g 940 The beam AB has a negligible mass and thickness and is subjected to a triangular distributed loading. It is supported at one end by a pin and at the other end by a post having a mass of 50 kg and negligible thickness. Determine the two coefficients of static friction at B and at C so that when the magnitude of the applied force is increased to P

Answer:

Andy's glasses are more likely to fog up because the cold air is being conducted by the hot air somthing like that

Answer:

The normal stress is 0.7821 MPa

Explanation:

The external diameter D = 160 mm

The thickness t = 3.8 mm = 3.8 x 10^-3 m

gauge pressure P = 78 kPa = 78 x 10^3 Pa

The maximum shear stress τmax = ?

The external radius of the shell from the external surface R = D/2 = 160/2 = 80 mm

The internal radius of the shell r = R - t

==> 80 - 3.8 = 76.2 mm

Therefore the internal diameter d = 2r = 2 x 76.2 = 152.4 mm

==> d = 152.4 x 10^-3 m

The normal stress σ = [tex]\frac{Pd}{4t}[/tex] = [tex]\frac{78*10^{3}*152.4*10^{-3} }{4*3.8*10^{-3} }[/tex] = 782052.63 Pa

==>  σ = 0.7821 MPa

Answer:

Explanation:

If the museum is running short of funds, and you decide to increase revenue. An increase or decrease in the price of admission into the museum depends on the following:

1. If demand for admission into the museum is elastic there are two possible outcomes

a. An increase in the price of admission leads to a decrease in the quantity demand of admission into the museum

b. A decrease in price of admission into the museum leads to an increase in the quantity demand of admission into the museum.

This follows the law of demand which states that "the higher the price, the lower the quantity demanded and the lower the price, the higher the quantity demanded".

2. If the demand for admission into the museum is inelastic, then an increase in price will lead to an increase in revenue of the museum.

Therefore, before the curator increase the price of admission into the museum, he should first determine the price elasticity of demand of the museum.

Answer:

59°C

Explanation:

Given that, Cc = McCp,c = 5000 /3600 × 4178 = 5803.2(W/K)

and Ch = MhCp,h = 10000 / 3600 × 4188 = 11634.3(W/K)

Therefore the minimum and maximum heat capacities are:

Cmin = Cc = 5803.2(W/K)

Cmax = Ch = 11634.3(W/K)

The capacity ratio is:

Cr = Cmin / Cmax = 0.499 = 0.5

The maximum possible heat transfer rate is:

Qmax = Cmin (Th,i - Tc,i) = 5803.2 (80 - 20) = 348192(W)

And the number of transfer units is: NTU = UA / Cmin = 11600 / 5803.2 = 1.99

Given that from the appropriate graph in the handouts we can read  = 0.7. So the actual heat transfer rate is: Qact = Qmax = 0.7 × 348192 = 243734.4(W)

Hence, the outlet hot temperature is: Th,o = Th,i - Qact / Ch = 59°C

Answer:

0.285 lbm per cubic feet

Explanation:

length of tank = 50 inches = 50/12 feet = 4.17 feet

diameter of tank = 16 inches = 16/12 feet = 1.33 feet

weight of water = 1.65 lbm

density of water = ?

We know that the density of a substance is given as

ρ = w/v

where ρ is the density in Ibm per cubic feet

w is the weight in lbm

v is the volume in cubic feet

Volume of a cylinder = [tex]\frac{\pi d^{2} l}{4}[/tex]

where d is the diameter

l is the length

volume = [tex]\frac{3.142*1.33^{2}* 4.17}{4}[/tex] = 5.79 cubic feet

Therefore, the density of water will be

ρ = w/v = 1.65/5.79 = 0.285 lbm per cubic feet

The density of water will be "0.285 lbm per cubic feet".

According to the question :

Length of tank = 50 inches or,

                        = [tex]\frac{50}{12}[/tex]

                        = 4.17 feet

Diameter of tank = 16 inches or,

                            = [tex]\frac{16}{12}[/tex]

                            = 1.33 feet

Weight of water = 1.65 lbm

We know the formula,

Volume of cylinder,

= [tex]\frac{\pi d^2 l}{2}[/tex]

By substituting the values, we het

= [tex]\frac{3.142\times (1.33)^2\times 4.17}{4}[/tex]

= 5.79 cubic feet

hence,

The density of water will be:

→ ρ = [tex]\frac{w}{v}[/tex]

By substituting the values,

      = [tex]\frac{1.65}{5.79}[/tex]

      = 0.285 lbm/cubic feet

Thus the above answer is correct.  

