A 400 kg machine is placed at the mid-span of a 3.2-m simply supported steel (E = 200 x 10^9 N/m^2) beam. The machine is observed to vibrate with a natural frequency of 9.3 HZ. What is the moment of inertia of the beam's cross section about its neutral axis?

Answer:

Explanation:

The only vertical force that can be applied at joint D is that of link CD. Since joint D is stationary, there must be no vertical force. Hence the force in link CD must be zero, as must the force in link DE.

At joint E, the only horizontal force is that applied by link EF, so it, too, must be zero.

Then link CE has 15 N tension.

The downward force in CE must be balanced by an upward force in CF. Of that force, only 1/√2 of it will be vertical, so the force in CF is a compression of 15√2 N.

In order for the horizontal forces at C to be balanced the 15 N horizontal compression in CF must be balanced by a 15 N tension in BC.

At joint F, the 15 N horizontal compression in CF must be balanced by a 15 N compression in FA. CF contributes a downward force of 15 N at joint F. Together with the external load of 10 N, the total downward force at F is 25 N. Then the tension in BF must be 25 N to balance that.

At joint B, the 25 N downward vertical force in BF must be balanced by the vertical component of the compressive force in AB. That component is 2/√5 of the total force in AB, which must be a compression of 25√5/2 N.

The horizontal forces at joint B include the 15 N tension in BC and the 25/2 N compression in AB. These are balanced by a (25/2+15) N = 55/2 N tension in BG.

In summary, the link forces are ...

_____

Note that the forces at the pins of G and A are in accordance with those that give a net torque about those point of 0, serving as a check on the above calculations.

Answer:

X = 1 (1st digit in the code)

Y = 0 (2nd digit)

Z = 4 (3rd multiplier digit)

104 → 10 × 10^4 Ω

→ 10 × 10000Ω

→ 100 kΩ

resistors are marked 104, 105, 205, 751, and 754. The resistor marked with 104 should be 100kΩ (10x10^4), 105 would be 1MΩ (10x10^5), and 205 is 2MΩ (20x10^5). 751 is 750Ω (75x10^1), and 754 is 750kΩ (75x10^4).

Here we need to understand how a code in a resistor gives us information on the resistor. Here we will see that the code means that the resistance is 100,000 Ω.

When we use numbers, let's assume that we have 3 single-digit numbers abc.

So if the code in our resistor is abc, this will mean that the resistance of the resistor is:

ab×10^c Ω

Using this general rule we can see that if the code is 104, then the resistance will be:

r = 10×10^4 Ω

 = 100,000 Ω

Then we can conclude that the resistive value of this resistor is  100,000 Ω

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Answer:

A) 251.8 kj/kg

B) 0.9150 kj/kg-k

C) 155.4 kj/kg

F) 1.50

G) 3.95 kw

H) 2.6 kw

Explanation:

Given conditions :

air conditioner : R -134a

compressor efficiency (nc) = 90%.

T1 = 9⁰c,  T3 = 39⁰c, mass flow rate = 0.027 kg/s

A) Specific enthalpy at the compressor inlet

at T = 9⁰c the saturated vapor (x) = 1

from the R-134a property table

h1 = 251.8 kj/kg

B ) specific entropy ( kj/kg-k) at the compressor inlet

at T = 9⁰c the saturated vapor (x) = 1

s = 0.9150 kj/kg-k ( from the R-134a property table )

C) specific enthalpy at the compressor exit

at T3 = 39⁰c , s2 = s1

has = 165.12 kj/kg

h2 = 155.4 kj/kg

attached below is the remaining solution to some of the problems

Answer:

C) a winding connected in series with the armature.

Explanation:

In a series motor, an electromagnet is used as a stator to generate its magnetic field. The field coil of this stator are connected through a commutator in series with the rotor windings. This stator which is the armature windings will conduct AC even on a DC machine, due to the  periodically reverses current direction (commutation) or due to electronic commutation (as in brushless DC motors).

Answer: the half-angle "alpha" of the Mach cone = 30⁰

Explanation:

To calculate the  half-angle "alpha" of the Mach cone.

we say ;

Sin∝ = 1 / Ma

given that Ma = 2

now we substitute

Sin∝ = 1 / 2

Sin∝ = 0.5

∝ = Sin⁻¹ 0.5

∝ = 30⁰

Therefore, the half-angle "alpha" of the Mach cone is 30⁰

Answer:

D) is due to an increase of the armature current.

Explanation:

Option D is correct because on the DC motor, when the load increases, it leads to an increase in the armature current.

The armature current then sets up a magnetic flux which opposes the main field flux. The net field flux gets reduced. It is at this point, the armature reaction occurs.

Armature reaction is seen as the effect of magnetic flux which is usually set up by an armature current. This occurs when there is the distribution of flux under the main poles.

There are two effects the armature flux causes on the main field flux.

