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The uniform sign has a weight of 1500 lb and is supported by the pipe AB, which has an inner radius of 2.75 in. and anouter radius of 3.00 in. If the face of the sign is subjected to a uniform wind pressure of p = 150lb/ft2, determine the state of stress at points C and D. Show the results on a differential volume element located at each of these points. Neglect the thickness of the sign, and assume that it is supported along the outside edge of the pipe.The uniform sign has a weight of 1500 lb and is supported bythe pipe AB, which has an inner radius of 2.75 in. and anouter radius of 3.00 in.. If the face of the sign issubjected to a uniform wind pressure of p = 150lb/ft2, determine the state of stress at pointsC and D. Show the results on a differentialvolume element located at each of these points. Neglect the thickness of the sign, and assume that it issupported along the outside edge of the pipe.
Answer:
See explanation
Explanation:
See the document for the complete FBD and the introductory part of the solution.
Static Balance ( Sum of Forces = 0 ) in all three directions
∑[tex]F_G_X = W - G_x = 0[/tex]
[tex]G_X = W = 1500 lb[/tex]
∑[tex]F_G_Y = P - G_Y = 0[/tex]
[tex]G_Y = P = -10,800 lb[/tex]
∑[tex]F_G_Z = - G_Z = 0[/tex]
Where, ( [tex]G_X, G_Y, G_Z[/tex] ) are internal forces at section ( G ) along the defined coordinate axes.
Static Balance ( Sum of Moments about G = 0 ) in all three directions
[tex]M_G = r_O_G x F_O[/tex]
Where,
r_OG: The vector from point O to point G
F_OG: The force vector at point O
- The vector ( r_OG ) and ( F_OG ) can be written as follows:
[tex]r_O_G = [ -( 3 + \frac{H}{2} ) i + (\frac{r_o}{12})j - ( \frac{r_o}{12} + \frac{L}{2})k ] \\\\r_O_G = [ -( 6 ) i + (0.25)j - (6)k ] \\[/tex]
[tex]F_O_G = [ ( W ) i + ( P ) k ]\\\\F_O_G = [ (1500) i - ( 10,800 ) k ] lb[/tex]
- Then perform the cross product of the two vectors ( r_OG ) and ( F_OG ):
[tex]( M_G_X )i + (M_G_Y)j+ (M_G_Z)k = \left[\begin{array}{ccc}i&j&k\\-6&0.25&-6\\1500&-10,800&0\end{array}\right] \\\\\\( M_G_X )i + (M_G_Y)j+ (M_G_Z)k = -( 6*10,800 ) i - ( 6*1500 ) j + [ ( 10,800*6) - ( 0.25*1500) ] k\\\\( M_G_X )i + (M_G_Y)j+ (M_G_Z)k = - (64,800)i - (9,000)j + (64,425)k[/tex]
- The internal torque ( T ) and shear force ( V ) that act on slice ( G ) are due to pressure force ( P ) as follows:
[tex]T = P*[\frac{L}{2}] = (10,800)*(6) = 64,800 lb.ft[/tex]
[tex]V = P = -10,800 lb[/tex]
- For the state of stress at point "C" we need to determine the the normal stress along x direction ( σ_x ) and planar stress ( τ_xy ) as follows:-
σ_x = [tex]-\frac{G_x}{A} - \frac{M_G_Y. z*}{I_Y_Y} + \frac{M_G_Z. y*}{I_Z_Z}[/tex]
Where,
A: The area of pipe cross section
[tex]A = \pi * [ ( \frac{r_o}{12})^2 - ( \frac{r_i}{12})^2 ] = \pi * [ ( \frac{3}{12})^2 - ( \frac{2.75}{12})^2 ] = 0.03136 ft^2[/tex]
z*: The distance of point "C" along z-direction from central axis ( x )
[tex]z*= [\frac{r_i}{12} ] = [\frac{2.75}{12} ] = 0.22916 ft[/tex]
I_YY: The second area moment of pipe along and about "y" axis:
[tex]I_Y_Y = \frac{\pi }{4} * [ (\frac{r_o}{12})^4 - (\frac{r_i}{12})^4 ]=\frac{\pi }{4} * [ (\frac{3}{12})^4 - (\frac{2.75}{12})^4 ] \\\\I_Y_Y = 0.00090 ft^4[/tex]
y*: The distance of point "C" along y-direction from central axis ( x )
[tex]y* = 0[/tex]
- The normal stress ( σ_x ) becomes:
σ_x = [tex][-\frac{1500}{0.03136} - \frac{-9,000*0.22916}{0.00090} + \frac{64,425*0}{0.00090} ] * (\frac{1}{12})^2 = 15.5 ksi[/tex]
- The planar stress is ( τ_xy ) is a contribution of torsion ( T ) and shear force ( V ):
τ_xy = [tex]- \frac{T.c}{J} + \frac{V.Q}{I.t}[/tex]
Where,
c: The radial distance from central axis ( x ) and point "C".
