Answer:
the answer would be B motor
A flat plate is subjected to a load of 5KN as shown in figure. The plate material is grey cast iron FFG 200 and the factor of safety is 2.5. Determine the thickness of the plate
Answer:
57.14 N/mm2
Explanation:
if the output of a signal is 36% on and 64% off and repeats itself is it considered periodic
helppppp
Answer:
No, it is not a periodic Signal
Explanation:
No, it is not a periodic Signal
This signal is repeating itself with the fixed on and off values but the major point to note here is that is this signal repeating after a fixed length of time every time. No, such information is provided in the question and hence, this signal cannot be termed as periodic.
Which two technologies were combined to create product life cycle management (PLM) software?
CAD and a database
spreadsheets and graphics
a database and spreadsheets
CAD and spreadsheets
Answer:
CAD and a database
Explanation:
The correct answer is CAD and a database. When American Motors Corportation introduced the Jeep Cherokee, it implemented CAD to increase engineering productivity and combined that with a new communications system.
three ways to advertise for AVID
Answer:
newspaper, radio, televison
Explanation:
had avid in 7th :)
Answer:
1. will help for colcollares.
2. gives you money for studies
3. helps you choose the perfect collage
Help meeeeeeeee plzzzzz need explanation
the picture is blank for me what does it say i can comment the answer plz mark brainlyist
a tensile specimen with a 12mm initial diameter and 50mm gage length reaches maximum load at 90KN and fractures at 70KN
the minimum diameter at fracture is 10mm
determine the engineering stress at maximum load and the true fracture stress.
Answer:
i) 796.18 N/mm^2
ii) 1111.11 N/mm^2
Explanation:
Initial diameter ( D ) = 12 mm
Gage Length = 50 mm
maximum load ( P ) = 90 KN
Fractures at = 70 KN
minimum diameter at fracture = 10mm
Calculate the engineering stress at Maximum load and the True fracture stress
i) Engineering stress at maximum load = P/ A
= P / [tex]\pi \frac{D^2}{4}[/tex] = 90 * 10^3 / ( 3.14 * 12^2 ) / 4
= 90,000 / 113.04 = 796.18 N/mm^2
ii) True Fracture stress = P/A
= 90 * 10^3 / ( 3.24 * 10^2) / 4
= 90000 / 81 = 1111.11 N/mm^2
How should backing plates, struts, levers, and other metal brake parts be cleaned?
Answer: Cleaning of mechanical parts is necessary to remove contaminants, and to avoid clogging of wastes which could restrict the functioning of the machine.
Explanation:
There are different agents used for cleaning different machine instruments to prevent their corrosion and experience proper cleaning.
Backing plates must be dry cleaned using a cotton cloth to remove the dirt, dust or any other dry contaminant.
Struts can be wet cleaned by applying alcoholic solvent.
Levers can be cleaned using a mineral spirit.
Metallic plates can be cleaned using water based solution or water.
The criminal and traffic code requires that a driver must have a valid driver's license in his/her
immediate possession at any time when operating a motor vehicle.
True
False
Answer:
true
Explanation:
the answer is true because if u don't have a valid license when operating a vehicle and you get pulled over you will get in trouble i know this because my parents got in trouble for it once
MAPP gas, natural gas, propane, and acetylene can be used with oxygen to cut metal.
