Fluid flowing through a constriction or partially opened valve in a pipe system may develop localized low pressures due to increased velocity. If a pipe contains water at 210 °F, what is the minimum absolute pressure that the water can be reduced to and not have cavitation occur?

Answers

Answer 1

Answer:

P=41.84psi

Explanation:

Cavitation is a phenomenon that arises when the pressure of a liquid fluid reaches the vapor pressure, this means that if at any point in a system of pipes and pumps the water pressure is equal to or less than the vapor pressure it begins to evaporate, producing explosive bubbles that can cause damage and noise in the system.

Therefore, for this problem, all we have to do is find the water vapor pressure at a temperature of 270F, using  thermodynamics tables

Pv@270F=41.84psi


Related Questions

What is an isochoric process? b) Can heat be exchanged in an isochoric process? c) A 100L container holding an ideal gas at an initial pressure of 10MPa is raised to a pressure of 15MPa. How much work is done?

Answers

Answer:

a)A constant volume process is called isochoric process.

b)Yes

c)Work =0

Explanation:

Isochoric process:

 A constant volume process is called isochoric process.

In constant volume process work done on the system or work done by the system will remain zero .Because we know that work done give as

work = PΔV

Where P is pressure and ΔV is the change in volume.

For constant volume process ΔV = 0⇒ Work =0

Yes heat transfer can be take place in isochoric process.Because we know that temperature difference leads to transfer of heat.

Given that

Initial P=10 MPa

Final pressure =15 MPa

Volume = 100 L

Here volume of gas is constant so the work work done will be zero.

As shown, a load of mass 10 kg is situated on a piston of diameter D1 = 140 mm. The piston rides on a reservoir of oil of depth h1 = 42 mm and specific gravity S = 0.8. The reservoir is connected to a round tube of diameter D2 = 5 mm and oil rises in the tube to height h2. Find h2. Assume the oil in the tube is open to atmosphere and neglect the mass of the piston.

Answers

Final answer:

Use Pascal's law and the hydrostatic pressure formula to solve for the height of the oil column, considering the specific gravity of the oil and the mass applied on the piston.

Explanation:

The question requires the application of principles from fluid mechanics to calculate the height h2 to which an oil rises in a tube, using the specific gravity of the oil and the mass of a load on a piston. Assuming the system is in equilibrium and by using Pascal's law, the pressure applied by the mass onto the piston is transmitted equally throughout the fluid. Therefore, the pressure exerted by the mass on the larger piston is balanced by the hydrostatic pressure of the oil column of height h2 in the smaller tube.

To solve for h2, we can use the formula for hydrostatic pressure P = ρgh, where ρ is the density of the fluid, g is the acceleration due to gravity, and h is the height of the fluid column. Given the specific gravity S of the oil, we can find its density (ρ) by multiplying S by the density of water (1000 kg/m³). Then, by setting the hydrostatic pressure of the oil column equal to the pressure exerted by the mass and solving for h2, we find the height of the oil in the tube.

If the dry-bulb temperature is 95°F and the wet-bulb temperature is also 78°F, what is the relative humid- ity? What is the dew point? What is the humidity ratio? What is the enthalpy?

Answers

Answer:

Relative humidity 48%.

Dew point 74°F

humidity ratio 118 g of moisture/pound of dry air

enthalpy 41,8 BTU per pound of dry air

Explanation:

You can get this information from a Psychrometric chart for water, like the one attached.

You enter the chart with dry-bulb and wet-bulb temperatures (red point in the attachment) and following the relative humidity curves you get approximately 48%.

To get the dew point you need to follow the horizontal lines to the left scale (marked with blue): 74°F

for the humidity ratio you need to follow the horizontal lines but to the rigth scale (marked with green): 118 g of moisture/pound of dry air

For enthalpy follow the diagonal lines to the far left scale (marked with yellow): 41,8 BTU per pound of dry air

A spherical tank is being designed to hold 10 moles of carbon dioxide gas at an absolute pressure of 5 bar and a temperature of 80°F. What diameter spherical tank should be used? The molecular weight of methane is 44 g/mole.