[tex]c = \frac{r_i}{12} = \frac{2.75}{12} = 0.22916 ft[/tex]
J: The polar moment of inertia of the annular cross section of pipe:
[tex]J = \frac{\pi }{2}* [ ( \frac{r_o}{12})^4 - ( \frac{r_i}{12})^4 ] = \frac{\pi }{2}* [ ( \frac{3}{12})^4 - ( \frac{2.75}{12})^4 ] = 0.00180 ft^4[/tex]
Q: The first moment of area for point "C" = semi-circle
[tex]Q = Y_c*A_c = \frac{4*( r_m)}{3\pi } * \frac{\pi*( r_m)^2 }{2} = \frac{2. ( r_m)^3}{3} \\\\Q = \frac{2. [ ( \frac{r_o}{12})^3 - ( \frac{r_i}{12})^3] }{3} = \frac{2. [ ( \frac{3}{12})^3 - ( \frac{2.75}{12})^3] }{3} = 0.00239ft^3[/tex]
I: The second area moment of pipe along and about "y" axis:
[tex]I_Y_Y = \frac{\pi }{4} * [ (\frac{r_o}{12})^4 - (\frac{r_i}{12})^4 ]=\frac{\pi }{4} * [ (\frac{3}{12})^4 - (\frac{2.75}{12})^4 ] \\\\I_Y_Y = 0.00090 ft^4[/tex]
t: The effective thickness of thin walled pipe:
[tex]t = 2* [ \frac{r_o}{12} - \frac{r_i}{12} ] = 2* [ \frac{3}{12} - \frac{2.75}{12} ] = 0.04166 ft[/tex]
- The planar stress is ( τ_xy ) becomes:
τ_xy = [tex][ - \frac{-64,800*0.22916}{0.0018} + \frac{-10,800*0.00239}{0.0009*0.04166} ] * [ \frac{1}{12}]^2 = 52.4 ksi[/tex]
- The principal stresses at point "C" can be determined from the following formula:-
σ_x = 15.55 ksi, σ_y = 0 ksi , τ_xy = 52.4 ksi
σ_1 =[tex]\frac{sigma_x+sigma_y}{2} + \sqrt{(\frac{sigma_x+sigma_y}{2})^2 + (tow_x_y)^2 }[/tex]
σ_2 = [tex]\frac{sigma_x+sigma_y}{2} - \sqrt{(\frac{sigma_x+sigma_y}{2})^2 + (tow_x_y)^2 }[/tex]
σ_1 = [tex]\frac{15.55+0}{2} + \sqrt{(\frac{15.55+0}{2})^2 + (52.4)^2 } = 60.75 ksi[/tex]
σ_2 =[tex]-\sqrt{\left(\frac{15.55+0}{2}\right)^2\:+\:\left(52.4\right)^2\:}+\frac{15.55+0}{2} = -45.20 ksi[/tex]
- The angle of maximum plane stress ( θ ):
θ = [tex]0.5*arctan ( \frac{tow_x_y}{\frac{sigma_x-sigma_y}{2} } )= 0.5*arctan*( \frac{52.4}{7.8} ) = 40.8 deg[/tex]
Note: The plane stresses at point D are evaluated using the following procedure given above. Due to 5,000 character limit at Brainly, i'm unable to post here.
