Answer:
Under the normal sign convention, the distributed load on a beam is equal to the: O The second derivative of the bending moment with respect to the length of the beam O Negative of the rate of change of the shear force with respect to the length of the beam.
Sorry if the answer is wrong
The ABC Corporation manufactures and sells two products: T1 and T2. 20XX budget for the company is given below:
Projected Sales Units Price T1 60,000 $165 T2 40,000 $250
Inventories in Units January 1, 20XX December 31, 20XX T1 20,000 25,000 T2 8,000 9,000
The following direct materials are used in the two products: Amount used per unit Direct Material Unit T1 T2 A pound 4 5 B pound 2 3 C each 0 1
Anticipated Inventories Direct Material Purchase Price January 1, 2012 December 31, 2012 A $12 32,000 lb. 36,000 lb. B $5 29,000 lb. 32,000 lb. C $3 6,000 units 7,000 units
Projected direct manufacturing labor requirements and rates for 20XX are as follows: Hours per Unit Rate per Hour T1 2 $12 T2 3 $16
4
Manufacturing overhead is allocated at the rate of $20 per direct manufacturing labor-hour. Marketing and distribution costs are projected to be $100,000 and $ 300,000, respectively.
a. What is the total expected revenue (in dollars) for 20XX? b. What is the expected production level (in units) both for T1 and T2? c. What is the total direct material purchases (in dollars) for each type of direct material? d. What is the total direct manufacturing labor cost (in dollars)? e. What is the total overhead cost (in dollars)? f. What is the total cost of goods sold (in dollars)? g. What is the total expected operating income (in dollars) for 20XX?
Answer:
The ABC Corporation
a) Total Expected Revenue (in dollars) for 20XX:
Revenue from T1 = 60,000 x $165 = $26,400,000
Revenue from T2 = 40,000 x $250 = $10,000,000
Total Revenue from T1 and T2 = $36,400,000
b) Production Level (in units) for T1 and T2
T1 T2
Total Units sold 160,000 40,000
Add Closing Inventory 25,000 9,000
Units Available for sale 185,000 49,000
less opening inventory 20,000 8,000
Production Level 165,000 units 41,000 units
c) Total Direct Material Purchases (in dollars):
Cost of direct materials used T1 T2
A: (165,000 x 4 x $12) $7,920,000 $2,460,000 (41,000 x 5 x $12)
B: (165,000 x 2 x $5) 1,650,000 615,000 (41,000 x 3 x $5)
C: 0 123,000 (41,000 x 1 x$3)
Total cost $9,570,000 $3,198,000 Total = $12,768,000
Cost of direct per unit = $58 ($9,570,000/165,000) for T1 and $78 ($3,198,000/41,000) for T2
Cost of direct materials used for production $12,768,000
Cost of closing direct materials:
A (36,000 x $12) $432,000
B (32,000 x $5) 160,000
C (7,000 x $3) 21,000 $613,000
Cost of direct materials available for prodn $13,381,000
Less cost of beginning direct materials:
A (32,000 x $12) $384,000
B (29,000 x $5) 145,000
C (6,000 x $3) 18,000 $547,000
Cost of direct materials purchases $12,834,000
d) The Total Direct Manufacturing Labor Cost (in dollars):
T1 T2
Direct labor per unit 2 hours 3 hours
Direct labor rate per hour $12 $16
Direct labor cost per unit $24 $48
Production level 165,000 units 41,000 units
Labor Cost ($) $3,960,000 $1,968,000
Total labor cost $5,928,000 ($3,960,000 + $1,968,000)
e) Total Overhead cost (in dollars):
Overhead rate = $20 per labor hour
Overhead cost per unit: T1 = $40 ($20 x 2) and T2 = $60 ($20 x 3)
T1 overhead = $20 x 2 x 165,000) = $6,600,000
T2 overhead = $20 x 3 x 41,000) = $2,460,000
Total Overhead cost = $9,060,000
Cost of goods produced:
Cost of opening inventory of materials = $547,000
Purchases of directials materials 12,834,000
less closing inventory of materials = $613,000
Cost of materials used for production 12,768,000
add Labor cost 5,928,000
add Overhead cost 9,060,000
Total production cost $27,756,000
f) Total cost of goods sold (in dollars):
Cost of opening inventory = $3,928,000
Total Production cost = $27,756,000
Cost of goods available for sale $31,684,000
Less cost of closing inventory $4,724,000
Total cost of goods sold $26,960,000
g) Total expected operating income (in dollars)
Sales Revenue: T1 and T2 $36,400,000
Cost of goods sold 26,960,000
Gross profit $9,440,000
less marketing & distribution 400,000
Total Expected Operating Income = $9,040,000
Explanation:
a) Cost of beginning inventory of finished goods:
T1, (Direct materials + Labor + Overhead) X inventory units =
T1 = 20,000 x ($58 + 24 + 40) = $2,440,000
T2 = 8,000 ($78 + 48 + 60) = $1,488,000
Total cost of beginning inventory = $3,928,000
b) Cost of closing Inventory of finished goods:
T1 = 25,000 x ($58 + 24 + 40) = $3,050,000
T2 = 9,000 ($78 + 48 + 60) = $1,674,000
Total cost of closing inventory = $4,724,000
A motor vehicle has a mass of 1.8 tonnes and its wheelbase is 3 m. The centre of gravity of the vehicle is situated in the central plane 0.9 m above the ground and 1.7 m behind the front axle. When moving on the level at 90 km/h the brakes applied and it comes to a rest in a distance of 50 m.
