Explain the two advantages and the two disadvantages of fission as an energy source.

Answer:

Maximum temperature of the cycle is 2231.3 K

Explanation:

See table (values there do not assume constant specific heat) and figure attached.

Assuming ideal gas behaviour, p1*v1 = p2*v2, rearranging p2/p1 = v1/v2

Data

[tex]p_1 = 1 bar [/tex]

[tex]T_1 = 300 K [/tex]

[tex] \frac{v_1}{v_2} = 8.5 [/tex] (compression ratio)

[tex] \frac{Q_{23}}{m} = 1400 kJ/kg [/tex]  (heat addition)

We can use the following relationship  for air

[tex] \frac{v_1}{v_2} = \frac{v_{r1}}{v_{r2}} [/tex]

[tex] v_{r1} [/tex] is only function of temperature and can be taken from table. In this case:

[tex] v_{r1} = 621.2 [/tex]  

Rearranging previous equation

[tex] v_{r2} = v_{r1} \times \frac{v_2}{v_1} [/tex]

[tex] v_{r2} = 621.2 \times \frac{1}{8}[/tex]

[tex] v_{r2} = 73.082 [/tex]

Interpolating from table

[tex] u_2 = 503.06 kJ/kg [/tex]

Energy balance in the process 2-3 gives

[tex] \frac{Q_{23}}{m} = u_3 - u_2 [/tex]  

[tex] u_3 = \frac{Q_{23}}{m} + u_2 [/tex]  

[tex] u_3 = 1400 kJ/kg + 503.06 kJ/kg [/tex]  

[tex] u_3 = 1903.06 kJ/kg [/tex]  

Interpolating from table

[tex] T_3 = 2231.3 K [/tex]

Answer:

False

Explanation:

In electric heater electric energy is converted into heat energy. In heater wires are present which have resistance and current is flow in heater when we connect the heater to supply.

And we know that whenever current is flow in any resistance then heat is produced so in electric heaters electric energy is converted into heat energy

So this is a false statement

Answer:

The micrometer with vernier graduation can measure reading to the nearest 0.0001 inches.  So, the statement is true.

Explanation:

Micrometer is the measuring device that used to measure length with more accuracy. Micrometer can measure the length in metric as well as in English unit. Micrometer is generally used to measure diameter and length of the mechanical component.

Working:

Micrometer is a screwed device. It contains spindle, anvil and thimble. Object is placed between spindle and anvil. Thimble is rotated that rotates the spindle till it touches the component completely. Two types of scales are used to measure the reading of micrometer, one is sleeve scale and other is thimble scale. Spindle moves toward component by 0.5 mm in or 0.025 in on every one rotation of spindle. There are three types of micrometer

Least count of micrometer:

Minimum measurement of any measuring device is the least count of that device. So, the least count for normal micrometer is 0.01 mm or 0.001 inches.

The micrometer is called vernier micrometer if the micrometer is provided with the vernier scale. The least count of vernier micrometer scale is 0.0001 inches.

Hence the micrometer with vernier graduation can measure reading to the nearest 0.0001 inches.

Thu, the statement is true.

Answer:

The rate of heat supplied to the engine is 71.7 kJ/min

Explanation:

Data

Engine hot temperature, [tex] T_H [/tex] = 900 K

Engine cold temperature, [tex] T_C [/tex] = 300 K

Refrigerator cold temperature, [tex] T'_C [/tex] = -15 C + 273 =  258 K

Refrigerator hot temperature, [tex] T'_H [/tex] = 300 K

Heat removed by refrigerator, [tex] Q'_{in} [/tex] = 295 kJ/min

Rate of heat supplied to the heat engine, [tex] Q_{in} [/tex] = ? kJ/min

See figure

From Carnot refrigerator coefficient of performance definition

[tex] COP_{ref} = \frac{T'_C}{T'_H - T'_C} [/tex]

[tex] COP_{ref} = \frac{258}{300 - 258} [/tex]

[tex] COP_{ref} = 6.14 [/tex]

Refrigerator coefficient of performance is defined as

[tex] COP_{ref} = \frac{Q'_{in}}{W} [/tex]

[tex] W = \frac{Q'_{in}}{COP_{ref}} [/tex]

[tex] W = \frac{295 kJ/min}{6.14} [/tex]

[tex] W = 48.04 kJ/min [/tex]

Carnot engine efficiency is expressed as

[tex] \eta = 1 - \frac{T_C}{T_H}[/tex]

[tex] \eta = 1 - \frac{300 K}{900 K}[/tex]

[tex] \eta = 0.67[/tex]

Engine efficiency is defined as

[tex] \eta = \frac{W}{Q_{in}} [/tex]

[tex] Q_{in} = \frac{W}{\eta} [/tex]

[tex] Q_{in} = \frac{48.04 kJ/min}{0.67} [/tex]

[tex] Q_{in} = 71.7 kJ/min [/tex]

Rounding to one decimal place, the rate of heat supplied to the engine is 147.5   kJ/min.