Answers

Answer:

r=0.228m

Explanation:

The equation that defines the states of a gas according to its thermodynamic properties is given by the general equation of ideal gases

PV=nRT

where

P=pressure =5bar=500.000Pa

V=volume

n=moles=10

R = universal constant for ideal gases = 8.31J / (K.mol)

T=temperature=80F=299.8K

solvig For V

V=(nRT)/P

[tex]V=(\frac{(10)(8.31)(299.8)}{500000} )\\V=0.0498m^3[/tex]

we know that the volume of a sphere is

[tex]V=\frac{4\pi r^3}{3} \\[/tex]

solving for r

[tex]r=\sqrt[3]{ \frac{3 V}{4\pi } }[/tex]

solving

[tex]r=\sqrt[3]{ \frac{3 (0.049)}{4\pi } }\\r=0.228m[/tex]

A large water jet with a discharge of 2m^3 /s rises 90m above the ground. The exit nozzle diameter to achieve this must be. (a) 0.246m (b) 0.318m (c) 1.3m (d) 0.052m (e) None of the above

Answers

Answer:

The correct answer is option 'a': 0.046 meters.

Explanation:

We know that the exit velocity of a jet of water is given by Torricelli's law as

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

where

'v' is velocity of head

'g' is acceleration due to gravity

'h' is the head under which the jet falls

Now since the jet rises to a head of 90 meters above ground thus from conservation of energy principle it must have fallen through a head of 90 meters.

Applying the values in above equation we get the exit velocity as

[tex]v=\sqrt{2\times 9.81\times 90}=42.02m/s[/tex]

now we know the relation between discharge and velocity as dictated by contuinity equation is

[tex]Q=V\times Area[/tex]

Applying values in the above equation and solving for area we get

[tex]Area=\frac{Q}{v}=\frac{2}{42.02}=0.0476m^{2}[/tex]

The circular area is related to diameter as

[tex]Area=\frac{\pi D^{2}}{4}\\\\\therefore D=\sqrt{\frac{4\cdot A}{\pi }}=\sqrt{\frac{4\times 0.0476}{\pi }}=0.246m[/tex]

Thus the diameter of the nozzle is 0.246 meters

Domestic units have power input ratings of between (a) 35 W and 375 W (b) 3.5 and 37.5 W (c) 350W and 375 W

Answers

Answer:

The answer is "b" 3.5W and 37.5w.

Explanation:

Domestic use normally refers to refrigeration equipment that is not subject to high thermal loads, for example, refrigerators should only extract heat from food, and domestic air conditioners should only extract heat from small spaces and a number reduced people.

A heating cable is embedded in a concrete slab for snow melting. The heating cable is heated electrically with joule heating to provide the concrete slab with a uniform heat of 1200 W/ m2. The concrete has a thermal conductivity of 1.4 W/m⋅K. To minimize thermal stress in the concrete, the temperature difference between the heater surface (T1) and the slab surface (T2) should not exceed 16°C (2015 ASHRAE Handbook—HVAC Applications, Chapter. 51). Formulate the temperature profile in the concrete slab, and determine the thickness of the concrete slab (L) so that T1 − T2 ≤ 16°C.

Answers

Answer:

18.7 mm

Explanation:

Fourier's law for heat conduction on a plate is:

q = -k/t * ΔT

Where

q: heat conducted per unit of time and surface

k: thermal conductivity

t: thickness of the plate

ΔT: temperature difference

Rearranging:

t = -k/q * ΔT

t = -1.4/(-1200) * 16 = 0.0187 m = 18.7 mm (q is negative because it is heat leaving the plate)

The maximum thickness of the concrete slab is 18.7 mm.

A piston-cylinder apparatus has a piston of mass 2kg and diameterof
10cm resting on a body of water 30 cm high atmospheric pressureis
101 kpa, and the temperature of water 50 degrees Celsius. What is
the mass of water in the container.

Answers

Answer:

M =2.33 kg

Explanation:

given data:

mass of piston - 2kg

diameter of piston is 10 cm

height of water 30 cm

atmospheric pressure 101 kPa

water temperature = 50°C

Density of water at 50 degree celcius is 988kg/m^3

volume of cylinder is  [tex]V = A \times h[/tex]

                                       [tex]= \pi r^2 \times h[/tex]

                                       [tex]= \pi 0.05^2\times 0.3[/tex]

mass of available in the given container is

[tex]M = V\times d[/tex]

  [tex] = volume \times density[/tex]

[tex]= \pi 0.05^2\times 0.3 \times 988[/tex]

M =2.33 kg

The information on a can of pop indicates that the can contains 360 mL. The mass of a full can of pop is 0.369 kg, while an empty can weighs 0.153 N. Determine the specific weight, density, and specific gravity of the pop and compare your results with the corresponding values for water at Express your results in SI units.