Calculate the normal reactions at the front and rear wheels during the braking period and the least coefficient of friction required between the tyres and the road. (Assume g = 10 m/s2)
Answer:
1) The normal reactions at the front wheel is 9909.375 N
The normal reactions at the rear wheel is 8090.625 N
2) The least coefficient of friction required between the tyres and the road is 0.625
Explanation:
1) The parameters given are as follows;
Speed, u = 90 km/h = 25 m/s
Distance, s it takes to come to rest = 50 m
Mass, m = 1.8 tonnes = 1,800 kg
From the equation of motion, we have;
v² - u² = 2·a·s
Where:
v = Final velocity = 0 m/s
a = acceleration
∴ 0² - 25² = 2 × a × 50
a = -6.25 m/s²
Force, F = mass, m × a = 1,800 × (-6.25) = -11,250 N
The coefficient of friction, μ, is given as follows;
[tex]\mu =\dfrac{u^2}{2 \times g \times s} = \dfrac{25^2}{2 \times 10 \times 50} = 0.625[/tex]
Weight transfer is given as follows;
[tex]W_{t}=\dfrac{0.625 \times 0.9}{3}\times \dfrac{6.25}{10}\times 18000 = 2109.375 \, N[/tex]
Therefore, we have for the car at rest;
Taking moment about the Center of Gravity CG;
[tex]F_R[/tex] × 1.3 = 1.7 × [tex]F_F[/tex]
[tex]F_R[/tex] + [tex]F_F[/tex] = 18000
[tex]F_R + \dfrac{1.3 }{1.7} \times F_R = 18000[/tex]
[tex]F_R[/tex] = 18000*17/30 = 10200 N
[tex]F_F[/tex] = 18000 N - 10200 N = 7800 N
Hence with the weight transfer, we have;
The normal reactions at the rear wheel [tex]F_R[/tex] = 10200 N - 2109.375 N = 8090.625 N
The normal reactions at the front wheel [tex]F_F[/tex] = 7800 N + 2109.375 N = 9909.375 N
2) The least coefficient of friction, μ, is given as follows;
[tex]\mu = \dfrac{F}{R} = \dfrac{11250}{18000} = 0.625[/tex]
The least coefficient of friction, μ = 0.625.
A(n) 78-hp compressor in a facility that operates at full load for 2500 h a year is powered by an electric motor that has an efficiency of 93 percent. If the unit cost of electricity is $0.11/kWh, what is the annual electricity cost of this compressor
Answer: $17,206.13
Explanation:
Hi, to answer this question we have to apply the next formula:
Annual electricity cost = (P x 0.746 x Ckwh x h) /η
P = compressor power = 78 hp
0.746 kw/hp= constant (conversion to kw)
Ckwh = Cost per kilowatt hour = $0.11/kWh
h = operating hours per year = 2500 h
η = efficiency = 93% = 0.93 (decimal form)
Replacing with the values given :
C = ( 78 hp x 0.746 kw/hp x 0.11 $/kwh x 2500 h ) / 0.93 = $17,206.13
Describe what you have been taught about the relationship between basic science research, and technological innovation before this class. Have you been told that it is similar to the linear model? Is your view of this relationship different after studying this unit's lectures and readings? Explain why in 3-4 sentences
Answer:
With the Breakthrough of Technology, the rate at which things are done are becoming much more easy. but without basic science, innovation towards technology cannot occur, so the both work hand in hand in the world of technology today.