First, we need to calculate the coefficient of performance (COP) of the Carnot refrigerator using the formula:

[tex]\[ \text{COP} = \frac{T_C}{T_H - T_C} \][/tex]

where:

[tex]- \( T_C \)[/tex]  is the absolute temperature of the cold sink (300 K)

[tex]- \( T_H \)[/tex]  is the absolute temperature of the heat source (900 K)

Substituting the given values, we get:

[tex]\[ \text{COP} = \frac{300}{900 - 300} = \frac{300}{600} = 0.5 \][/tex]

Next, we use the COP of the refrigerator to find the rate of heat supplied to the engine:

[tex]\[ \text{Rate of heat supplied to the engine} = \text{COP} \times \text{Rate of heat removed by the refrigerator} \][/tex]

Given that the rate of heat removed by the refrigerator is 295 kJ/min, we can calculate the rate of heat supplied to the engine:

[tex]\[ \text{Rate of heat supplied to the engine} = 0.5 \times 295 = 147.5 \, \text{kW} \][/tex]

Rounding to one decimal place, the rate of heat supplied to the engine is 147.5 kJ/min.

The complete question is here.

A carnot heat engine receives heat at 900K and rejects the waste heat to the enviroment at 300K. The entire work output of the heat engine is used to drive a carnot refrigerator that removes heat from the cooled space at -150C at a rate of 250 kJ/min and rejects it to the same enviroment at 300 K. Determine (a) the rate of heat supplied to the heat engine and (b) the total rate of heat rejection to the enviroment.

Answer:

The change in specific internal energy is 3.5 kj.

Explanation:

Step1

Given:

Total change in energy is 15.5 kj.

Change in kinetic energy is –3.5 kj.

Change in potential energy is 0 kj.

Mass is 5.4 kg.

Step2

Calculation:

Change in internal energy is calculated as follows:

[tex]\bigtriangleup E=\bigtriangleup KE+\bigtriangleup PE+\bigtriangleup U[/tex][tex]15.5=-3.5+0+\bigtriangleup U[/tex]

[tex]\bigtriangleup U=19[/tex] kj.

Step3

Specific internal energy is calculated as follows:

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

[tex]\bigtriangleup u=\frac{19}{5.4}[/tex]

[tex]\bigtriangleup u=3.5[/tex] kj/kg.

Thus, the change in specific internal energy is 3.5 kj/kg.

Answer:

the value of conductivity in IP is [tex]8.406\dfrac{Btu}{ft.hr.F}[/tex]

Explanation:

Given that

Thermal conductivity K=14.54 W/m.K

This above given conductivity is in SI unit.

     SI unit                                           IP unit              Conversion factor

    m                                                      ft                      0.3048

   W                                                       Btu/hr               0.293          

The unit of conductivity in IP is Btu./ft.hr.F.

Now convert into IP divided by 1.73 factor.

[tex]0.57\dfrac{Btu}{ft.hr.F}=1 \dfrac{W}{m.K}[/tex]

So

[tex]0.57\times 14.54\dfrac{Btu}{ft.hr.F}=14.54 \dfrac{W}{m.K}[/tex]

[tex]8.406\dfrac{Btu}{ft.hr.F}=14.54 \dfrac{W}{m.K}[/tex]

So the value of conductivity in IP is [tex]8.406\dfrac{Btu}{ft.hr.F}[/tex]

Answer:

shaft power 0.2034 MW

Explanation:

given details

radius of turbine = 25 m

average wind velocity = 12 m/s

density of air = 1.20 kg/m^2

Total power is calculated as

[tex]P = \frac{1}{2} \rho AV^3[/tex]

  [tex]= \frac{1}{1} \rho \pir^2 V^3[/tex]