True
False
As part of a heat treatment process, cylindrical, 304 stainless steel rods of 100-mm diameter are cooled from an initial temperature of 500 C by suspending them in an oil bath at 30 C. If a convection coefficient of 500 W/m2 K is maintained by circulation of the oil, how long does it take for the centerline of a rod to reach a temperature of 50 C, at which point it is withdrawn from the bath
Answer:
Explanation:
Given that:
diameter = 100 mm
initial temperature = 500 ° C
Conventional coefficient = 500 W/m^2 K
length = 1 m
We obtain the following data from the tables A-1;
For the stainless steel of the rod [tex]\overline T = 548 \ K[/tex]
[tex]\rho = 7900 \ kg/m^3[/tex]
[tex]K = 19.0 \ W/mk \\ \\ C_p = 545 \ J/kg.K[/tex]
[tex]\alpha = 4.40 \times 10^{-6} \ m^2/s \\ \\ B_i = \dfrac{h(\rho/4)}{K} \\ \\ =0.657[/tex]
Here, we can't apply the lumped capacitance method, since Bi > 0.1
[tex]\theta_o = \dfrac{T_o-T_{\infty}}{T_i -T_\infty}} \\ \\ \theta_o = \dfrac{50-30}{500 -30}} \\ \\ \theta_o = 0.0426\\[/tex]
[tex]0.0426 = c_1 \ exp (- E^2_1 F_o_)\\ \\ \\ 0.0426 = 1.1382 \ exp (-10.9287)^2 \ f_o \\ \\ = f_o = \dfrac{In(0.0374)}{0.863} \\ \\ f_o = 3.81[/tex]
[tex]t_f = \dfrac{f_o r^2}{\alpha} \\ \\ t_f = \dfrac{3.81 \times (0.05)^2}{4.40 \times 10^{-6}} \\ \\ t_f= 2162.5 \\ \\ t_f = 36 mins[/tex]
However, on a single rod, the energy extracted is:
[tex]\theta = pcv (T_i - T_{\infty} )(1 - \dfrac{2 \theta}{c} J_1 (\zeta) ) \\ \\ = 7900 \\times 546 \times 0.007854 \times (500 -300) (1 - \dfrac{2 \times 0.0426}{1.3643}) \\ \\ \theta = 1.54 \times 10^7 \ J[/tex]
Hence, for centerline temperature at 50 °C;
The surface temperature is:
[tex]T(r_o,t) = T_{\infty} +(T_1 -T_{\infty}) \theta_o \ J_o(\zeta_1) \\ \\ = 30 + (500-30) \times 0.0426 \times 0.5386 \\ \\ \mathbf{T(r_o,t) = 41.69 ^0 \ C}[/tex]
Some project managers prefer the PERT chart over the Gantt chart because it clearly illustrates task dependencies. A PERT chart, however, can be much more difficult to interpret, especially on complex projects. Alternatively, some project managers may choose to use both techniques. If you are the project manager of a residential construction project, will you prefer PERT chart to Gantt chart? Explain why?
Answer:
PERT Chart and GANTT Chart
As the project manager of a residential construction project, I will prefer the PERT chart to the GANTT chart because a PERT chart displays task dependencies unlike a Gantt chart. With the PERT chart, the sequence of tasks is clearly mapped out. Dependent tasks are carried out when other tasks that they depend on have been executed.
Explanation:
By definition, a Gantt chart is like a bar chart that lays out project tasks and timelines using bars. On the other hand, a PERT chart follows a structure in the form of flow charts or network diagrams. It displays all the project tasks in separate boxes. The boxes are then connected with arrows which clearly show the task dependencies.
Nate needs to replace the cable to his lamp. He is stripping it to connect it to the termils. What should he remember to do with the knife
Answer: i got you its d
Explanation:had the smae question as you
5. The pin support at A allows _______. Select the one that applies. (a) displacement in the x direction (b) rotation about its central axis (c) displacement in the y direction (d) none of the above 6. The support at B does not allow _______. Select the one that applies. (a) displacement in the x direction (b) rotation about its central axis (c) displacement in the y direction
Answer: Diagram associated with your question is attached below
5) B
6) C
Explanation:
5) The pin support at A allows ; Rotation about its central axis
This is because pin supports does not allow the translation of its structural member in any direction i.e. y or x but only rotation about its axis
6) The support at B does not allow displacement in y direction
This is because roller support allows displacement only in the direction that they are situated and in this case it is the x - direction
1)Saturated steam at 1.20bar (absolute)is condensed on the outside ofahorizontal steel pipe with an inside and outside diameter of 0.620 inches and 0.750 inches, respectively. Cooling water enters the tubes at 60.0°F and leaves at 75.0°F at a velocity of 6.00ft/s. (HINT: You may assume laminar condensate flow.You many also assume that the mean bulk temperature of the cooling water is equal to the wall temperature on the outside of the pipe, T".You may also neglect the viscosity correction in your calculations.)a)What are the inside
Answer:
hi = 7026.8 W/m^2.k
Explanation:
Given data :
pressure of saturated steam = 1.2 bar
Horizontal steel pipe : inside diameter = 0.620 , outside diameter = 0.750 inches
temperature of water at entry = 60°F
temperature of water at exit = 75°F
velocity of water = 6 ft/s
Calculate the Inside convective heat transfer coefficient ( hi )
mean temperature ( Tm ) = 60 + 75 / 2 = 67.5°F ≈ 292.877 K ≈ 19.727°C
next : find the properties of water at this temperature ( 19.727°C )
thermal conductivity = 0.598 w/m.k
density = 1000 kg/m^3
specific heat ( Cp ) = 4.18 KJ/kg.k
viscosity = 0.001 pa.s
velocity of water = 6 ft/s ≈ 1.8288 m/s
∴ Re ( Reynolds number ) = 28712.16
and Prandtl number ( Pr ) = (4180 * 0.001) / 0.598 = 6.989
finally to determine the inside convective heat transfer coefficient we will apply the Dittos - Bolter equation
hi = 7026.8 w/m^2.k
attached below is the remaining solution
Can some one help me with this plumbing question. Even just a guess.