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Answer:

(a) The magnitude of force is 116.6 lb, as exerted by the rod CD

(b) The reaction at A is (-72.7j-38.1k) lb and at B it is (37.5j) lb.

Explanation:

Step by step working is shown in the images attached herewith.

For this given system, the coordinates are the following:

A(0, 0, 0)

B(26, 0, 0)

And the value of angle alpha is 20.95°

Hope that answers the question, have a great day!

Answer:

(a) The value of the resistance is 3.431 Ω

(b) The value of the inductance is 0.0264 H

Explanation:

Given;

frequency of the source, f = 60 Hz

rms voltage, V-rms = 220 V

real power, Pr = 1500 W

apparent power, Pa = 4600 VA

(a). Determine the value of the resistance

[tex]P_r = I_{rms}^2R[/tex]

where;

R is resistance

[tex]I_{rms} = \frac{Apparent \ Power}{V_{rms}} \\\\I_{rms} = \frac{P_a}{V_{rms}}\\\\I_{rms}= \frac{4600}{220} \\\\I_{rms}= 20.91 \ A[/tex]

Resistance is calculated as;

[tex]R = \frac{P_r}{I_{rms}^2} \\\\R = \frac{1500}{(20.91)^2} \\\\R = 3.431 \ ohms[/tex]

(b). Determine the value of the inductance.

[tex]Q_L = I_{rms}^2 X_L[/tex]

where;

[tex]Q_L[/tex] is reactive power

[tex]X_L[/tex] is inductive reactance

[tex]Apparent \ power = \sqrt{Q_L^2 + P_r^2} \\\\P_a^2 = Q_L^2 + P_r^2\\\\Q_L^2 = P_a^2 - P_r^2\\\\Q_L^2 = 4600^2 - 1500^2\\\\Q_L^2 = 18910000\\\\Q_L = \sqrt{18910000}\\\\Q_L = 4348.56 \ VA[/tex]

inductive reactance is calculated as;

[tex]X_L = \frac{Q_L}{I_{rms}^2} \\\\X_L = \frac{4348.56}{(20.91)^2} \\\\X_L = 9.95 \ ohms[/tex]

inductance is calculated as;

[tex]X_L = \omega L\\\\X_L = 2\pi f L\\\\L = \frac{X_L}{2\pi f} \\\\L = \frac{9.95}{2\pi *60} \\\\L = 0.0264 \ H\\\\L = 26.4 \ mH[/tex]

Answer:

Explanation:

30 we know that radius is 18 and the circumference is 36pi and the time to go around is is 36pi/30=1.2pi≈3.76991118

Answer:

See calculation below

Explanation:

Given:

W = 30 kN = 30x10³ N

d = 75 mm

p = 6 mm

D = 300 mm

μ = tan Φ = 0.2

1. Force required at the rim of handwheel

Let P₁ = Force required at the rim of handwheel

Inner diameter or core diameter of the screw = dc = do - p = 75 - 6 = 69 mm

Mean diameter of screw:    *d = [tex]\frac{do + dc}{2}[/tex] = (75 + 69) / 2 = 72 mm

and

tan α = p / πd  =  6 / (π x 72)  =  0.0265

∴ Torque required to overcome friction at he threads is  T = P x d/2

T = W tan (α + Ф) d/2

T =  [tex]W(\frac{tan \alpha + tan \theta}{1 - tan \alpha + tan \theta } ) * \frac{d}{2}[/tex]

T = 30x10³ * ((0.0265 + 0.2) / (1 - 0.0265 x 0.2)) x 72/2

T = 245,400 N-mm

We know that the torque required at the rim of handwheel (T)

245,400 = P1 x D/2 = P1 x (300/2) = 150 P1

P1 = 245,400 / 150

P1 = 1636 N

2. Maximum compressive stress in the screw

                         30x10³

Qc = W / Ac = -------------- = 8.02 N/mm²

                      π/4 * 69²

Qc = 8.02 MPa

Bearing pressure on the threads (we know that number of threads in contact with the nut)

n = height of nut / pitch of threads = 150 / 6 = 25 threads

thickness of threads, t = p/2 = 6/2 = 3 mm

bearing pressure on the threads = Pb = W / (π d t n)

Pb = 30 x 10³ / (π * 72 * 3 * 25)