  [tex]= \frac{1}{2} 1.20\times \pi \times 625\times 12^3 = 2034,720 watt[/tex]

P = 2.034 MW

shaft power [tex] = \eta \times P[/tex]

                    [tex]= 0.10 \times 2.034[/tex]

                    = 0.2034 MW

Answer:

The volume of the extra water is [tex]2.195 ft^{3}[/tex]

Solution:

As per the question:

Mass of the canoe, [tex]m_{c} = 175 lb + w[/tex]

Height of the canoe, h = 21.5 ft

Mass of the kevlar canoe, [tex]m_{Kc} = 38 lb + w[/tex]

Now, we know that, bouyant force equals the weight of the fluid displaced:

Now,

[tex]V\rho g = mg[/tex]

[tex]V = \frac{m}{\rho}[/tex]                                  (1)

where

V = volume

[tex]\rho = 62.41 lb/ft^{3}[/tex] = density

m = mass

Now, for the canoe,

Using eqn (1):

[tex]V_{c} = \frac{m_{c} + w}{\rho}[/tex]

[tex]V_{c} = \frac{175 + w}{62.41}[/tex]

Similarly, for Kevlar canoe:

[tex]V_{Kc} = \frac{38 + w}{62.41}[/tex]

Now, for the excess volume:

V = [tex]V_{c} - V_{Kc}[/tex]

V = [tex]\frac{175 + w}{62.41} - \frac{38 + w}{62.41} = 2.195 ft^{3}[/tex]

Explanation:

From Stefan's formula

P=A&T^4

T=(P/A&)^1/4

T=(632000W/1.6m^2 x 5.67E-8W/m^2K^4)^1/4

T=

Answer:

Velocity in inch per second will be 52.11 inch/sec

Explanation:

We have given velocity = 7943 cm/min

We have to convert this velocity into inches/sec

We know that 1 cm = 0.3937 inch

So 7943 cm = 7943×0.3937=3127.1193inch

And 1 minute = 60 sec

So [tex]7943cm/min =\frac{7943\times 0.3937inch}{60sec}=52.11inch/sec[/tex]

So velocity in inch per second will be 52.11 inch/sec

Answer:

Shape factor

Explanation:

Shape factor is the ratio of maximum plastic moment to maximum elastic moment.Shape factor is denoted by K.

Shape factor can be given as

[tex]K=\dfrac{M_p}{M_y}[/tex]

[tex]K=\dfrac{\sigma _yZ_p}{\sigma _y Z}[/tex]

[tex]K=\dfrac{Z_p}{ Z}[/tex]

For a solid rectangular section made from ductile material shape factor is 1.5 .

Answer:

stress  = 50MPa

Explanation:

given data:

Length of strain guage is 5mm

displacement[tex] \delta = 1.25 \mu m =\frac{1.25}{1000} =  0.00125 mm[/tex]

stress due to displacement in structural steel can be determined by using following relation

[tex]E =\frac{stress}{strain}[/tex]

[tex]stress = E \times strain[/tex]

where E is young's modulus of elasticity

E for steel is 200 GPa

[tex]stress = 200\times 10^3 *\frac{1.25*10^{-3}}{5}[/tex]

stress  = 50MPa

The density of water in SI units, converted from 1.94 slug/ft^3, is approximately 998.847 kg/m^3 when expressed to three significant figures.

The student has asked to convert the density of water from slug/ft3 to SI units. To convert from slug/ft3 to kg/m3, we need to use the appropriate conversion factors. One slug is equivalent to 14.5939 kilograms, and there are 0.3048 meters in a foot. Therefore, the conversion is as follows:

(1.94 slug/ft3)
* (14.5939 kg/slug)
* ((1 ft/0.3048 m)3)

This equals 1.94 * 14.5939 * (1/0.3048)3 kg/m3, which simplifies to 998.847 kg/m3 when rounded to three significant figures. This is the density of water in SI units.

Water's density conversion to SI units is 1000 kg/m³.

The density of water in SI units can be expressed as 1000 kg/m³. This conversion is based on the fact that the density of water is exactly 1 g/cm³, equivalent to 1000 kg/m³.

Water's density conversion to SI units is 1000 kg/m³.

The density of water in SI units can be expressed as 1000 kg/m³. This conversion is based on the fact that the density of water is exactly 1 g/cm³, equivalent to 1000 kg/m³.