Plz no shady links
Answer:
true
Explanation:
For a bronze alloy, the stress at which plastic deformation begins is 284 MPa and the modulus of elasticity is 106 GPa. (a) What is the maximum load that can be applied to a specimen having a cross-sectional area of 310 mm2 without plastic deformation? (b) If the original specimen length is 120 mm, what is the maximum length to which it may be stretched without causing plastic deformation?
Answer:
a) the maximum load is 88,040 N
b)
the maximum length to which the specimen may be stretched is 0.12032148 mm
Explanation:
Given the data in the question;
the stress at which plastic deformation begins σ = 284 MPa = 2.84 × 10⁸ Pa
modulus of elasticity E = 106 GPa = 1.06 × 10¹¹ Pa
a)
Area A = 310 mm² = 310 × 10⁻⁶ m ( without plastic deformation )
now, lets consider the equation relating to stress and cross sectional area.
σ = F / A₀
hence, maximum load F = σA₀
so we substitute
F = (2.84 × 10⁸) × (310 × 10⁻⁶)
F = 88,040 N
Therefore, the maximum load is 88,040 N
b)
Initial length specimen l₀ = 120 mm = 120 × 10⁻³ m
using engineering strain, ε = (l₁ - l₀)/l₀
Also from Hooke's law, σ = Eε
so from the equation above;
l₁ = l₀( ε + 1 )
l₁ = l₀( σ/E + 1 )
so we substitute
l₁ = (120 × 10⁻³)( (2.84 × 10⁸)/(1.06 × 10¹¹)) + 1 )
l₁ = (120 × 10⁻³) ( 1.002679 )
l₁ = 0.12032148 mm
Therefore, the maximum length to which the specimen may be stretched is 0.12032148 mm
Technician A says when you push the horn button, electromagnetism moves an iron bar inside the horn, which opens and closes contacts in the horn circuit. Technician B says most vehicle horn circuits use a relay. Which technician is correct?
Answer: Both technicians A and B
Explanation:
I took the pf test
Consider a mild steel specimen with yield strength of 43.5 ksi and Young's modulus of 29,000 ksi. It is stretched up to a point where the strain in the specimen is 0.2% (or 0.002). If the specimen is unloaded (i.e. load reduces to zero), the residual strain (or permanent set) is: 0.05% 0.1% 0% 0.2%
Answer:
0.05%
Explanation:
From the question, we have;
The yield strength of the mild steel, [tex]\sigma _c[/tex] = 43.5 ksi
Young's modulus of elasticity, ∈ = 29,000 ksi
The total strain, [tex]\epsilon _c[/tex] = 0.2% = 0.002
The inelatic strain [tex]\epsilon_c^{in}[/tex] is given as follows;
[tex]\epsilon_c^{in}[/tex] = [tex]\epsilon _c[/tex] - [tex]\sigma _c[/tex]/∈
Therefore, we have;
[tex]\epsilon_c^{in}[/tex] = 0.002 - 43.5/(29,000) = 0.0005
Therefore, the inelastic strain, [tex]\epsilon_c^{in}[/tex] = 0.0005 = 0.05%
Taking the inelastic strain as the residual strain, we have;
The residual strain = 0.05%
pacing pieces of information into groups to remember them better is called
a.
Visualizing
c.
Rhyming
b.
Keywording
d.
Categorizing
Please select the best answer from the choices provided
Answer:
D. Categorizing
Explanation:
pls mark me as your brainlist
Answer:
D
Explanation:
You are performing a machining operation that approximates orthogonal cutting. Given that the chip thickness prior to chip formation is 0.5 inches and the chip thickness after separation is 1.125 inches, calculate the shear plane angle and shear strain. Use a rake angle of 10 degrees. 21. Suppose in the prior problem that the cutting force and thrust force were measured as 1559 N and 1271 N, respectively. The width of the cut is 3.0mm. Using this new information, calculate the shear strength of the material.