Explanation:
Technological innovation and Basic science research plays a major role in the world of science and technology today, while we all want technology innovation the more, without basic science, innovation cannot come in place,
Just as we are going further in technology, breakthroughs and growth are been made which helps on the long run in science research which in turn has made things to be done much better and easily.
cubical tank 1 meter on each edge is filled with water at 20 degrees C. A cubical pure copper block 0.46 meters on each edge with an initial temperature of 100 degrees C is quickly submerged in the water, causing an amount of water equal to the volume of the smaller cube to spill from the tank. An insulated cover is placed on the tank. The tank is adiabatic. Estimate the equilibrium temperature of the system (block + water). Be sure to state all applicable assumptions.
Answer:
final temperature = 26.5°
Explanation:
Initial volume of water is 1 x 1 x 1 = 1 [tex]m^{3}[/tex]
Initial temperature of water = 20° C
Density of water = 1000 kg/[tex]m^{3}[/tex]
volume of copper block = 0.46 x 0.46 x 0.46 = 0.097 [tex]m^{3}[/tex]
Initial temperature of copper block = 100° C
Density of copper = 8960 kg/[tex]m^{3}[/tex]
Final volume of water = 1 - 0.097 = 0.903 [tex]m^{3}[/tex]
Assumptions:
since tank is adiabatic, there's no heat gain or loss through the wallsthe tank is perfectly full, leaving no room for cooling airtotal heat energy within the tank will be the summation of the heat energy of the copper and the water remaining in the tank.mass of water remaining in the tank will be density x volume = 1000 x 0.903 = 903 kg
specific heat capacity of water c = 4186 J/K-kg
heat content of water left Hw = mcT = 903 x 4186 x 20 = 75.59 Mega-joules
mass of copper will be density x volume = 8960 x 0.097 = 869.12 kg
specific heat capacity of copper is 385 J/K-kg
heat content of copper Hc = mcT = 869.12 x 385 x 100 = 33.46 Mega-joules
total heat in the system = 75.59 + 33.46 = 109.05 Mega-joules
this heat will be distributed in the entire system
heat energy of water within the system = mcT
where T is the final temperature
= 903 x 4186 x T = 3779958T
for copper, heat will be
mcT = 869.12 x 385 = 334611.2T
these component heats will sum up to the final heat of the system, i.e
3779958T + 334611.2T = 109.05 x [tex]10^{6}[/tex]
4114569.2T = 109.05 x [tex]10^{6}[/tex]
final temperature T = (109.05 x [tex]10^{6}[/tex])/4114569.2 = 26.5°
An aluminium bar 600mm long with a diameter 40mm has a hole drilled in the centre of which 30mm in diameter and 100mm long if the modulus of elasticity is 85GN/M2 calculate the total contraction oon the bar due to comprehensive load of 160KN.
Answer:
Total contraction on the bar = 1.238 mm
Explanation:
Modulus of Elasticity, E = 85 GN/m²
Diameter of the aluminium bar, [tex]d_{Al} = 40 mm = 0.04 m[/tex]
Load, P = 160 kN
Cross sectional area of the aluminium bar without hole:
[tex]A_1 = \frac{\pi d_{Al}^2 }{4} \\A_1 = \frac{\pi 0.04^2 }{4}\\A_1 = 0.00126 m^2[/tex]
Diameter of hole, [tex]d_h = 30 mm = 0.03 m[/tex]
Cross sectional area of the aluminium bar with hole:
[tex]A_2 = \frac{\pi( d_{Al}^2 - d_{h}^2)}{4} \\A_2 = \frac{\pi (0.04^2 - 0.03^2) }{4}\\A_2 = 0.00055 m^2[/tex]
Length of the aluminium bar, [tex]L_{Al} = 600 mm = 0.6 m[/tex]
Length of the hole, [tex]L_h = 100mm = 0.1 m[/tex]
Contraction in aluminium bar without hole [tex]= \frac{P * L_{Al}}{A_1 E}[/tex]
Contraction in aluminium bar without hole [tex]= \frac{160*10^3 * 0.6}{0.00126 * 85 * 10^9 }[/tex]
Contraction in aluminium bar without hole = 96000/107100000
Contraction in aluminium bar without hole = 0.000896
Contraction in aluminium bar with hole [tex]= \frac{P * L_{h}}{A_2 E}[/tex]
Contraction in aluminium bar without hole [tex]= \frac{160*10^3 * 0.1}{0.00055 * 85 * 10^9 }[/tex]
Contraction in aluminium bar without hole = 16000/46750000
Contraction in aluminium bar without hole = 0.000342
Total contraction = 0.000896 + 0.000342
Total contraction = 0.001238 m = 1.238 mm
Design a decimal arithmetic unit with two selection variables, V1, and Vo, and two BCD digits, A and B. The unit should have four arithmetic operations which depend on the values of the selection variables as shown below. V1=0011, V0=0101 and output functions are as follows;
1- A+9's complement of B
2- A+B
3- A+10's complement of B
4- A+1 (add 1 to A)
(You can see question number 3 in the attached file)
A walrus loses heat by conduction through its blubber at the rate of 220 W when immersed in −1.00°C water. Its internal core temperature is 37.0°C, and it has a surface area of 2.23 m2. What is the average thickness of its blubber? The conductivity of fatty tissue without blood is 0.20 (J/s · m · °C).