The student has asked to convert the density of water from slug/ft3 to SI units. To convert from slug/ft3 to kg/m3, we need to use the appropriate conversion factors. One slug is equivalent to 14.5939 kilograms, and there are 0.3048 meters in a foot. Therefore, the conversion is as follows:

(1.94 slug/ft3) * (14.5939 kg/slug) * ((1 ft/0.3048 m)3)

This equals 1.94 * 14.5939 * (1/0.3048)3 kg/m3, which simplifies to 998.847 kg/m3 when rounded to three significant figures. This is the density of water in SI units.

The student has asked to convert the density of water from slug/ft3 to SI units. To convert from slug/ft3 to kg/m3, we need to use the appropriate conversion factors. One slug is equivalent to 14.5939 kilograms, and there are 0.3048 meters in a foot. Therefore, the conversion is as follows:

(1.94 slug/ft3)
* (14.5939 kg/slug)
* ((1 ft/0.3048 m)3)

This equals 1.94 * 14.5939 * (1/0.3048)3 kg/m3, which simplifies to 998.847 kg/m3 when rounded to three significant figures. This is the density of water in SI units.

Answer:

The radius of curvature is 979 meter

Explanation:

We have given velocity of the canon ball v = 98 m/sec

Acceleration due to gravity [tex]g=9.81m/sec^2[/tex]

We know that at highest point of trajectory angular acceleration is equal to acceleration due to gravity

Acceleration due to gravity is given by [tex]a_c=\frac{v^2}{r}[/tex], here v is velocity and r is radius of curvature

So [tex]\frac{98^2}{r}=9.81[/tex]

r = 979 meter

So the radius of curvature is 979 meter

Answer:

C)185,500 KJ

Explanation:

Given that

Latent heat fusion = 333.23 KJ/kg

Latent heat vaporisation = 333.23 KJ/kg

Mass of ice = 100 kg

Mass of water = 40 kg

Mass of vapor=60 kg

Ice at 0°C ,first it will take latent heat of vaporisation and remain at constant temperature 0°C and it will convert in to water.After this water which at 0°C will take sensible heat and gets heat up to 100°C.After that at 100°C vapor will take heat as heat of  vaporisation .

Sensible heat for water Q

[tex]Q=mC_p\Delta T[/tex]

For water

[tex]C_p=4.178\ KJ/Kg.K[/tex]

Q=4.178 x 40 x 100 KJ

Q=16,712 KJ

So total heat

Total heat =100 x 333.23+16,712 + 60 x 2257 KJ

Total heat =185,455 KJ

Approx Total heat = 185,500 KJ

So the answer C is correct.

Answer:

Option ‘a’ is the cheapest for this house.

Explanation:

Cheapest method of heating must have least cost per kj of energy. So, convert all the energy in the same unit (say kj) and take select the cheapest method to heat the house.

Given:

Three methods are given to heat a particular house are as follows:

Method (a)

Through Gas, this gives energy of amount $1.33/therm.

Method (b)

Through electric resistance, this gives energy of amount $0.12/KWh.

Method (c)

Through oil, this gives energy of amount $2.30/gallon.

Calculation:

Step1

Change therm to kj in method ‘a’ as follows:

[tex]C_{1}=\frac{\$ 1.33}{therm}\times(\frac{1therm}{105500kj})[/tex]

[tex]C_{1}=1.2606\times10^{-5}[/tex] $/kj.

Step2

Change kWh to kj in method ‘b’ as follows:

[tex]C_{2}=\frac{\$ 0.12}{kWh}\times(\frac{1 kWh }{3600kj})[/tex]

[tex]C_{2}=3.334\times10^{-5}[/tex] $/kj.

Step3

Change kWh to kj in method ‘c’ as follows:

[tex]C_{3}=\frac{\$ 2.30}{gallon}\times(\frac{1 gallon }{138500kj})[/tex]

[tex]C_{3}=1.66\times10^{-5}[/tex] $/kj.

Thus, the method ‘a’ has least cost as compare to method b and c.

So, option ‘a’ is the cheapest for this house.

Answer:

a) 2.18 m/s^2

b) 9.83 m/s

Explanation:

The flywheel has a moment of inertia

J = m * k^2

Where

J: moment of inertia

k: radius of gyration

In this case:

J = 144 * 0.45^2 = 29.2 kg*m^2

The block is attached through a wire that is wrapped around the wheel. The weight of the block causes a torque.