An automobile travels along a straight road at 15.65 m/s through a 11.18 m/s speed zone. A police car observed the automobile. At the instant that the two vehicles are abreast of each other, the police car starts to pursue the automobile at a constant acceleration of 1.96 m/s2. The motorist noticed the police car in his rear view mirror 12 s after the police car started the pursuit and applied his brakes and decelerates at 3.05 m/s2. (Hint: The police will not go against the law.) a) Find the total time required for the police car to overtake the automobile. (12 marks) b) Find the total distance travelled by the police car while overtaking the automobile. (2 marks) c) Find the speed of the police car at the time it overtakes the automobile
Answer:
A.) Time = 17.13 seconds
B.) Distance = 31.9 m
C.) V = 11.18 m/s
D.) V = 7.1 m/s
Explanation:
The initial velocity U of the automobile is 15.65 m/s.
At the instant that the two vehicles are abreast of each other, the police car starts to pursue the automobile with initial velocity U = 0 at a constant acceleration of 1.96 m/s². Because the police is starting from rest.
For the automobile, let us use first equation of motion
V = U - at.
Acceleration a is negative since it is decelerating with a = 3.05 m/s² . And
V = 0.
Substitute U and a into the formula
0 = 15.65 - 3.05t
15.65 = 3.05t
t = 15.65/3.05
t = 5.13 seconds
But the motorist noticed the police car in his rear view mirror 12 s after the police car started the pursuit and applied his brakes and decelerates at 3.05 m/s².
The total time required for the police car to overtake the automobile will be
12 + 5.13 = 17.13 seconds.
b.) Using the third equation of motion formula for the police car at V = 11.18 m/s and a = 1.96 m/s²
V^2 = U^2 + 2aS
Where S = distance travelled.
Substitute V and a into the formula
11.18^2 = 0 + 2 × 1.96 ×S
124.99 = 3.92S
S = 124.99/3.92
S = 31.88 m
c.) The speed of the police car at the time it overtakes the automobile will be in line with the speed zone which is 11.18 m/s
d.) That will be the final velocity V of the automobile car.
We will use third equation of motion to solve that.
V^2 = U^2 + 2as
V^2 = 15.65^2 - 2 × 3.05 × 31.88
V^2 = 244.9225 - 194.468
V = sqrt( 50.4545)
V = 7.1 m/s
A piston–cylinder device contains 0.85 kg of refrigerant- 134a at 2108C. The piston that is free to move has a mass of 12 kg and a diameter of 25 cm. The local atmospheric pressure is 88 kPa. Now, heat is transferred to refrigerant-134a until the temperature is 158C. Determine (a) the final pressure, (b) the change in the volume of the cylinder, and (c) the change in the enthalpy of the refrigerant-134a.
Question:
A piston–cylinder device contains 0.85 kg of refrigerant- 134a at -10°C. The piston that is free to move has a mass of 12 kg and a diameter of 25 cm. The local atmospheric pressure is 88 kPa. Now, heat is transferred to refrigerant-134a until the temperature is 15°C. Determine (a) the final pressure, (b) the change in the volume of the cylinder, and (c) the change in the enthalpy of the refrigerant-134a.
Answer:
a) 90.4 kPa
b) 0.0205 m³
c) 17.4 kJ/kg
Explanation:
Given:
Mass, m = 0.85 kg
a) The final pressure here is equal to the initial pressure. Let's use the formula:
[tex] P_2 = P_1 = P_a_t_m + \frac{mg}{\pi D^2 / 4}[/tex]
[tex] = 88*10^3 + \frac{12kg * 9.81}{\pi (0.25)^2 / 4} [/tex]
= 90398 Pa
≈ 90.4 KPa
Final pressure = 90.4 kPa
b) Change in volume of the cylinder:
To find the initial and final volume, let's use the values from the A-13 table for refrigerant-134a, at initial values of 90.4 kPa and -10°C and final values of 90.4 kPa and 15°C
v1 = 0.2302m³/kg
h1 = 247.76 kJ/kg
v2 = 0.2544 m³/kg
h2 = 268.2 kJ/kg
Change in volume is calculated as:
Δv = m(v2 - v1)
Δv = 0.85(0.2544 - 0.2302)
= 0.0205 m³
Change in volume = 0.0205 m³
c) Change in enthalpy
Let's use the formula:
Δh = m(h2 - h1)
= 0.85(268.2 - 247.76)
= 17.4 kJ/kg
Change in enthalpy = 17.4 kJ/kg
Increase the sampling time by a factor of 10 (to 0.1 seconds), keeping the frequency of the square wave the same, and observe the delay. Discuss relationship between sampling time and delay from one board to another.