Answers

Answer:

Specific weight of the pop, [tex]w_{s} = 8619.45 N/m^{3}[/tex]

Density of the pop, [tex]\rho_{p} = 8790.76 kg/m^{3}[/tex]

[tex]g_{s} = 8.79076[/tex]

[tex]w_{w} = 9782.36 N/m^{3}[/tex]

Given:

Volume of pop, V = 360 mL = 0.36 L = [tex]0.36\times 10^{-3} m^{3}[/tex]

Mass of a can of pop , m = 0.369 kg

Weight of an empty can, W = 0.153 N

Solution:

Now, weight of a full can pop, W

W' = mg = [tex]0.369\times 9.8 = 3.616 N[/tex]

Now weight of the pop in can is given by:

w = W' - W = 3.616 - 0.513 = 3.103 N

Now,

The specific weight of the pop, [tex]w_{s} = \frac{weight of pop}{volume of pop}[/tex]

[tex]w_{s} = \frac{3.103}{0.36\times 10^{- 3}} = 8619.45 N/m^{3}[/tex]

Now, density of the pop:

[tex]\rho_{p} = \frac{w_{s}}{g}[/tex]

[tex]\rho_{p} = \frac{86149.45}{9.8} = 8790.76 kg/m^{3}[/tex]

Now,

Specific gravity, [tex]g_{s} = \frac{\rho_{p}}{density of water, \rho_{w}}[/tex]

where

[tex]g_{s} = \frac{8790.76}{1000} = 8.79076[/tex]

Now, for water at [tex]20^{\circ}c[/tex]:

Specific density of water = [tex]998.2 kg/m^{3}[/tex]

Specific gravity of water = [tex]0.998 kg/m^{3}[/tex]

Specific weight of water at [tex]20^{\circ}c[/tex]:

[tex]w_{w} = \rho_{20^{\circ}}\times g = 998.2\times 9.8 = 9782.36 N/m^{3}[/tex]

When all network cables connect to one central point the network topology is typically referred to as a(n)_______

Answers

Answer:

Star

Explanation:

When all hosts are connected to a central hub it is known as a star topology.

The advantages of the star network is that it is very easy to add new devices, a failure in a host will not cause problems on the rest of the network, it is appropriate for large networks and works well under heavy loads,

The disadvantages are that a failure of the central hub brings the whole networks down and it is expensive due to the amount of cables needed.

What is mass flow of 200 lbm/min in kg/sec?

Answers

Answer:

Mass flow in kg /sec will be 1.511 kg/sec

Explanation:

We have given mass flow = 200 lbm/min

We have to convert lbm/min into kg /sec

We know that 1 lbm = 0.4535 kg

So for converting lbm to kg we have multiply with 0.4535

So 200 lbm = 200×0.4535 =90.7 kg

Now we know that 1 minute = 60 sec

So 200 lbm/min [tex]=\frac{200\times 0.4535kg}{60sec}=1.511kg/sec[/tex]

So mass flow in kg /sec will be 1.511 kg/sec

A brittle intermetallics specimen is tested with a bending test. The specimen's width 0.45 in and thickness 0.20 in. The length of the specimen between supports 2.5 in. Determine the transverse rupture strength if failure occurs at a load 1200 lb.

Answers

Answer:

250 kpsi

Explanation:

Given:

Width of the specimen, w = 0.45 in

Thickness of the specimen, t = 0.20 in

length of the specimen between supports, L = 2.5 in

Failure load, F = 1200 lb

Now,

The transverse rupture strength [tex]\sigma_t=\frac{1.5FL}{wt^2}[/tex]

on substituting the respective values, we get

[tex]\sigma_t=\frac{1.5\times1200\times2.5}{0.45\times0.2^2}[/tex]

or

[tex]\sigma_t=250,000\ psi\ =\ 250 kpsi[/tex]

An electric motor supplies 200 N·m of torque to a load. What is the mechanical power supplied to the load if the shaft speed is 1000 rpm? Express the result in watts and horsepower.

Answers

Answer:

power = 20943.95 watts

power = 28.086 horsepower

Explanation:

given data

torque = 200 N

speed = 1000 rpm

to find out

What is the mechanical power in watts and horse power

solution

we know that mechanical power formula that is

power = torque × speed   ...................1

here we have given both torque and speed

we know speed = 1000 rpm = [tex]\frac{2* \pi *1000}{60}[/tex] = 104.66 rad/s

so put here value in equation 1

power = 200 × 104.719

power = 20943.95 watts

and

power = [tex]\frac{20943.95}{745.7}[/tex]

power = 28.086 horsepower

Answer:

Part 1) Power required for motor = 20944 watts.