T = p * r

r is the radius of the wheel.

T = m1 * g * r

T = 18 * 9.81 * 0.6 = 106 N*m

The torque will cause an acceleration on the flywheel:

T = J * γ

γ = T/J

γ = 106/29.2 = 3.63 rad/s^2

SInce the block is attached to the wheel the acceleration of the block is the same as the tangential acceleration at the eddge of the wheel:

at = γ * r

at = 3.63 * 0.6 = 2.81 m/s^2

Now that we know the acceleration of the block we can forget about the flywheel.

The equation for uniformly accelerated movement is:

X(t) = X0 + V0*t + 1/2*a*t^2

We can set a frame of reference that has X0 = 0, V0 = 0 and the X axis points in the direction the block will move. Then:

X(t) = 1/2*a*t^2

Rearranging

t^2 = 2*X(t)/a

[tex]t = \sqrt{\frac{2*X(t)}{a}}[/tex]

[tex]t = \sqrt{\frac{2*18}{2.81}} = 3.6 s[/tex]

It will reach the 1.8 m in 3.6 s.

Now we use the equation for speed under constant acceleration:

V(t) = V0 + a*t

V(3.6) = 2.81 * 3.6 = 9.83 m/s

Answer:

No we cannot carry 1 cubic meter of liquid water.

Explanation:

As we know that density of water is 1000 kilograms per cubic meter of water hence we infer that 1 cubic meter of water will have a weight of 1000 kilograms of 1 metric tonnes which is beyond the lifting capability of strongest man on earth let alone a normal human being who can just lift a weight of 100 kilograms thus we conclude that we cannot lift 1 cubic meter of liquid water.

No, I cannot carry 1 cubic meter (1 m³) of liquid water. To understand why, let's calculate the weight of 1 cubic meter of water.

1 cubic meter (m³) of water is equivalent to 1000 liters (L). The density of water is approximately 1 kilogram per liter (kg/L). Therefore, the weight of 1 cubic meter of water can be calculated as:

[tex]\[ 1 \, \text{m}^3 \times 1000 \, \text{L/m}^3 \times 1 \, \text{kg/L} = 1000 \, \text{kg} \][/tex]

So, 1 cubic meter of water weighs 1000 kilograms, or about 2204.62 pounds.

This weight is far beyond the carrying capacity of an average human. For comparison, most people can carry only a few tens of kilograms comfortably for a short period, so carrying 1000 kilograms is not feasible for any human.

Answer:

Mass of oil will be 7.176 pound

Explanation:

We have given specific gravity of oil  = 0.86

We know that specific gravity is given by [tex]specific\ gravity=\frac{density\ of\ oil}{density\ of\ water}[/tex]

[tex]0.86=\frac{density\ of\ oil}{1000}[/tex]

Density of oil = [tex]860kg/m^3[/tex]

We have given volume of oil = 1 gallon

We know that 1 gallon = 0.003785 [tex]m^3[/tex]

So mass of oil = volume ×density

mass = 0.003785×860 = 3.2551 kg

We know that 1 kg = 2.2046 pound

So 3.2551 kg = 3.2551×2.2046 = 7.176 pound

Answer:

Impulse =14937.9 N

tangential force =14937.9 N

Explanation:

Given that

Mass of car m= 800 kg

initial velocity u=0

Final velocity v=390 km/hr

Final velocity v=108.3 m/s

So change in linear momentum P= m x v

           P= 800 x 108.3

 P=86640 kg.m/s

We know that impulse force F= P/t

So F= 86640/5.8 N

F=14937.9 N

Impulse force F= 14937.9 N

We know that

v=u + at

108.3 = 0 + a x 5.8

[tex]a=18.66\ m/s^2[/tex]

So tangential force F= m x a

F=18.66 x 800

F=14937.9 N

It is convenient to model a structural element like a beam when a significant amount of forces produce the stress called flexion.

Flexion occurs when an element is supported on one or more supports and a force is presented between them, driving a bending moment in the element.

Answer:

Free convection:

   When heat transfer occurs due to density difference between fluid then this type of heat transfer is know as free convection.The velocity of fluid is zero or we can say that fluid is not moving.

Force convection:

   When heat transfer occurs due to some external force then this type of heat transfer is know as force convection.The velocity of fluid is not zero or we can say that fluid is moving in force convection.