Pb = 1.77 N/mm²

Max shear stress on the threads = τ = 16 T / (π dc³)

τ = (16 * 245,400) / ( π * 69³ )

τ = 3.8 M/mm²

*the mean dia of the screw (d) = d = do - p/2 = 75 - 6/2 = 72

∴max shear stress in the threads τmax = 1/2 * sqrt(8.02² + (4 * 3.8²))

τmax = 5.5 Mpa

3. efficiency of the straightener

To = W tan α x d/2 = 30x10³ * 0.0265 * 72/2 = 28,620 N-mm

∴Efficiency of the straightener is η =  To / T = 28,620 / 245,400

η = 0.116 or 11.6%

By assuming the coefficient of friction for the collar as 0.2. efficiency of straightner is 11.6%.

Wheels are circular frames or disks that are mounted on machines or vehicles and are designed to rotate around an axis.

Given:

W = 30 kN = 30x10³ N

d = 75 mm

p = 6 mm

D = 300 mm

μ = tan Φ = 0.2

1. Force required at the rim of handwheel :

Let P₁ = Force required at the rim of handwheel

Inner diameter or core diameter of the screw = dc = do - p = 75 - 6 = 69 mm

Mean diameter of screw:    *d =  = (75 + 69) / 2 = 72 mm and

tan α = p / πd  =  6 / (π x 72)  =  0.0265

Torque required to overcome friction at he threads is  T = P x d/2

T = W tan (α + Ф) d/2

T = 30x10³ * ((0.0265 + 0.2) / (1 - 0.0265 x 0.2)) x 72/2

T = 245,400 N-mm

We know that the torque required at the rim of handwheel (T)

245,400 = P1 x D/2 = P1 x (300/2) = 150 P1

P1 = 245,400 / 150

P1 = 1636 N

2. Maximum compressive stress in the screw :

30x10³

Qc = W / Ac = -------------- = 8.02 N/mm²

π/4 * 69²

Qc = 8.02 MPa

Bearing pressure on the threads (we know that number of threads in contact with the nut)

n = height of nut / pitch of threads = 150 / 6 = 25 threads

thickness of threads, t = p/2 = 6/2 = 3 mm

bearing pressure on the threads = Pb = W / (π d t n)

Pb = 30 x 10³ / (π * 72 * 3 * 25)

Pb = 1.77 N/mm²

Max shear stress on the threads = τ = 16 T / (π dc³)

τ = (16 * 245,400) / ( π * 69³ )

τ = 3.8 M/mm²

*the mean dia of the screw (d) = d = do - p/2 = 75 - 6/2 = 72

max shear stress in the threads τmax = 1/2 * sqrt(8.02² + (4 * 3.8²))

τmax = 5.5 Mpa

Therefore, efficiency of the straightener :

To = W tan α x d/2 = 30x10³ * 0.0265 * 72/2 = 28,620 N-mm

Efficiency of the straightener is η =  To / T = 28,620 / 245,400

η = 0.116 or 11.6%

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Answer:

24 hours

Explanation:

you must exchange insurance details after a collision if someone is injured. Otherwise you must report the collision to us as soon as possible (and no later than 24 hours). Although you must report such a collision straight away you should always seek medical help in the first instance.

The red and green lights are sidelights that are positioned on the port side (red) (left as facing the bow) and starboard (green) (right as facing the bow) side of the boat. Various white lights are required depending on the size of the boat, but generally, a white masthead light and stern light are required. See the US Coast Guard site in the link below for more specific information.

Hope this helps

Answer:

Hello the needed relation is missing below is the required relation

[tex]X_{p} = \frac{x_{s} }{1+x_{s} }[/tex]   composition : propane = 0.70, butane = 0.3

Answer : ≈ 5.75 hrs

Explanation:

Applying the data given in regards to the material balance

Butane balance input into the still = 5 mole feed/hr | 0.30 mol butane/molfeed

since the total volume of the liquid in the still is constant

The output from the still is = 5mol condensed/hr | x[tex]_{p}[/tex] mol butane/mol condensed

unsteady state equation = [tex]\frac{dx_{s} }{dt}[/tex] = 0.15 - [tex]0.5X_{p}[/tex]

note : to reduce the equation a single dependent variable we have to substitute for [tex]x_{p}[/tex]

[tex]\frac{dx_{s} }{dt}[/tex] [tex]= 0.15 + x_{s} / 1 + (0.5)x_{s}[/tex]