Answer:
A)
shear plane angle = 31.98°
shear strain = cot ( 31.98° ) + tan ( 31.98 - 10 )
B) shear strength = 7339.78
Explanation:
a) Determine the shear plane angle and shear strain
Given data :
Chip thickness before chip formation = 0.5 inches
Chip thickness after separation = 1.125 inches
rake angle ( ∝ ) = 10°
shear plane angle : Tan ∅ = [tex]\frac{rcos\alpha }{1-sin\alpha }[/tex] ----- ( 1 )
r = chip thickness ratio = 0.5 / 1.125 = 0.4444
back to equation 1 : Tan ∅ = ( 0.444 ) * cos 10 / 1 - sin 10
Tan ∅ = 0.4444 * 0.9848 / 1 - 0.1736 = 0.5296
hence ∅ = tan^-1 ( 0.5296 ) = 31.98°
shear strain : R = cot ∅ + tan ( ∅ - ∝ ) ---------- ( 1 )
R = cot ( 31.98° ) + tan ( 31.98 - 10 )
B) determine the shear strength of the material
cutting force = 1559 N
thrust force = 1271 N
width of cut ( diameter ) = 3.0 mm
shear strength = c + σ.tan ∅
c = cohesion force = 1271 * 3 = 3813
σ = normal stress = F / A = 1559 / π/4 * ( 0.5 )^2 = 1559 / 0.1963 = 7941.94
hence : shear strength of material = 3813 + 7941.94 * 0.6244 = 7339.78
When an arbitrary substance undergoes an ideal throttling process through a valve at steady state, (SELECT ALL THAT APPLY) inlet and outlet mass flowrates will be equal. inlet and outlet specific enthalpies will be equal. inlet and outlet pressures will be equal. inlet and outlet mass flowrates will be equal. inlet and outlet specific enthalpies will be equal. inlet and outlet temperatures will be equal.
Answer:
15x
Explanation:
A cylindrical specimen of this alloy 12 mm in diameter and 188 mm long is to be pulled in tension. Assume a value of 0.34 for Poisson's ratio.Calculate the stress (in MPa) necessary to cause a 0.0105 mm reduction in diameter.
This question is incomplete, the missing image in uploaded along this answer below.
Answer:
The required stress is 200 Mpa
Explanation:
Given the data in the question;
diameter D = 12 mm = 12 × 10⁻³ m
Length L = 188 mm = 188 × 10⁻³ m
Poisson's ratio v = 0.34
Reduction in diameter Δd = 0.0105 mm = 0.0105 × 10⁻³ m
The transverse strain will;
εˣ = Δd / D
εˣ = -0.0105 × 10⁻³ / 12 × 10⁻³ m
εˣ = -0.00088
The longitudinal strain will be;
[tex]E^z[/tex] = - ( εˣ / v )
[tex]E^z[/tex] = - ( -0.00088 / 0.34 )
[tex]E^z[/tex] = - ( - 0.002588 )
[tex]E^z[/tex] = 0.0026
Now, Using the values for strain, we get the value of stress from the graph provided in the question, ( first image uploaded below.
From the graph, in the Second image;
The stress is 200 Mpa
Therefore, The required stress is 200 Mpa
A rectangular block of 1m by 0.6m by 0.4m floats in water with 1/5th of its volume being out of water. Find the weight of the block.
Answer:
Weight of block is 191.424 Kg
Explanation:
The volume of rectangular block = [tex]1*0.6*0.4 = 0.24[/tex] cubic meter
1/5th of its volume being out of water which means water of volume nearly 4/5 th of the volume of rectangular block is replaced
Volume of replaced water = [tex]\frac{4}{5} * 0.24 = 0.192[/tex] cubic meter
Weight of replaced water = weight of rectangular block = [tex]0.192 * 997[/tex] Kg/M3
= 191.424 Kg
what is the answer to life the universe and everything
(worth 95 points!)
Answer:
In the absence of dark energy, a flat universe expands forever but at a continually decelerating rate, with expansion asymptotically approaching zero; with dark energy, the expansion rate of the universe initially slows down, due to the effects of gravity, but eventually increases, and the ultimate fate of the universe ...