Answer:
The average thickness of the blubber is 0.077 m
Explanation:
Here, we want to calculate the average thickness of the Walrus blubber.
We employ a mathematical formula to calculate this;
The rate of heat transfer(H) through the Walrus blubber = dQ/dT = KA(T2-T1)/L
Where dQ is the change in amount of heat transferred
dT is the temperature gradient(change in temperature) i.e T2-T1
dQ/dT = 220 W
K is the conductivity of fatty tissue without blood = 0.20 (J/s · m · °C)
A is the surface area which is 2.23 m^2
T2 = 37.0 °C
T1 = -1.0 °C
L is ?
We can rewrite the equation in terms of L as follows;
L × dQ/dT = KA(T2-T1)
L = KA(T2-T1) ÷ dQ/dT
Imputing the values listed above;
L = (0.2 * 2.23)(37-(-1))/220
L = (0.2 * 2.23 * 38)/220 = 16.948/220 = 0.077 m
Find the minimum diameter of an alloy, tensile strength 75 MPa, needed to support a 30 kN load.
Answer:
The minimum diameter to withstand such tensile strength is 22.568 mm.
Explanation:
The allow is experimenting an axial load, so that stress formula for cylidrical sample is:
[tex]\sigma = \frac{P}{A_{c}}[/tex]
[tex]\sigma = \frac{4\cdot P}{\pi \cdot D^{2}}[/tex]
Where:
[tex]\sigma[/tex] - Normal stress, measured in kilopascals.
[tex]P[/tex] - Axial load, measured in kilonewtons.
[tex]A_{c}[/tex] - Cross section area, measured in square meters.
[tex]D[/tex] - Diameter, measured in meters.
Given that [tex]\sigma = 75\times 10^{3}\,kPa[/tex] and [tex]P = 30\,kN[/tex], diameter is now cleared and computed at last:
[tex]D^{2} = \frac{4\cdot P}{\pi \cdot \sigma}[/tex]
[tex]D = 2\sqrt{\frac{P}{\pi \cdot \sigma} }[/tex]
[tex]D = 2 \sqrt{\frac{30\,kN}{\pi \cdot (75\times 10^{3}\,kPa)} }[/tex]
[tex]D = 0.0225\,m[/tex]
[tex]D = 22.568\,mm[/tex]
The minimum diameter to withstand such tensile strength is 22.568 mm.
The lower half of a 7-m-high cylindrical container is filled with water (rho = 1000 kg/m3) and the upper half with oil that has a specific gravity of 0.85. Determine the pressure difference between the top and the bottom of the cylinder. (Round the final answer to one decimal place.)
Answer:
Pressure difference (ΔP) = 63,519.75 kpa
Explanation:
Given:
ρ = 1,000 kg/m³
Height of cylindrical container used (h) = 7m / 2 = 3.5m
Specific gravity (sg) = 0.85
Find:
Pressure difference (ΔP).