Part 2) Power required for motor in Horsepower equals = 28.075H.P

Explanation:

Power is defined as the rate of consumption of energy. For rotational motion power is calculated as

[tex]Power=Torque\times \omega[/tex]

where,

[tex]\omega [/tex] is the angular speed of the motor.

Since the rotational speed of the motor is given as 1000 rpm, the angular speed is calculated as

[tex]\omega =\frac{N}{60}\times 2\pi[/tex]

where,

'N' is the speed in rpm

Applying the given values we get

[tex]\omega =\frac{1000}{60}\times 2\pi=104.72rad/sec[/tex]

hence the power equals

[tex]Power=200\times 104.72=20944Watts[/tex]

Now since we know that 1 Horse power equals 746 Watts hence 20944 Watts equals

[tex]Power_{H.P}=\frac{20944}{746}=28.075H.P[/tex]

There are two questions about SolidWorks.
1. Why is it good practice to fully define
sketchgeometry?
2. Describe what happens if you do not select a planeprior to
clicking on the Sketch icon when creating a sketch.

Answers

Answer:

1. It is a good practice to fully define a sketch to avoid having erroneous dimensions on the faces of a solid, this avoids that when it is required to make an assembly with the drawn part appear assembly errors.

2. The 2D sketch should always be done on a plane, so solidworks would ask you to select a plane on which you want to make the sketch, on the other hand, if it is a 3D sketch, solidworks allows you to do it without the need for Select any plane.

You want a potof water to boil at 105 celcius. How heavy a
lid should youput on the 15 cm diameter pot when Patm =
101kPa?

Answers

Answer:

35.7 kg lid we put

Explanation:

given data

temperature = 105 celcius

diameter = 15 cm

Patm = 101 kPa

to find out

How heavy a  lid should you put

solution

we know Psaturated from table for temperature is 105 celcius is

Psat = 120.8 kPa

so

area will be here

area = [tex]\frac{\pi }{4} d^2[/tex]    ..................1

here d is diameter

put the value in equation 1

area = [tex]\frac{\pi }{4} 0.15^2[/tex]

area = 0.01767 m²

so net force is

Fnet = ( Psat - Patm ) × area

Fnet = ( 120.8 - 101 ) × 0.01767

Fnet = 0.3498 KN = 350 N

we know

Fnet = mg

mass = [tex]\frac{Fnet}{g}[/tex]

mass  = [tex]\frac{350}{9.8}[/tex]

mass = 35.7 kg

so 35.7 kg lid we put

What is the origin of the "horsepower"? Why would anyone wish to express power in the unit of horsepower? How many watts are in one horsepower?

Answers

Answer:

Horsepower unit was first time used by James Watt in 1782. The  story refers that James Watt uses worked with pony to charge coal from the mines. According to that story,  he have the need of a unit to measure the force from one of this animals. He founds that they can move 22.000 lbs per minute, so he (arbitrarily) increase this measure in 50% been the unit Horsepower in 33.000 lb/feet per minute.

Explanation:

This measure unit can measure "work" or "force". In the SI correspond to move up 75 Kg, to 1 meter high,  in one second.

1 HP = (330 lb) x (100 feet)/1min = 33000 lb x feet/min

Why would anyone wish to express power in HP?

Is a practical unit, because reduce the amount of digits in a specific value. Also, his use is very spread specially in mechanical applications.

1 HP= 746 W (0,746 kW).

Various factors to be considered in deciding the factor of safety?

Answers

Answer with Explanation:

There are various factors that needed to be taken into account while deciding the factor of safety some of which are summarized below as:

1) Importance of the structure: When we design any structure different structures have different importance in our society. Take an example of hospital, in case a natural disaster struck's a place the hospital should be the designed to withstand the disaster as it's role in the crisis management following a disaster is well understood. Thus while designing it we need it to have a higher factor of safety against failure when compared to a local building.

2) Errors involved in estimation of strength of materials: when we design any component of any machine or a structure we need to have an exact idea of the behavior of the material and know the value of the strength of the material. But many materials that we use in structure such as concrete in buildings have a very complex behavior and we cannot estimate the strength of the concrete absolutely, thus we tend to decrease the strength of the concrete more if errors involved in the estimation of strength are more to give much safety to the structure.