Heat transfer coefficient of force convection is high as compare to the natural convection.That is why heat force convection reach a steady-state faster than an object subjected to free-convection.

We know that convective heat transfer given as

 q = h  A ΔT

h=Heat transfer coefficient

A= Surface area

ΔT = Temperature difference

Answer:

Principal Plane: It is that plane in a stressed body over which no shearing stresses act. As we know that in a stressed body on different planes 2 different kind of stresses act normal stresses acting normal to the plane ans shearing stresses acting in the plane. The special planes over which no shearing stresses act and only normal stresses are present are termed as principal planes.

Principal Stress: The stresses in the principal planes are termed as normal stresses.

Anelasticity: It is the behavior of a material in which no definite relation can found to exist between stress and strain at any point in the stressed body.

Yield Point: It is the point in the stress-strain curve of a body at which the stress in the body reaches it's yield value or the object is just about to undergo plastic deformation if we just increase value of stress above this value. It is often not well defined in high strength materials or in some materials such as mild steel 2 yield points are observed.

Ultimate tensile strength: It is the maximum value of stress that a body can develop prior to fracture.

Hardness: it is defined as the ability of the body to resist scratches or indentation or abrasion.

Toughness: It is the ability of the body to absorb energy and deform without fracture when it is loaded. The area under the stress strain curve is taken as a measure of toughness of the body.

Elastic limit: The stress limit upto  which the body regains it's original shape upon removal of the stresses is termed as elastic limit of the body.

Answer:

22800 tonne

Explanation:

Given:

Length of the container, L = 240 m

Width of the container, B = 22 m

Depth inside the water, H = 10 m

Mass of the ship, m = 30000 tonnes

Now,

Total immersed volume of the ship = LBH = 240 × 22 × 10 = 52800 m³

From the Archimedes principle, we have

Total mass of the ship (i.e mass of the ship along with the load carried)

= Mass of the volume of water displaced by ship

= 52800 × Density of water

also,

Density of water = 1000 kg/m³

thus,

Total mass of the ship (i.e mass of the ship along with the load carried)

= 52800 × 1000 kg

also,

1 tonne = 1000 kg

thus,

Total mass of the ship (i.e mass of the ship along with the load carried)

= 52800 tonne

Therefore,

the load carried by the ship = Total mass of the ship  - mass of ship

or

the load carried by the ship = 52800 - 30000 = 22800 tonne

Answer:

the evaluation in SI unit will be [tex]2.69\times 10^{-5}sec^{2}[/tex]

Explanation:

We have evaluate [tex]\frac{(204mm\times 0.00457kg)}{34.6N}[/tex]

We know that 1 mm [tex]=10^{-3}m[/tex]

So 240 mm [tex]=204\times 10^{-3}m[/tex]

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

So [tex]\frac{(204\times 10^{-3}m)\times 0.00457kg}{34.6kgm/sec^2}=2.69\times 10^{-5}sec^{2}[/tex]

So the evaluation in SI unit will be [tex]2.69\times 10^{-5}sec^{2}[/tex]

Answer:

Boiling point of HF is higher as compared to HCl because of presence of hydrogen bonding in it.

Explanation:

In HF, intermolecular force of attraction is hydrogen bonding.

Hydrogen bonding is a type of electrostatic force of attraction existing between H atom and electronegative atom.

For a molecule to have hydrogen bonding, H atom must be bonded to electronegative atom, O, N and F.

Hydrogen bonding can be intermolecular and intramolecular.

So, in HF hydrogen bonding present.

In HCl, only van der Waals force exists. van der Waals forces are weak as compared to hydrogen bonding.

Because of presence of hydrogen bonding, HF molecules are held tightly and so requires more heat to boil.

Therefore, boiling point of HF is more as compared to HCl.

Answer:50 , 20

Explanation:

Given

Diametrical Pitch[tex]\left ( P_D\right )=\frac{T}{D}[/tex]

where T= no of teeths

D=diameter

module(m) of gears must be same

[tex]m=\frac{D}{T}=\frac{1}{P_D}=0.1[/tex]

Let [tex]T_1 & T_2[/tex]be the gears on two gears

Therefore Center distance is given by

[tex]m\frac{\left ( T_1+T_2\right )}{2}=3.5[/tex]

thus

[tex]0.1\frac{\left ( T_1+T_2\right )}{2}=3.5[/tex]