1. The main field flux is distorted by the armature reaction.

2. The magnitude of the main field flux is reduced by the armature flux.

Answer:

There are five (5) unknown support reactions in this problem.

Explanation:

A roller joint rotates and translates along the surface on which the roller rests. The resulting reaction force is always a single force that is perpendicular to, and away from, the surface. This allows the roller to move in a single plane along the surface where it rests.

A cable support provides support in one direction, parallel, and in opposite direction to the load on it. There exists a single reaction from the cable pointed upwards.

A ball-and-socket joint have  reaction forces in all 3 cardinal  directions. This allows it to move in the x-y-z plane.

The total unknown reactions on the member are five in number.

Answer:

A) 13.80 ft/s^2

B) 1.714 Ib

Explanation:

Magnitude of acceleration center G

mass = W / g = 8 / 32.2 = 0.2484 Ib.s^2/ft

calculate the acceleration along x direction

A = ra

r = radius

a = angular acceleration

A = 6 in [tex]\frac{1 ft}{12 in}[/tex] * a

a= 2A

equation of the plane along the x-direction

w sin∅ - F = ma

8* sin40 - F = 0.2484 * a

hence F = 5.1423 - 0.2484 a

next find the moment of inertia along the z axis

I = 1/2 mr^2

  = 1/2 * 0.2484 * (6/12)^2  = 0.03105 Ib.ft.s^2

Applying moment balance equation

F * r = inertia * a

(5.1423 - 0.2484 a)*0.5 = 0.03105 * 2A

2.57115 = 0.1863 A      hence

A = 13.80 ft/s^2  ( acceleration of the cylinder )

B) Calculate the friction force exerted by the ramp on the cylinder

F = 5.1423 - 0.2484 A

  = 5.1423 - 0.2484 ( 13.80 )

  = 1.714 Ib

The magnitudes of the acceleration and the friction force are;

Acceleration = 13.8 ft/s²

Friction Force = 1.714 lb

The image of the solid homogeneous cylinder is missing and so i have attached it.

From the image we see that;

We are given;

      We know that formula for weight is; W = mg

Thus; m = W/g

where g is acceleration due to gravity = 32.2 ft/s²

m = 8/32.2

mass; m = 0.2484 lb.s²/ft

     Now, to get the acceleration along the x-axis, we will use the formula;

a = rα

where α is angular acceleration. Thus;

a = 0.5α

α = 2a   ----- (eq 1)

    Now, resolving forces along the x-direction gives;

W*sinθ - F = ma

Plugging in the relevant values;

8*sin 40 - F = 0.2484a

F = 8*sin 40 - 0.2484a    -----(eq 2)

     Now, moment of inertia of the cylinder along the z-axis is gotten from;

I = ¹/₂mr²

I = ¹/₂ × 0.2484 × 0.5²

I = 0.03105 lb.ft/s²

     Taking equilibrium of moments we have;

F*r = I*α

Thus;

(8*sin 40 - 0.2484a)0.5 = 0.03105α

⇒ 2.57115 - 0.1242a = 0.03105α

⇒ 0.03105α + 0.1242a = 2.57115

From eq 1, α = 2a. Thus;

0.03105(2a) + 0.1242a = 2.57115

0.1863a = 2.57115

a = 2.57115/0.1863

a = 13.8 ft/s²

F = 8*sin 40 - 0.2484a

F = 5.1423 - 0.2484(13.8)

F = 1.714 lb

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Answer:

pay off the parking tickets

Explanation:

In the scenario being described, the best thing to do would be to pay off the parking tickets. The parking tickets stay under your name, and if they are not paid in time can cause problems down the road. For starters, if they are not paid in time the amount will increase largely which will be harder to pay. If that increased amount is also not paid, then the government will suspend your licence indefinitely which can later lead to higher insurance rates.

The building-construction materials are materials with a relatively large bulk modulus, would the choose of the constructor. Thus, option (e) is correct.

The term building refers to the large structure of the bricks. The building was the large structure made by the engineer and the architecture. There was the building are the people are the stay. The building was the used in the permanent basis. There are the different types of the building.

According to the building-construction materials are the engineers to the order of the bulk in the material. The bulk level of the material is the high discount rate of the order. The large amount of the material to the used in the building-construction, that are the order in the bulk of the production.

Therefore, option (e) is correct.

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Your question is incomplete, but most probably the full question was.

Answer:

current is measured in Ampere (A)

Answer:

Ampere (A)

Explanation:

The ampere is defined so the elementary charge e is 1.602 176 634 × 10−19 C or A•s.