In order to find the time it will take for X to change from 0.3 to 0.35

integrate the above equation using the limits : t = 0, x[tex]_{s}[/tex] = 0.3 and t = Ф,

x[tex]_{s} = 0.35[/tex]

= [tex][ - (x_{s} /0.35 - (1/(0.35)^2)* In(0.15 - 0.35x_{s} ) ]_{0.3} ^{0.35}[/tex]

hence t = Ф ≈ 5.75 hrs

Answer: 164.2253 MPa

Explanation:

First we find the half internal crack which is  = length of surface crack /2

so α = 8.6 /2 = 4.3mm ( 4.3×10⁻³m )

Now we find the dimensionless parameter using the critical stress crack propagation equation

∝ = K / Y√πα

stress level ∝ = 112Mpa

fracture toughness K = 26Mpa

dimensionless parameter Y = ?

SO working the formula

Y = K / ∝√πα

Y = 26 / 112 (√π × 4.3× 10⁻³)

Y = 1.9973

We are asked to find stress level for internal crack length of  4m

so half internal crack is  = length of surface crack /2

4/2 = 2mm ( 2 × 10⁻³)

from the previous formula critical stress crack propagation equation

∝ = K / Y√πα

∝ = 26 / 1.9973 √(π × 2 × 10⁻³)

∝ = 164.2253 Mpa

Answer:

a) cross sectional area of the core = 0.0187 m²

b) The secondary voltage on no-load = 413 V

c) The primary currency and power factor on no load is 1.21 A and 0.168 lagging respectively.

Explanation:

See attached solution.

Answer:

at  t = 45 s :  

To = 61.7⁰c,  T1 = 55.6⁰c, T2 = 49.5⁰c, T3 = 34.8⁰C

Explanation:

Wall thickness = 0.12 m

thermal diffusivity = 1.5 * 10^-6 m^2/s

Δt ( time increment ) = 300 s

Δ x   = 0.03 m ( dividing wall thickness into 4 parts assuming the system to be one dimensional )

using the explicit finite-difference technique

Detailed solution is attached below  

Answer:

A. Forces that act perpendicular to the surface and pull an object apart exert a tensile stress on the object.

Explanation:

Tensile stress is due to tension forces on a material. Tensile force acts perpendicularly away from the surface of the substance. The pull on the material due to the tensile force exerts tensile stress on the material, that tends to pull the material apart. The magnitude of the tensile stress is given as

σ = [tex]\frac{P}{A}[/tex]

where σ is the tensile stress

P is the tensile force pulling the material apart

A is the cross-sectional area through which the tensile force acts perpendicularly.

Answer:

26 feet and longer boats that have garbage dumping placard must be prominently posted and the boats which are 40 feet and longer must have the written waste management plan.

Answer:

Voltage across resistor = 2 v

Explanation:

Given data

pulse height = 2 v

pulse width = 4s

calculate voltage across resistor ( the free hand sketch attached below explains more )

pulse height is also = amplitude of voltage ) = 2v

The voltage across the resistor = 2v  Since  the voltage from the source of the circuit is equal to the amplitude voltage in the circuit ( assuming no loss of voltage )

also the graphical representation of the problem is attached below

Answer:

Maneuvering speed.

Explanation:

The speed above which an airplane will experience structural damage when a load is applied, instead of stalling, is called the maneuvering speed and varies with weight.

In aeronautical engineering, the maneuvering speed (Va) of an aircraft such as an aeroplane, helicopter, or jet is an airspeed limitation which is mainly selected by an aircraft designer.

Generally, at speeds higher or greater than the manoeuvring speed, aircraft pilots are advised not to attempt a full deflection of any flight control surface because it's capable of resulting in a damage to the structure of an aircraft.

If you're a pilot, to find the maneuvering speed of an aircraft, you should look at the flight manual of the aircraft or on the cockpit placard in the aircraft. The maneuvering speed of an aircraft is a calibrated speed and should not be exceeded by any pilot.

Answer:

Assuming steady state condition the temperature distribution is calculated as expressed in the attached solution below

Explanation:

Given data :

thickness : L , inner surface (x) : 0,  uniform flux : q"o

fraction : β  

volumetric heat generation :  q˙(x)=(1−β)q''oα^e−αx

determine the temperature distribution in the quartz

attached below is the detailed solution

Hey,

Who plays a role in the financial activities of a company?