Explanation:
I think it goes on forever.
What regulates the car engines temperature
Answer:
car’s thermostat is used to regulate the temperature of the engine to help the engine stay cool.
Explanation:
Answer:
Coolant
Explanation:
4.54 Saturated liquid nitrogen at 600 kPa enters a boiler at a rate of 0.008 kg/s and exits as saturated vapor (see Fig. P4.54). It then flows into a superheater also at 600 kPa, where it exits at 600 kPa, 280 K. Find the rate of heat transfer in the boiler and the superheater.
Answer:
hello the figure attached to your question is missing attached below is the missing diagram
answer :
i) 1.347 kW
ii) 1.6192 kW
Explanation:
Attached below is the detailed solution to the problem above
First step : Calculate for Enthalpy
h1 - hf = -3909.9 kJ/kg ( For saturated liquid nitrogen at 600 kPa )
h2- hg = -222.5 kJ/kg ( For saturated vapor nitrogen at 600 kPa )
second step : Calculate the rate of heat transfer in boiler
Q1-2 = m( h2 - h1 ) = 0.008( -222.5 -(-390.9) = 1.347 kW
step 3 : find the enthalpy of superheated Nitrogen at 600 Kpa and 280 K
from the super heated Nitrogen table
h3 = -20.1 kJ/kg
step 4 : calculate the rate of heat transfer in the super heater
Q2-3 = m ( h3 - h2 )
= 0.008 ( -20.1 -(-222.5 ) = 1.6192 kW
A brass alloy is known to have a yield strength of 275 MPa (40,000 psi), a tensile strength of 380 MPa (55,000 psi), and an elastic modulus of 103 GPa (15.0×106 psi). A cylindrical specimen of this alloy 5.4 mm (0.21 in.) in diameter and 225 mm (8.87 in.) long is stressed in tension and found to elongate 6.8 mm (0.27 in.). On the basis of the information given, is it possible to compute the magnitude of the load that is necessary to produce this change in length? If not, explain why.
Answer:
The magnitude of the load can be computed because it is mandatory in order to produce the change in length ( elongation )
Explanation:
Yield strength = 275 Mpa
Tensile strength = 380 Mpa
elastic modulus = 103 GPa
The magnitude of the load can be computed because it is mandatory in order to produce the change in length ( elongation ) .
Given that the yield strength, elastic modulus and strain that is experienced by the test spectrum are given
strain = yield strength / elastic modulus
= 0.0027
Assume, X Company Limited (XCL) is one of the leading 4th generation Life Insurance
Companies in Bangladesh. The Company is fully customer focused. This Life insurance company are
experimenting with analysis of consumer profiles (to determine whether a person eats healthy food,
exercises, smokes or drinks too much, has high-risk hobbies, and so on) to estimate life expectancy.
Companies might use the analysis to find populations to market policies to. From the perspective of
privacy, what are some of the key ethical or social issues raised? Evaluate some of them.
Answer:
The issues related to the privacy are:
1. Informational privacy
2. Discrimination factors
3. Biased grouping on the basis of Data mining
4. Lack of consent
5. Morally wrong
6. Illegal distribution of information risks
7. Possibility of threat to life
Let's look at some major concerns:
1. Informational privacy : The concept of privacy of the personal information is totally nullified when the information is being used for a purpose other than the intended one for which it was given. This unethical use of information even for general purposes is not correct and is a matter of concern. It is more like using the sensitive data of others for personal benefit which is purely objectionable and raises security issues. Sometimes the data is also shared with the potential employers which might have certain impacts we are unaware of.
2. Data mining issues : The process of using a certain information to arrive and understand the trend and outcomes is called data mining. In this case, the consumer's data undergoes grouping and might get placed in the wrong group rather than the actual one. Also, there can be a case of biasing towards the groups which are not be focused on, or are not a part of the intended audience. This leads to the discrimination factors if we see it from a social point of view.
3. Lack of consent : Use of information without the consent or awareness of the consumers raises concern over the business ethics followed by the company. No one deserves the right to misuse information for his personal benefits without any of its information to the consumer. It is morally wrong and againt the work ethics. Moreover, it raises trust issues between the two involved, and hence is socially unacceptable.Explanation:
Two ball bearings from different manufacturers are being considered for a certain application. Bearing A has a catalog rating of 2.12 kN based on a catalog rating system of 3000 hours at 500 rev/min. Bearing B has a catalog rating of 7.5 kN based on a catalog that rates at 106 cycles. For a given application, determine which bearing can carry the larger load.