[tex]T_1+T_2=70----1[/tex]

and Velocity ratio is given by

[tex]VR=\frac{No\ of\ teeths\ on\ Driver\ gear}{No.\ of\ teeths\ on\ Driven\ gear} [/tex]

[tex]2.5=\frac{T_1}{T_2}----2[/tex]

From 1 & 2 we get

[tex]T_1=50, T_2=20[/tex]

Answer:

At highest point:

y1 = 10.4 ft

v1 = (26.5*i + 0*j) ft/s

When he lands:

x2 = 31.5 ft (distance he travels)

t2 = 1.19 s

V2 = (26.5*i - 25.9*j) ft/s

a2 = -44.3°

Explanation:

Since he let go of the tow rope upon leaving the ramp he is in free fall from that moment on. In free fall he is affected only by the acceleration of gravity. Gravity has a vertical component only, so the movement will be at constant acceleration in the vertical component and at constant speed in the horizontal component.

20 mi / h = 29.3 ft/s

If the ramp has an angle of 25 degrees, the speed is

v0 = (29.3 * cos(25) * i + 29.3 * sin(25) * j) ft/s

v0 = (26.5*i + 12.4*j) ft/s

I set up the coordinate system with the origin at the base of the ramp under its end, so:

R0 = (0*i + 8*j) ft

The equation for the horizontal position is:

X(t) = X0 + Vx0 * t

The equation for horizontal speed is:

Vx(t) = Vx0

The equation for vertical position is:

Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2

The equation for vertical speed is:

Vy(t) = Vy0 + a * t

In this frame of reference a is the acceleration of gravity and its values is -32.2 ft/s^2.

In the heighest point of the trajectory the vertical speed will be zero because that is the point where it transitions form going upwards (positive vertical speed) to going down (negative vertical speed), and it crosses zero.

0 = Vy0 + a * t1

a * t1 = -Vy0

t1 = -Vy0 / a

t1 = -12.4 / -32.2 = 0.38 s

y1 = y(0.38) = 8 + 12.4 * 0.38 + 1/2 * (-32.2) * (0.38)^2 = 10.4 ft

The velocity at that moment will be:

v1 = (26.5*i + 0*j) ft/s

When he lands in the water his height is zero.

0 = 8 + 12.4 * t2 + 1/2 * (-32.2) * t2^2

-16.1 * t2^2 + 12.4 * t2 + 8 = 0

Solving this equation electronically:

t2 = 1.19 s

Replacing this time on the position equation:

X(1.19) = 26.5 * 1.19 = 31.5 ft

The speed is:

Vx2 = 26.5 ft/s

Vy2 = 12.4 - 32.2 * 1.19 = -25.9 ft/s

V2 = (26.5*i - 25.9*j) ft/s

a2 = arctg(-25.9 / 26.5) = -44.3

Answer:

Explanation:

The pressures given are relative

p1 = 2000 psi

P1 = 2014 psi = 13.9 MPa

p2 = 4 psi

P2 = 18.6 psi = 128 kPa

Values are taken from the steam pressure-enthalpy diagram

h2 = 2500 kJ/kg

If the output of the turbine has a quality of 85%:

t2 = 106 C

I consider the expansion in the turbine to adiabatic and reversible,  therefore, isentropic

s1 = s2 = 6.4 kJ/(kg K)

h1 = 3500 kJ/kg

t2 = 550 C

The work in the turbine is of

w = h1 - h2 = 3500 - 2500 = 1000 kJ/kg

The thermal efficiency of the cycle depends on the input heat.

η = w/q1

q1 is  not a given, so it cannot be calculated.

Answer with Explanation:

The acidity of an aqueous solution is a term used to identify how acidic the solution is. An acidic solution is a solution in which the concentration of hydrogen ions is greater than the concentration of hydroxide ions. In the other case around if  the concentration of hydrogen ions is lesser than the concentration of hydroxide ions the solution is termed to be basic or alkaline. For a solution with equal concentration of hydrogen and hydroxide ions the solution is termed to be neutral.

The acidity of solutions is compared on the basis of the concentration of the hydrogen ions reduced to log of base 10 to ease calculations. The comparison is made in terms of 'pH' value which is defined as

[tex]pH=-log[H^+][/tex]

where

 [tex][H^+][/tex] is the hydrogen ion concentration of the solution in moles per liter of solution.

If the pH is < 7 the solution is acidic and the closer the pH value to 1 the higher is the acidity of the solution.