Answer:
Time delay increases
Explanation:
Time delay is the delay between occurance of signal. If sampling time that is time between two samples is increased, the delay in the occurance of regenerated samples is also increased.
The temperature of a flowing gas is to be measured with a thermocouple junction and wire stretched between two legs of a sting, a wind tunnel test fixture. The junction is formed by butt-welding two wires of different material. For wires of diameter D = 125 m and a convection coefficient of h = 700 W/m^2 K, determine the minimum separation distance between the two legs of the sting, L=L1+L2, to ensure that the sting temperature does not influence the junction temperature and, in turn, invalidate the gas temperature measurement. Consider two different types of thermocouple junctions consisting of (i) copper and constantan wires and (ii) chromel and aluminel wires. Evaluate the thermal conductivity of copper and constantan at T300 K. Use kCh =19 W/mK and kA = l29 W/mK for the thermal conductivities of the chromel and alumel wires, respectively.
Answer:
minimum separation distance between the two legs of the sting L = L 1 + L 2 therefore L = 9.48 + 4.68 = 14.16 mL = 1.14 mExplanation:
D ( diameter ) = 125 m
convection coefficient of h = 700 W/m^2
Calculate THE CROSS SECTIONAL AREA
Ac = [tex]\frac{\pi }{4} * D^2[/tex] = [tex]\frac{\pi }{4} * ( 125 )^2[/tex] = 0.79 * 15625 = 12343.75 m^2
perimeter
p = [tex]\pi * D[/tex] = 3.14 * 125 = 392.5 m
at 300k temperature the thermal conductivity of copper and constantan from the thermodynamic property table are :
Kcu = 401 w/m.k
Kconstantan = 23 W/m.k
To calculate the length of copper wire of the thermocouple junction
L 1 = 4.6 ([tex]\frac{Kcv Ac}{h P}[/tex]) ^ 1/2 = 4.6 [tex](\frac{401 *12343.75 }{700 *392.5})^\frac{1}{2}[/tex]
L 1 = 4.6 ( 4949843.75 / 274750 )^1/2
L 1 = 9.48 m
calculate length of constantan wire
L 2 = 4.6 [tex](\frac{kcons*Ac}{hp} )^\frac{1}{2}[/tex]
= 4.6 ( (23 * 12343.75) / ( 700 * 392.5) ) ^1/2
L 2 = 4.6 ( 283906.25 / 274750 ) ^ 1/2
L 2 = 4.68 m
I) therefore the minimum separation distance between the two legs of the sting L = L 1 + L 2
L = 9.48 + 4.68 = 14.16 m
ii) Evaluating the thermal conductivity of copper and constantan
Kc ( thermal conductivity of chromel) = 19 w/m.k
Ka ( thermal conductivity of alumel ) = 29 W/m.k
distance between the legs L = L 1 + L 2
THEREFORE
L = 4.6 ( (Kcn * Ac ) / ( hp ) )^1/2 + 4.6 ( (Kac * Ac)/(hp) )^1/2
L = 4.6 [tex](\frac{Ac}{hp} )^\frac{1}{2} [ (Kcn)^\frac{1}{2} + (Kal)^\frac{1}{2} ][/tex]
L = 4.6 ( 12343.75 /( 700 * 392.5) )^1/2 * [ 19^1/2 + 29^1/2 ]
L = 4.6 ( 12343.75 / 274750 ) ^1/2 * 5.39
L = 1.14 m
g a heat engine is located between thermal reservoirs at 400k and 1600k. the heat engine operates with an efficiency that is 70% of the carnot effieciency. if 2kj of work are produced, how much heat is rejected to the low temperature reservior
Answer:
Heat rejected to cold body = 3.81 kJ
Explanation:
Temperature of hot thermal reservoir Th = 1600 K
Temperature of cold thermal reservoir Tc = 400 K
efficiency of the Carnot's engine = 1 - [tex]\frac{Tc}{Th}[/tex]
eff. of the Carnot's engine = 1 - [tex]\frac{400}{600}[/tex]
eff = 1 - 0.25 = 0.75
efficiency of the heat engine = 70% of 0.75 = 0.525
work done by heat engine = 2 kJ
eff. of heat engine is gotten as = W/Q
where W = work done by heat engine
Q = heat rejected by heat engine to lower temperature reservoir
from the equation, we can derive that
heat rejected Q = W/eff = 2/0.525 = 3.81 kJ
A fully recrystallized sheet of metal with a thickness of 11 mm is to be cold worked by 40% in rolling. Estimate the necessary roll force if the sheet was 0.5 m wide and there was no lateral spreading during rolling. The strength coefficient is 200 MPa, the work hardening exponent of 0.1 and the roll contact length is 40 mm. Assume no friction.