3) Variability of the loads that may act on the structure: If the loads that act on the structure are highly variable such as earthquake loads amd dynamic loads then we tend to increase the factor of safety while estimating the loads on the structure while designing it.

4) Economic consideration: If our project has abundant funds then we can choose a higher factor of safety while designing the project.

Yield and tensile strengths and modulus of elasticity . with increasing temperature. (increase/decrease/independent)

Answers

Answer:

Yield strength, tensile strength decreases with increasing temperature and modulus of elasticity decreases with increasing in temperature.

Explanation:

The modulus of elasticity of a material is theoretically a function of the shape of curve plotted between the potential energy stored in the material as it is loaded versus the inter atomic distance in the material. The temperature distrots the molecular structure of the metal and hence it has an effect on the modulus of elasticity of a material.

Mathematically we can write,

[tex]E(t)=E_o[1-a\frac{T}{T_m}][/tex]

where,

E(t) is the modulus of elasticity at any temperature 'T'

[tex]E_o[/tex] is the modulus of elasticity at absolute zero.

[tex]T_{m}[/tex] is the mean melting point of the material

Hence we can see that with increasing temperature modulus of elasticity decreases.

In the case of yield strength and the tensile strength as we know that heating causes softening of a material thus we can physically conclude that in general the strength of the material decreases at elevated temperatures.

What is the range of a 32-bit unsigned integer?

Answers

Answer:

0 to 4294967295

Explanation:

Unsigned integers have only positive numbers and zero. Their range goes from zero to (2^n)-1.

In the case of a 32-bit unsigned integer this would be.

(2^32) - 1 = 4294967296 - 1 = 4294967295

So the range goes from 0 to 4294967295 or form zero to about 4.3 billion.

The minus one term is because the zero takes one of the values.

A Carnot heat engine operates between 1000 deg F and 50 deg F, producing 120 BTU of work. What is the heat input to the engine?

Answers

Answer:

184.6 BTU

Explanation:

The thermal efficiency for a Carnot cycle follows this equation:

η = 1 - T2/T1

Where

η: thermal efficiency

T1: temperature of the heat source

T2: temperature of the heat sink

These temperatures must be in absolute scale:

1000 F = 1460 R

50 F = 510 R

Then

η = 1 - 510/1460 = 0.65

We also know that for any heat engine:

η = L / Q1

Where

L: useful work

Q1: heat taken from the source

Rearranging:

Q1 = L / η

Q1 = 120 / 0.65 = 184.6 BTU

In this exercise we will deal with the Carnot cycle and calculate how much heat was initially placed, so we have that:

The heat input to the engine was 184.6 BTU

How does the Carnot Cycle work?

The theoretical Carnot cycle is formed by two isothermal transformations and two adiabatic transformations. One of the isothermal transformations is used for the temperature of the hot source, where the expansion process takes place, and the other for the cold source, where the compression process takes place.

The thermal efficiency for a Carnot cycle follows this equation:

[tex]\eta= 1 - T_2/T_1[/tex]

Where:

η: thermal efficiencyT1: temperature of the heat sourceT2: temperature of the heat sink

These temperatures must be in absolute scale:

[tex]\eta = 1 - 510/1460 = 0.65[/tex]

Rearranging:

[tex]Q_1 = L / \eta\\Q_1 = 120 / 0.65 = 184.6 BTU[/tex]

See more about carnot cycle at brainly.com/question/21126569

A tow truck is using a cable to pull a car up a 15 degree hill. If the car weighs 4000 lbs and the cable has a diameter of .75 inches, find the stress in the cable when the truck comes to a stop while on the hill. Ignore friction between the car and the pavement.

Answers

Answer:40.603 MPa

Explanation:

Given

Car weighs (m)4000 lbs [tex]\approx 1814.37 kg[/tex]

diameter of cable(d)[tex]=0.75 in.\approx 19.05 mm[/tex]

Hill angle [tex]\theta =15^{\circ}[/tex]

Now

tension in cable will bear the weight of car acting parallel to rope which is [tex]mgsin\theta [/tex]

Thus

[tex]T=mgsin\theta [/tex]

[tex]T=1814.37\times 9.81\times sin(15)=11,574.45 N[/tex]

T=11.57 kN

thus stress([tex]\sigma [/tex])=[tex]\frac{T}{A}[/tex]

where A=cross section of wire[tex]=285.059 mm^2[/tex]

[tex]\sigma =\frac{11574.45}{285.059}=40.603 MPa[/tex]

An automobile tire with a volume of 0.8 m3 is inflated to a gage pressure of 200 kPa. Calculate the mass of air in the tire if the temperature is 23 °C.