Hope this helps <3

Answer:

[tex]w_{out}=319.1\frac{BTU}{lbm}[/tex]

Explanation:

Hello,

In this case, for the inlet stream, from the steam table, the specific enthalpy and entropy are:

[tex]h_1=1456.0\frac{BTU}{lbm} \ \ \ s_1=1.6413\frac{BTU}{lbm*R}[/tex]

Next, for the liquid-vapor mixture at the outlet stream we need to compute its quality by taking into account that since the turbine is adiabatic, the entropy remains the same:

[tex]s_2=s_1[/tex]

Thus, the liquid and liquid-vapor entropies are included to compute the quality:

[tex]x_2=\frac{s_2-s_f}{s_{fg}}=\frac{1.6313-0.39213}{1.28448}=0.965[/tex]

Next, we compute the outlet enthalpy by considering the liquid and liquid-vapor enthalpies:

[tex]h_2=h_f+x_2h_f_g=236.14+0.965*933.69=1136.9\frac{BTU}{lbm}[/tex]

Then, by using the first law of thermodynamics, the maximum specific work is computed via:

[tex]h_1=w_{out}+h_2\\\\w_{out}=h_1-h_2=1456.0\frac{BTU}{lbm}-1136.9\frac{BTU}{lbm}\\\\w_{out}=319.1\frac{BTU}{lbm}[/tex]

Best regards.

The question is incomplete. The complete question is :

The solid rod shown is fixed to a wall, and a torque T = 85N?m is applied to the end of the rod. The diameter of the rod is 46mm .

When the rod is circular, radial lines remain straight and sections perpendicular to the axis do not warp. In this case, the strains vary linearly along radial lines. Within the proportional limit, the stress also varies linearly along radial lines. If point A is located 12 mm from the center of the rod, what is the magnitude of the shear stress at that point?

Solution :

Given data :

Diameter of the rod : 46 mm

Torque, T = 85 Nm

The polar moment of inertia of the shaft is given by :

[tex]$J=\frac{\pi}{32}d^4$[/tex]

[tex]$J=\frac{\pi}{32}\times (46)^4$[/tex]

J = 207.6 [tex]mm^4[/tex]

So the shear stress at point  A is :

[tex]$\tau_A =\frac{Tc_A}{J}$[/tex]

[tex]$\tau_A =\frac{85 \times 10^3\times 12 }{207.6}$[/tex]

[tex]$\tau_A = 4913.29 \ MPa$[/tex]

Therefore, the magnitude of the shear stress at point A is 4913.29 MPa.

Answer:

Correct Answer:

A. water pump

Explanation:

Timing belt in a vehicle helps to ensure that crankshaft, pistons and valves operate together in proper sequence. Timing belts are lighter, quieter and more efficient than chains that was previously used in vehicles.

Most car manufacturers recommended that, when replacing timing belt, tension assembly, water pump, camshaft oil seal should also be replaced with it at same time.

Answer: hello attached below is the diagram which is part of your question

Total entropy change  = entropy change in cold reservoir + entropy change in hot reservoir = -0.166 + 0.083 = -0.0837 kj/k  it violates Clausius increase of entropy which is Sgen > 0

Explanation:

Clausius statement states that it is impossible to transfer heat energy from a cooler body to a hotter body in a cycle or region without any other external factors affecting it .  

applying the increase in entropy principle to prove this

temp of cold reservoir (t hot)= 600 k

temp of hot reservoir(t cold) = 1220 k

energy (q) = 100 kj

total entropy change  = entropy change in cold reservoir + entropy change in hot reservoir = -0.166 + 0.083 = -0.0837 kj/k

entropy change in cold reservoir = Q/t cold = 100 / 600 = -0.166 kj/k

entropy change in hot reservoir = Q / t hot = 100 / 1220 = 0.083 kj/k

hence it violates  Clausius inequality of increase of entropy principle which is states that generated entropy has to be > 0

Answer:

Clay bricks are manufactured by mining and clay moulded blocks. There are 20,000 brick klins in Pakistan.

Explanation:

In Pakistan, the clay bricks are manufactured by mining and baking the clay moulded blocks in brick kilns. According to an estimate, the baking process emits about  1.4 pounds of carbon per brick made, but in Pakistan, because the systems are outdated, brick kilns are used, which is producing more than the average amount of the pollution.

There are about 20,000 brick kilns in Pakistan. The traditional brick production in Pakistan is consists of hand-made bricks which are first baked in Fixed Chimney Bull's Trench Kilns (FCBTK), this is the most widely used brick firing technology in South Asia.