O D. Everyone at the company, including managers and employees

Answer:

Everyone at the company, including managers and employees

Explanation:

Q:What velocity does the boy attain if he throws the bricks one at a time?

Answer:Linear velocity since it moves back and firth and does not rotate like angular velocity.

Answer:

no need for that

Explanation:

they are not the same at all

Answer: the thermal conductivity of the sample is 22.4 W/m . °C

Explanation:

We already know that the thermal conductivity of a material is to be determined by ensuring one-dimensional heat conduction, and by measuring temperatures when steady operating conditions are reached.

ASSUMPTIONS

1. Steady operating conditions exist since the temperature readings do not change with time.

2. Heat losses through lateral surfaces are well insulated, and thus the entire heat generated by the heater is conducted through the samples.

3. The apparatus possess thermal symmetry

ANALYSIS

The electrical power consumed by resistance heater and converted to heat is:

Wₐ = VI = ( 110 V ) ( 0.4 A ) = 44 W

Q = 1/2Wₐ = 1/2 ( 44 A )

Now since only half of the heat generated flows through each samples because of symmetry. Reading the same temperature difference across the same distance in each sample also confirms that the apparatus possess thermal symmetry. The heat transfer area is the area normal to the direction of heat transfer. which is the cross- sectional area of the cylinder in this case; so

A = 1/4πD² = 1/3 × π × ( 0.05 m )² = 0.001963 m²

Now Note that, the temperature drops by 15 degree Celsius within 3 cm in the direction of heat flow, the thermal conductivity of the sample will be

Q = kA ( ΔT/L ) → k = QL / AΔT

k = ( 22 W × 0.03 m ) / (0.001963 m² × 15°C )

k = 22.4 W/m . °C

Answer:

A to C = 6.4 km

Explanation:

A to B = 4 km

B to C = 5 km

A to C =  using pythagorean theorem

a² + b² = c²

a = A to B = 4

b = B to C = 5

c = A to C

c² = 4² + 5²

c = 6.4 km (A to C)

According to the scenario, the distance between town C from town A is found to be 6.40 Km.

The background that this question depends on is known as the direction-based question. These types of questions completely depend on the distance of moving bodies like cars, persons, or any other objects as well with respect to the initial position.

According to the question,

The distance between town A to town B = 4 km.

The distance between town B to town C = 5 km.

Now, according to the Pythagoras theorem, the distance between town C to town A is as follows:

[tex]AC^2[/tex] = [tex]AB^2 +BC^2[/tex].

[tex]AC^2[/tex] = [tex]4^2+5^2[/tex]

[tex]AC^2[/tex] = 16 + 25 = 41.

AC = √41 = 6.40 km.

Therefore, the distance between town C from town A is found to be 6.40 Km.

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Answer:

19

Explanation:

when you plus it it will give this

the 8kpa and 11mpa are plused

Answer:

A. Breaking; forming; positive.

B. Twice; half.

Explanation:

A. The reaction involves breaking five blue-blue and twenty blue-red bonds and then forming twenty blue-red bonds. Enthalpies for bond breaking are always positive.

B. In the depicted reaction, both reactants and products are assumed to be in the gas phase. There are twice as many molecules of gas in the products, delta S is half for this reaction.

Bond enthalpy refers to the amount of energy that is required to break a mole of a particular bond and it's an endothermic reaction, therefore it is always positive.

However, bond forming is always an exothermic reaction because energy is released, therefore, it is always negative. This simply means that, the breaking of a bond is an endothermic reaction (positive) while the formation of a bond is an exothermic reaction (negative).

Answer:

a. True

Explanation:

Apply the principle of conservation of mass.

and also the expression for the steady flow energy equation.

kinetic and potential energy effects can be neglected.

The given statement by the inventor who is claiming the development of a device that requires no work or heat transfer input yet is able to produce hot and cold air streams at a steady state is definitely false.

Heat transfer may be characterized as a type of process which involves the migration of heat from one object or component to another by numerous mechanisms like conduction, convection, and/or radiation.

The process of heat transfer may occur where there is a temperature difference between two objects exist. It significantly utilizes the mechanism of exchanging thermal energy between two or more physical systems.

According to the concept of physics, no object or thing has the ability to perform its function without the utilization of any source of heat or energy. Then, how it is possible for that device to produce hot and cold air streams at a steady state.

Therefore, the given statement by the inventor is absolutely false.

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