Answers

The mass of the air in the tire is 1040.192 grams.

We must know the concept of the Ideal Gas Equation to solve this question.

What is Ideal Gas Equation?

The Ideal Gas Equation shows the empirical relationship between the Volume, Pressure, Temperature, and the number of moles in a given substance.

Using Ideal Gas Equation:

PV = nRT

From the given information:

The volume of the air O₂ = 0.8 m³ = 0.8 × 1000 L = 800 LThe pressure = 200 KPaThe temperature = 23°C = (273 + 23) K = 296 KThe rate constant = 8.31446 L.kPa.K⁻¹.mol⁻¹

[tex]\mathbf{200 kPa \times 800 \ L = n \times 8.31446 L.kPa.K^{-1}.mol^{-1} \times 296 \ K }[/tex]

[tex]\mathbf{n = \dfrac{200 kPa \times 800 \ L}{8.31446 \ L.kPa.K^{-1}.mol^{-1} \times 296 \ K}}[/tex]

[tex]\mathbf{n =65.012 \ mol}[/tex]

From the relation;
Number of moles = mass/molar mass.

The molar mass of O₂ = 16 g/mol

Mass of air (O₂) = number of moles × molar mass

Mass of air (O₂) = 65.012 mol × 16 g/mol

Mass of air (O₂) = 1040.192 grams

Learn more about ideal gas equation here:

https://brainly.com/question/25290815

The flow curve for a certain metal has parameters: strain-hardening
exponent is 0.22 and strength coefficient is54,000
lb/in2 . Determine:
a) the true stress at a true strain = 0.45
b) the true strain at a true stress = 40,000
lb/in2.

Answers

Answer:(a)[tex]45,300.24 lb/in/^2[/tex]

(b)0.255

Explanation:

Given

Strain hardening exponent(n)=0.22

Strength coefficient(k)[tex]=54000 lb/in/^2[/tex]

and we know

[tex]\sigma =k\left ( \epsilon \right )^n[/tex]

where

[tex]sigma =true\ stress[/tex]

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

(a)True strain=0.45

[tex]\sigma =54000\times 0.45^{0.22}[/tex]

[tex]\sigma =45,300.24 lb/in^2[/tex]

(b)true stress[tex]=40,000 lb/in^2[/tex]

[tex]40000=54000\times \epsilon ^{0.22}[/tex]

[tex]\epsilon ^{0.22}=0.7407[/tex]

[tex]\epsilon =0.7407^{4.5454}=0.255[/tex]

Calculate the rate at which body heat is conducted through the clothing of a skier in a steady- state process, given the following data: the body surface area is 1.80 m and the clothing is 1.00 cm thick; the skin surface temperature is 33.0 C and the outer surface of the clothing is at 1.00 C the thermal conductivity of the clothing is 0.040 W/m K

Answers

The rate of heat transfer through the skier's clothing is approximately 230.4 watts.

Given the information you provided, we can calculate the rate of heat transfer through the skier's clothing using the following formula for heat conduction:

Q = k * A * ΔT / L

where:

Q is the heat transfer rate (W)

k is the thermal conductivity of the clothing (W/m K)

A is the body surface area (m²)

ΔT is the temperature difference between the skin and the outer surface of the clothing (K)

L is the thickness of the clothing (m)

Let's plug in the values:

k = 0.040 W/m K

A = 1.80 m²

ΔT = 33.0°C - 1.0°C = 32.0 K (convert temperature difference from Celsius to Kelvin)

L = 0.01 m (convert centimeter to meter)

Q = 0.040 W/m K * 1.80 m² * 32.0 K / 0.01 m

Q = 230.40 W

a baseball is thrown downward from a 50 ft tower with an
initialspeed of 18 ft/s.
what is the speed when the ball hits the ground and the time
oftrave?

