know that mass flow rate = ρAv

[tex]P=mass\ flow\ rate\ \times v^2[/tex]

250 x 1000 = mass flow rate x 3600

mass flow rate = 69.44 kg/s

So the mass flow rate of jet =69.44 kg/s

Answer:n=0.973

Explanation:

Given

When True strain[tex]\left ( \epsilon _T_1\right )=0.171[/tex]

at [tex]\sigma _1=263.8 MPa[/tex]

When True stress[tex]\left ( \sigma _2\right )[/tex]=346.2 MPa

true strain [tex]\left ( \epsilon _T_2\right )[/tex]=0.226

We know

[tex]\sigma =k\epsilon ^n [/tex]

where [tex]\sigma [/tex]=True stress

[tex]\epsilon [/tex]=true strain

n=strain hardening exponent

k=constant

Substituting value

[tex]263.8=k\left ( 0.171\right )^n------1[/tex]

[tex]346.2=k\left ( 0.226\right )^n-----2[/tex]

Divide 1 & 2 to get

[tex]\frac{346.2}{263.8}=\left ( \frac{0.226}{0.171}\right )^n[/tex]

[tex]1.312=\left ( 1.3216\right )^n[/tex]

Taking Log both side

[tex]ln\left ( 1.312\right )=nln\left ( 1.3216\right )[/tex]

n=0.973

Answer:

133.276 Btu/lbm

Explanation:

1 kJ/kg = 0.429923 Btu/lbm

Therefore, 310 kJ/kg = 133.276 Btu/lbm

Specific internal energy can be defined as the internal energy (kJ or Btu) divided by a unit of mass (kg or lbm). Both sets of units (kJ/kg and Btu/lbm) are units for specific energy.

Energy is an object's ability to do work and is represented in this example in Joules (J) or British Thermal Units (Btu).

1 kJ = 1000 J

Mass tells us how much something weighs and is represented in this example in kg or lbm (pounds).