Answer:
F[tex]_D[/tex] for A > F[tex]_D[/tex] for B
Hence, Bearing A can carry the larger load
Explanation:
Given the data in the question,
First lets consider an application which requires desired speed of n₀ and a desired life of L₀.
Lets start with Bearing A
so we write the relation between desired load and life catalog load and life;
[tex]F_R(L_Rn_R60)^{1/a}[/tex] = [tex]F_D(L_Dn_D60)^{1/a}[/tex]
where F[tex]_R[/tex] is the catalog rating( 2.12 kN)
L[tex]_R[/tex] is the rating life ( 3000 hours )
n[tex]_R[/tex] is the rating speed ( 500 rev/min )
F[tex]_D[/tex] is the desired load
L[tex]_D[/tex] is the desired life ( L₀ )
n[tex]_D[/tex] is the the desired speed ( n₀ )
Now as we know, a = 3 for ball bearings
so we substitute
[tex]2.12( 3000 * 500 * 60 )^{1/3[/tex] = [tex]F_D( L_0n_060)^{1/3[/tex]
950.0578 = [tex]F_D( L_0n_0)^{1/3} 3.914867[/tex]
950.0578 / 3.914867 = [tex]F_D( L_0n_0)^{1/3}[/tex]
242.6794 = [tex]F_D( L_0n_0)^{1/3}[/tex]
F[tex]_D[/tex] for A = (242.6794 / [tex]( L_0n_0)^{1/3}[/tex] ) kN
Therefore the load that bearing A can carry is (242.6794 / [tex]( L_0n_0)^{1/3}[/tex] ) kN
Next is Bearing B
[tex]F_R(L_Rn_R60)^{1/a}[/tex] = [tex]F_D(L_Dn_D60)^{1/a}[/tex]
F[tex]_R[/tex] = 7.5 kN, [tex](L_Rn_R60) = 10^6[/tex]
Also, for ball bearings, a = 3
so we substitute
[tex]7.5(10^6)^{1/3[/tex] = [tex]F_D(L_0n_060)^{1/3}[/tex]
750 = [tex]F_D(L_0n_0)^{1/3} 3.914867[/tex]
750 / 3.914867 = [tex]F_D(L_0n_0)^{1/3}[/tex]
191.5773 = [tex]F_D(L_0n_0)^{1/3}[/tex]
F[tex]_D[/tex] for B = ( 191.5773 / [tex](L_0n_0)^{1/3}[/tex] ) kN
Therefore, the load that bearing B can carry is ( 191.5773 / [tex](L_0n_0)^{1/3}[/tex] ) kN
Now, comparing the Two results above,
we can say;
F[tex]_D[/tex] for A > F[tex]_D[/tex] for B
Hence, Bearing A can carry the larger load
An asphalt binder is mixed with aggregate and compacted into a sample. The mass of the dry sample is 1173.5 g, the mass of the sample submerged and then surface-dried with a damp towel is 1175.5 g, and the mass of the sample completely submerged in water is 652.5 g. Find the bulk specific gravity of the compacted sample.
Answer:
[tex]\mathbf{G_m = 2.25}[/tex]
Explanation:
From the given information:
Let the weight of the mix in the air be = [tex]W_{ma}[/tex]
Let the weight of the mix in water be = [tex]W_{mw}[/tex]; &
the bulk specific gravity be = [tex]G_m[/tex]
SO;
[tex]W_{mw} = W_{ma} - v \delta _{w} --- (1)[/tex]
Also;
[tex]G_m = \dfrac{W_{mw}}{v \delta_w} --- (2)[/tex]
From (2), make[tex]v \delta_w[/tex] the subject:
[tex]v \delta_w = \dfrac{W_{ma}}{G_m}[/tex]
Now, equation (1) can be rewritten as:
[tex]W_{mw} = W_{ma} - \dfrac{W_{ma}}{G_m}[/tex]
[tex]G_m = \dfrac{W_{ma}}{W_{ma} - W_{mw}}[/tex]
Replacing the values;
[tex]G_m = \dfrac{1173.5}{1173.5 -652.5}[/tex]
[tex]G_m = \dfrac{1173.5}{521}[/tex]
[tex]\mathbf{G_m = 2.25}[/tex]