Answers

Answer:

Final velocity will be 36.11 m/sec

Time required by ball to heat the ground = 1.297 sec

Explanation:

We have given height = 50 feet

Initial velocity u = 18 ft/sec

Acceleration due to gravity [tex]g=32.8ft/sec^2[/tex]

We have to find final velocity v

From third equation of motion

[tex]v^2=u^2+2gh[/tex]

So [tex]v^2=18^2+2\times 32.8\times 50=3543.82[/tex]

v = 59.53m/sec

From first equation of motion we know that

v= u+gt

So [tex]59.53=18+32.8\times t[/tex]

t = 1.297 sec

A spacecraft component occupies a volume of 8ft^3 and weighs 25 lb at a location where the acceleration of gravity is 31.0 ft/s^2. Determine its weight, in pounds, and its average density, in lbm/ft^3, on the moon, where g=5.57 ft/s^2.

Answers

Answer:

The weight is 4.492 lb

The density is [tex]0.5614 lbm/ft^{3}[/tex]

Solution:

As per the question:

Volume of spacecraft component, [tex]V_{s} = 8ft^{3}[/tex]

Mass of the component of spacecraft, [tex]m_{s} = 25 lb[/tex]

Acceleration of gravity at a point on Earth, [tex]g_{E} = 31.0 ft/s^{2}[/tex]

Acceleration of gravity on Moon, [tex]g_{M} = 5.57 ft/s^{2}[/tex]

Now,

The weight of the component, [tex]w_{c} = m_{c}\times \frac{g_{M}}{g_{E}}[/tex]

[tex]w_{c} = 25\times \frac{5.57}{31.0}[/tex]

[tex]w_{c} = 4.492 lb[/tex]

Now,

Average density, [tex]\rho_{avg}[/tex]

[tex]\rho_{avg} = \farc{w_{s}}{V_{s}}[/tex]

[tex]\rho_{avg} = \farc{4.492}{8} = 0.5614 lbm/ft^{3}[/tex]

Answer:

density=3.125pounds/ft^3

weight=4.35lbf

Explanation:

Density is a property of matter that indicates how much mass a body has in a given volume.

It is given by the following equation.

ρ=m/v   (1)

where

ρ=density

m=mass

v=volume

The weight on the other hand is the force which the earth (or the moon) attracts to a body with mass, this force is given by the following equation

W=mg (2)

W=weight

m=mass

g=gravity

to solve this problem we have to calculate the mass of the component using the ecuation number 2

W=mg

m=w/g

w=25lbf=804.35pound .ft/s^2

g=32.2ft/s^2

m=804.35/32.2=25 pounds

density

ρ=m/v   (1)

ρ=25pounds/8ft^3

ρ=3.125pounds/ft^3 =density

weight in the moon

W=mg

W=(25pounds)(5.57ft/s^2.)=139.25pound .ft/s^2=4.35lbf

An unknown immiscible liquid seeps into the bottom of an openoil
tank. Some measurements indicate that the depth of theunkown liquid
is 1.5m and the depth of the oil ( specific weight =8.5
kN/m3) floating on top is 5m. A pressure gageconnected
to the bottom of the tank reads 65 kPa. What is thespecific gravity
of the unkown liquid?

Answers

Answer:

1.53

Explanation:

Given:

depth of the unknown liquid  = 1.5m

the depth of the oil floating on top = 5m

specific weight of oil, γ = 8.5 kN/m³

Total pressure at the bottom = 65 kPa = 65 kN/m²

let the specific weight of the unknown liquid be " γ' "

Now,

The total pressure

= Pressure due to the unknown liquid + Pressure due to floating liquid

or

65 = γ' × 1.5 + γ × 5

or

65 = γ' × 1.5 + 8.5 × 5

or

22.5 = γ' × 1.5

or

γ' = 15 kN/m³

Also,

Specific gravity = [tex]\frac{\textup{Specific weight of unknown liquid}}{\textup{Specific weight of water}}[/tex]

specific weight of water = 9.81 kN/m³

or

Specific gravity = [tex]\frac{15}{\9.81}[/tex]  = 1.53

Final answer:

To calculate the specific gravity of the unknown liquid, you use the pressure reading from the tank, along with the height and specific weight of the oil, and then compare it to the specific weight of water.

Explanation:

The question revolves around finding the specific gravity of an unknown liquid using a pressure gauge reading at the bottom of a tank containing both oil and the unknown liquid. To find the specific gravity, we use the equation for pressure caused by a static fluid column, which is P = h⋅γ, where P is pressure, h is the height of the fluid column, and γ is the specific weight of the fluid. Based on the given pressure reading and the specific weight of the oil, we can determine the specific weight of the unknown liquid. Once we have that, the specific gravity is found by dividing the specific weight of the unknown liquid by the specific weight of water (9.81 kN/m3, since the specific weight of water is its density (1000 kg/m3) times the acceleration due to gravity (9.81 m/s2)).