Answer:

thread depth = 2.5 mm

thread width = 2.5 mm

mean diameter = 27.5 mm

root diameter = 25 mm

lead of screw = 5 mm

Explanation:

given data

power screw diameter D = 30 mm

thread pitch  P = 5 mm

solution

First, we get here thread depth fr square thread

thread depth = [tex]\frac{P}{2}[/tex]   ......................1

thread depth = [tex]\frac{5}{2}[/tex]

thread depth = 2.5 mm

and

thread width for square thread

thread width = [tex]\frac{P}{2}[/tex]   ......................2

thread width = [tex]\frac{5}{2}[/tex]

thread width = 2.5 mm

and

mean diameter is

mean diameter = D - [tex]\frac{P}{2}[/tex]    ................3

mean diameter = 30 - [tex]\frac{5}{2}[/tex]

mean diameter = 27.5 mm

and

root diameter is

root diameter = D - P   ....................4

root diameter = 30 - 5

root diameter = 25 mm

and

lead of screw for single thread so n = 1

so lead of screw = 1 × 5

lead of screw = 5 mm

Answer:

Explanation:

BCC structure can also be called Body-Centred Cubic. In BCC Structure, single atom is located at every corner of the cube, and one is also located at the centre of the structure.

Each of the four cube corners through which this (110) plane travels has an atom, which it shares with four neighboring unit The percentage of the entire crystallographic plane area that is occupied by atoms is known as planar density.

The packing density of crystals is measured by planar density. A face-centered cubic unit cell's planar density may be determined by following a few straightforward procedures. Determine how many atoms are centered on a certain plane. An FCC crystal, for instance, has 2 atoms on a (1 1 0) plane.

The amount of atoms per unit area on an interest plane defines the planar density, a crucial aspect of a crystal structure.

The percentage of the entire crystallographic plane area that is occupied by atoms is known as planar density. Atomic packing factor analogs in one and two dimensions are linear and planar densities.

Thus, The center atom, however, is totally contained within the unit cell. This means that this BCC (110) plane has an equivalent of 2 atoms.

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Answer:

flow chart

Explanation:

it's a flow chart right ? don't know the explanation cause I am not even sure I am right

Answer: answer provided in the explanation section.

Explanation:

Weather phenomenons that would impart Aviation Operations in Santa Barbara -

1. Although winters are cold, wet, and partly cloudy here. It is in general favorable for flying. But sometimes strong winds damage this pleasant weather.

2.  The Sundowner winds cause rapid warming and a decrease in relative humidity. The wind speed is very high surrounding this area for this type of wind.  

3. Cloud is an important factor that affects aviation operations. Starting from April, here the sky is clouded up to November. The sky is overcast (80 to 100 percent cloud cover) or mostly cloudy (60 to 80 percent) 44% on a yearly basis. Thus extra cloud cover can trouble aviation operations.

4. The average hourly wind speed can also be a factor. This also experiences seasonal variations, these variations are studied carefully in the aviation industry. The windier part of the year starts in January and ends in June. In April, the wind speed can reach 9.5 miles per hour.

This and more are some factors to look into when considering wheather conditions that would affect aviation operations.

I hope this was a bit helpful. cheers

Answer:

True

Explanation:

There are 16 in all and 4 are not mandatory

There are some sections of the SDS that are not mandatory. Thus, option A is correct.

A Safety Data Sheet, previously known as a Material Safety Data Sheet, is a thorough instructional report written by a dangerous product's producer or exporter. It gives a description of something like the product's chemical and physical characteristics.

Sections 1 through 11 and 16 must be used; sections 12 through 15 are optional but could be included. the section which will are not compulsory are:

Section 12: Environmental data

Section 13: Guidelines for Recycling

Section 14: Documentation regarding mobility

Section 15: Relevant data about regulations

In Safety Data Sheet, all the sections are not compulsory or mandatory. this statement is true. Therefore, option A is the correct option.

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Answer:

D.  Resistance = 30 ohms

Explanation:

Using Ohm's law

V = I times R

Given:

V = 60 V

I = 2 A

Resistance = V / I = 60 V / 20 A

Resistance = 30 ohms

Answer:

I don't know

............................

The output will be an error message since the input() function in Python always takes input as a string, even if the entered value is an integer like 12.

Here,

It cannot directly concatenate a string with an integer.

In order to fix this, we need to convert the input string into an integer using the int() function,

```

x = int(input())

x = x + "abc" # this will give an error message

```If we fix the code, the correct output will be:

`>>> x = int(input())`

`12`

`>>> x + "abc"`

`TypeError: unsupported operand type(s) for +: 'int' and 'str'`

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Answer:

hugdxvb huiiiuttyuj ftfrr

Answer:

Explanation:

Temperature of atmospheric air To = 25°C = 298 K

Free  stream velocity of air Vo = 25 m/s

Length and width of plate = 1m

Temperature of plate Tp = 125°C = 398 K

We know for air, Prandtl number Pr = 1

And for air, thermal conductivity K = 24.1×10?³ W/mK

Here, charectorestic dimension D = 1m

Given value of Reynolds number Re = 105

For laminar boundary layer flow over flat plate

= 3.402

Therefore, hx = 0.08199 W/m²K

So, heat transfer rate q = hx×A×(Tp – To)

                                          = 0.08199×1×(398 – 298)