Answer:
Roll force, F = 5.6 MN
Explanation:
Sheet width, b = 0.5 m
Roll contact length, [tex]l_{p} = 40 mm[/tex]
Strength coefficient, [tex]\sigma_{0} = 200 MPa[/tex]
Thickness, h = 11 mm
Since the sheet of metal is cold worked by 40%, the reduction in thickness will be:
Δh = 40% * 11 = 0.4 * 11 = 4.4 mm
Strain, e = (Δh)/h
e = 4.4/11 = 0.4
The roll force is calculated by the formula:
[tex]F = \sigma_{f} l_{p} b[/tex]
[tex]\sigma_{f} = \sigma_{0} (e+1)\\\sigma_{f} = 200 (0.4+1)\\\sigma_{f} = 200 *1.4\\\sigma_{f} = 280 MPa[/tex]
Substituting the value of [tex]\sigma_{f}[/tex], [tex]l_{p}[/tex], and b into the formula for the roll force:
[tex]F = \sigma_{f} l_{p} b\\F = 280 * 0.04 * 0.5\\F = 5.6 MN[/tex]
The drum has a mass of 50 kg and a radius of gyration about the pin at O of 0.23 o k m = . If the 15kg block is moving downward at 3 / m s , and a force of P N =100 is applied to the brake arm, determine how far the block descends from the instant the brake is applied until it stops. Neglect the thickness of the handle. The coefficient of kinetic friction at the brake pad is 0.5 k = .
Note: The diagram referred to in this question is attached as a file below.
Answer:
The block descended a distance of 9.75m from the instant the brake is applied until it stops.
Explanation:
For clarity and easiness of expression, the calculations and the Free Body Diagram are contained in the attached file. Check the attached file below.
The block descended a distance of 9.75 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.
For the pipe-fl ow-reducing section of Fig. P3.54, D 1 5 8 cm, D 2 5 5 cm, and p 2 5 1 atm. All fl uids are at 20 8 C. If V 1 5 5 m/s and the manometer reading is h 5 58 cm, estimate the total force resisted by the fl ange bolts.
Answer:
The total force resisted by the flange bolts is 163.98 N
Explanation:
Solution
The first step is to find the pipe cross section at the inlet section
Now,
A₁ = π /4 D₁²
D₁ = diameter of the pipe at the inlet section
Now we insert 8 cm for D₁ which gives us A₁ = π /4 D (8)²
=50.265 cm² * ( 1 m²/100² cm²)
= 5.0265 * 10^⁻³ m²
Secondly, we find cross section area of the pipe at the inlet section
A₂ = π /4 D₂²
D₂ = diameter of the pipe at the inlet section
Now we insert 5 cm for D₁ which gives us A₁ = π /4 D (5)²
= 19.63 cm² * ( 1 m²/100² cm²)
= 1.963 * 10^⁻³ m²
Now,
we write down the conversation mass relation which is stated as follows:
Q₁ = Q₂
Where Q₁ and Q₂ are both the flow rate at the exist and inlet.
We now insert A₁V₁ for Q₁ and A₂V₂ for Q₂
So,
V₁ and V₂ are defined as the velocities at the inlet and exit
We now insert 5.0265 * 10^⁻³ m² for A₁ 5 m/s for V₁ and 1.963 * 10^⁻³ m² for A₂
= 5.0265 * 5 = 1.963 * V₂
V₂ = 12.8 m/s
Note: Kindly find an attached copy of the part of the solution to the given question below
A piston-cylinder assembly contains 5kg of water that undergoes a series of processes to form a thermodynamic cycle. Process 1à 2: Constant pressure cooling from p1=20bar and T1=360°C to saturated vapor Process 2à 3: Constant volume cooling to p3=5 bar Process 3à 4: Constant pressure heating Process 4à 1: Polytropic process following Pv =constant back to the initial state Kinetic and potential energy effects are negligible. Calculate the net work for the cycle in kJ.