The equation to find the specific gravity (SG) of the unknown liquid is SG = γunknown / γwater, where γunknown is the specific weight of the unknown liquid calculated from the pressure reading at the bottom of the tank, and γwater is the specific weight of water.

A car is accelerated 5.5 ft/s^2. Calculate the initial velocity v, the car must have if it is to attain a final velocity v of 45 mph in a distance of 352 ft. Compute also the time t required to attain that final velocity.

Answers

Answer:

1) The initial velocity must be 22 feet/sec.

2) The time required to change the velocity is 8 seconds.

Explanation:

This problem can be solved using third equation of kinematics

According to the third equation of kinematics we have

[tex]v^{2}=u^{2}+2as[/tex]

where

'v' is the final velocity of object

'u' is the initial velocity of object

'a' is acceleration of the object

's' is the distance in which this velocity change is brought

Since the final velocity is 45 mph converting this to feet per seconds we get

45 mph =66 feet/sec

Applying Values in the above equation and solving for 'u' we get

[tex]66^{2}=u^{2}+2\times 5.5\times 352\\\\u^{2}=484\\\\\therefore u=22feet/sec[/tex]

2)

The time required to attain this velocity can be found using first equation of kinematics as

[tex]v=u+at[/tex]

where

't' is the time in which the velocity change occurs

'v', 'u', 'a' have the usual meaning

Thus applying given values we get

[tex]66=22+5.5\times t\\\\\therefore t=\frac{66-22}{5.5}=8seconds[/tex]

Applying the given values we get

A carpenter uses a hammer to strike a nail. Approximate the hammer's weight of 1.8lbs, as being concentrated at the head, and assume that at impact the head is traveling in the -j direction. If the hammer contacts the nail at 50mph and the impact occurs over 0.023 seconds, what is the magnitude of the average force exerted by the nail of the hammer?

Answers

Answer:

The average force F exherted by the nail over the hammer is 178.4 lbf.

Explanation:

The force F exherted by the nail over the hammer is defined as:

F = |I|/Δt

Where I and Δt are the magnitude of the impact and the period of time respectively. We know that the impact can be calculated as the difference in momentum:

I = ΔP = Pf - Pi

Where Pf and Pi are the momentum after and before the impact. Recalling for the definition for momentum:

P = m.v  

Where m and v are the mass and the velocity of the body respectively. Notice that final hummer's momentum is zero due to the hammers de-acelerate to zero velocity. Then the momentum variation will be expressed as:

ΔP =  - Pi = -m.vi

The initial velocity is given as 50 mph and we will expressed in ft/s:

vi = 50 mph * 1.47 ft/s/mph = 73.3 ft/s

By multiplyng by the mass of 1.8 lbs, we obtain the impulse I:

|I|= |ΔP|= |-m.vi| = 1.8 lb  * 73.3 ft/s = 132 lb.ft/s

Dividing the impulse by a duration of 0.023 seconds, we finally find the force F:

F =  132 lb.ft/s / 0.023 s = 5740 lb.ft/s^2  

Expressing in lbf:

F = 5740 lb.ft/s^2 * 0.031 lbf/lb.ft/s^2  =  178.4 lbf

Gas is kept in a 0.1 m diameter cylinder under the weight of a 100 kg piston that is held down by a spring with a stiffness k = 5 kN / m. If the gauge pressure of the gas is 300 kPa, how much is the spring compressed?

Answers

Answer:

The spring is compressed by 0.275 meters.

Explanation:

For equilibrium of the gas and the piston the pressure exerted by the gas on the piston should be equal to the sum of  weight of the piston and the force the spring exerts on the piston

Mathematically we can write

[tex]Force_{pressure}=Force_{spring}+Weight_{piston}[/tex]

we know that

[tex]Force_{pressure}=Pressure\times Area=300\times 10^{3}\times \frac{\pi \times 0.1^2}{4}=750\pi Newtons[/tex]

[tex]Weight_{piston}=mass\times g=100\times 9.81=981Newtons[/tex]

Now the force exerted by an spring compressed by a distance 'x' is given by [tex]Force_{spring}=k\cdot x=5\times 10^{3}\times x[/tex]

Using the above quatities in the above relation we get

[tex]5\times 10^{3}\times x+981=750\pi \\\\\therefore x=\frac{750\pi -981}{5\times 10^{3}}=0.275meters[/tex]

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