Computation:
⇒ Pressure difference (ΔP) = h g [ ρ(sg) + ρ] ∵ [ g = 9.81]
⇒ Pressure difference (ΔP) = (3.5)(9.81) [ 1,000(0.85) + 1,000]
⇒ Pressure difference (ΔP) = 34.335 [8,50 + 1,000]
⇒ Pressure difference (ΔP) = 34.335 [1,850]
⇒ Pressure difference (ΔP) = 63,519.75 kpa
A phone charger requires 0.5 A at 5V. It is connected to a transformer with 100 % of efficiency whose primary contains 2200 turns and is connected to 220-V household outlet.
(a) How many turns should there be in the secondary?
(b) What is the current in the primary?
(c) What would be the output current and output voltage values if number of secondary turns (N2) doubled of its initial value?
Answer:
Explanation:
a ) for transformer which steps down voltage , if V₁ and V₂ be voltage of primary and secondary coil and n₁ and n₂ be the no of turns of wire in them
V₁ /V₂ = n₁ / n₂
Here V₁ = 220 V , V₂ = 5V , n₁ = 2200 n₂ = ?
220 /5 = 2200 / n₂
n₂ = 2200 x 5 / 220
= 50
b )
for 100 % efficiency
input power = output power
V₁ I₁ = V₂I₂
I₁ and I₂ are current in primary and secondary coil
220 x I₁ = 5 x .5
I₁ = .01136 A .
c )
If n₂ = 100
V₁ /V₂ = n₁ / n₂
220 / V₂ = 2200 / 100
V₂ = 10 V
V₁ I₁ = V₂I₂
220 x .01136 = 10 I₂
I₂ = .25 A.
A solid square rod is cantilevered at one end. The rod is 0.6 m long and supports a completely reversing transverse load at the other end of 62 kN. The material is AISI 1080 hot-rolled steel. If the rod must support this load for 104 cycles with a design factor of 1.5, what dimension should the square cross section have
Answer:
The dimension of the square cross section is = 30mm * 30mm
Explanation:
Before proceeding with the calculations convert MPa to Kpsi
Sut ( ultimate strength ) = 770 MPa * 0.145 Kpsi/MPa = 111.65 Kpsi
the fatigue strength factor of Sut at 10^3 cycles for Se = Se' = 0.5 Sut
at 10^6 cycles" for 111.65 Kpsi = f ( fatigue strength factor) = 0.83
To calculate the endurance limit use Se' = 0.5 Sut since Sut < 1400 MPa
Se'( endurance limit ) = 0.5 * 770 = 385 Mpa
The surface condition modification factor
Ka = 57.7 ( Sut )^-0.718
Ka = 57.7 ( 770 ) ^-0.718
Ka = 0.488
Assuming the size modification factor (Kb) = 0.85 and also assuming all modifiers are equal to one
The endurance limit at the critical location of a machine part can be expressed as :
Se = Ka*Kb*Se'
Se = 0.488 * 0.85 * 385 = 160 MPa
Next:
Calculating the constants to find the number of cycles
α = [tex]\frac{(fSut)^2}{Se}[/tex]
α =[tex]\frac{(0.83*770)^2}{160}[/tex] = 2553 MPa
b = [tex]-\frac{1}{3} log(\frac{fSut}{Se} )[/tex]
b = [tex]-\frac{1}{3} log (\frac{0.83*770}{160} )[/tex] = -0.2005
Next :
calculating the fatigue strength using the relation
Sf = αN^b
N = number of cycles
Sf = 2553 ( 10^4) ^ -0.2005
Sf = 403 MPa
Calculate the maximum moment of the beam
M = 2000 * 0.6 = 1200 N-m
calculating the maximum stress in the beam
∝ = ∝max = [tex]\frac{Mc}{I}[/tex]
Where c = b/2 and I = b(b^3) / 12
hence ∝max = [tex]\frac{6M}{b^3}[/tex] = 6(1200) / b^3 = 7200/ b^3 Pa
note: b is in (meters)
The expression for the factor of safety is written as
n = [tex]\frac{Sf}{\alpha max }[/tex]
Sf = 403, n = 1.5 and ∝max = 7200 / b^3
= 1.5 = [tex]\frac{403(10^6 Pa/Mpa)}{7200 / B^3}[/tex] making b subject of the formula in other to get the value of b
b = 0.0299 m * 10^3 mm/m
b = 29.9 mm therefore b ≈ 30 mm
since the size factor assumed is near the calculated size factor using this relation : de = 0.808 ( hb)^1/2
the dimension = 30 mm by 30 mm