Answer:
[tex]W_{net} = - 1223 kJ[/tex]
Explanation:
State 1:
[tex]P_1 = 20 bar\\T_1 = 360^{0}C\\ h_1 = 3159.3 kJ/kg\\S_1 = 6.9917 kJ/kg[/tex]
State 2:
[tex]P_2 = 20 bar\\x_2 = 1 \\ h_2 = 2799.5 kJ/kg\\u_2 = 2600.3 kJ/kg\\v_2 = 0.09963m^3/kg[/tex]
State 3:
[tex]P_2 = 5 bar\\v_2 = v_3 \\v_3 = v_f + x_3 (v_g - v_f)\\0.09963 = (1.0926 * 10^{-3}) +x_3 (0.3749 - (1.0926 * 10^{-3}))\\x_3 = 0.263[/tex]
[tex]u_{3} = u_f + x_3 ( u_g - u_f)\\u_{3} = 639.68 + 0.263 (2561.2 - 639.68)\\u_{3} = 1146.2 kJ/kg[/tex]
State 4:
[tex]P_{4} = 5 bar\\T_4 = 360^0 C\\h_4 = 3188.4 kJ/kg\\S_4 = 7.660 kJ/kg-K\\Q_{12} = h_2 - h_1 = 2799.5-3159.3 = -359 kJ/kg\\Q_{23} = u_3 - h_2 =1146.2-2006.3 = -1454.1 kJ/kg\\Q_{34} = h_4 - h_3 = 3188.4-1196.04 = 1992.36 kJ/kg\\Q_{41} = T(S_1 - S_4) = (360 + 273) (6.9917 - 7.660) = -423.04 kJ/kg[/tex]
Calculate the network done for the cycle
[tex]W_{net} = m( Q_{12} + Q_{23} + Q_{34} + Q_{41})\\W_{net} = 5( -359.8 - 1454.1 + 1992.36 - 423.04)\\W_{net} = -1223 kJ[/tex]
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.
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
The yield strength for an alloy that has an average grain diameter, d1, is listed above as Yield Stress 1 . At a grain diameter of d2, the yield strength increases Yield Stress 2. At what grain diameter, in mm, will the yield strength be 217 MPa
Complete Question:
Grain diameter 1 (mm) = 4.4E-02
Yield stress 1 (MPa) = 131
Grain diameter 2 (mm) = 7.7E-03
Yield Stress 2 (MPa) = 268
The yield strength for an alloy that has an average grain diameter, d1, is listed above as Yield Stress 1 . At a grain diameter of d2, the yield strength increases Yield Stress 2. At what grain diameter, in mm, will the yield strength be 217 MPa
Answer:
d = 1.3 * 10⁻² m
Explanation:
According to the Hall Petch equation:
[tex]\sigma_y = \sigma_0 + k/\sqrt{d} \\[/tex]
At [tex]d_{1} = 4.4 * 10^{-2} mm[/tex], [tex]\sigma_{y1} = 131 MPa = 131 N/ mm^2[/tex]
[tex]131 = \sigma_0 + k/\sqrt{4.4 * 10^{-2}} \\k = 27.45 - 0.2096 \sigma_0[/tex]
At [tex]d_{2} = 7.7 * 10^{-3} mm[/tex], [tex]\sigma_{y2} = 131 MPa = 268 N/ mm^2[/tex]
[tex]268 = \sigma_0 + (27.45 - 0.2096 \sigma_0)/\sqrt{7.7 * 10^{-3}} \\23.5036 = 27.47 - 0.1219 \sigma_0\\ \sigma_0 = 32.45 N/mm^2[/tex]
k = 27.45 - 0.2096(32.45)
k = 20.64
At [tex]\sigma_y = 217 MPa[/tex], reapplying Hall Petch law:
[tex]\sigma_y = \sigma_0 + k/\sqrt{d} \\[/tex]
[tex]217 =32.45 + 20.64/\sqrt{d} \\217 - 32.45 = 20.64/\sqrt{d}\\184.55 = 20.64/ \sqrt{d} \\\sqrt{d} = 20.64/184.55\\\sqrt{d} = 0.11184\\d = 0.013 mm[/tex]
d = 1.3 * 10⁻² m
An insulated rigid tank is divided into two equal parts by a partition. Initially, one part contains 4 kg of an ideal gas at 750 kPa and 48°C, and the other part is evacuated. The partition is now removed, and the gas expands into the entire tank. Determine the final temperature and pressure in the tank. (Round the final answers to the nearest whole number.)
Answer:
The final temperature and pressure in the insulated rigid tank are [tex]48\,^{\circ}C[/tex] and [tex]375\,kPa[/tex].
Explanation:
An ideal gas is represented by the following model:
[tex]P\cdot V = \frac{m}{M}\cdot R_{u} \cdot T[/tex]
Where:
[tex]P[/tex] - Pressure, measured in kilopascals.
[tex]V[/tex] - Volume, measured in cubic meters.
[tex]m[/tex] - Mass of the ideal gas, measured in kilograms.
[tex]M[/tex] - Molar mass, measured in kilograms per kilomole.
[tex]T[/tex] - Temperature, measured in Kelvin.
[tex]R_{u}[/tex] - Universal constant of ideal gases, equal to [tex]8.314\,\frac{kPa\cdot m^{3}}{kmol\cdot K}[/tex]
As tank is rigid and insulated, it means that no volume deformations in tank, heat and mass interactions with surroundings occur during expansion process. Hence, final pressure is less that initial one, volume is doubled (due to equal partitioning) and temperature remains constant. Hence, the following relationship can be derived from model for ideal gases:
[tex]\frac{P_{1}\cdot V_{1}}{T_{1}} = \frac{P_{2}\cdot V_{2}}{T_{2}}[/tex]
Now, final pressure is cleared:
[tex]P_{2} = P_{1}\cdot \frac{T_{2}}{T_{1}}\cdot \frac{V_{1}}{V_{2}}[/tex]
[tex]P_{2} = (750\,kPa)\cdot 1 \cdot \frac{1}{2}[/tex]
[tex]P_{2} = 375\,kPa[/tex]
The final temperature and pressure in the insulated rigid tank are [tex]48\,^{\circ}C[/tex] and [tex]375\,kPa[/tex].
what is called periodic function give example? Plot the output which is started with zero degree for one coil rotating in the uniform magnetic field and name it. How can you represent this output as the periodic function?
Answer:
A periodic function is a function that returns to its value over a certain period at regular intervals an example is the wave form of flux density (B) = sin wt
Explanation:
A periodic function is a function that returns to its value over a certain period at regular intervals an example is the wave form of flux density (B) = sin wt
attached to the answer is a free plot of the output starting with zero degree for one coil rotating in a uniform magnetic field
B ( wave flux density ) = Bm sinwt and w = 2[tex]\pi[/tex]f = [tex]\frac{2\pi }{T}[/tex] rad/sec
The internal loadings at a critical section along the steel drive shaft of a ship are calculated to be a torque of 2300 lb⋅ft, a bending moment of 1500 lb⋅ft, and an axial thrust of 2500 lb. If the yield points for tension and shear are σY= 100 ksi and τY = 50 ksi, respectively, determine the required diameter of the shaft using the maximum-shear-stress theory
Answer:
Explanation:
Given that:
Torque T = 2300 lb - ft
Bending moment M = 1500 lb - ft
axial thrust P = 2500 lb
yield points for tension σY= 100 ksi
yield points for shear τY = 50 ksi
Using maximum-shear-stress theory
[tex]\sigma_A = \dfrac{P}{A}+\dfrac{Mc}{I}[/tex]
where;
[tex]A = \pi c^2[/tex]
[tex]I = \dfrac{\pi}{4}c^4[/tex]
[tex]\sigma_A = \dfrac{P}{\pi c^2}+\dfrac{Mc}{ \dfrac{\pi}{4}c^4}[/tex]
[tex]\sigma_A = \dfrac{2500}{\pi c^2}+\dfrac{1500*12c}{ \dfrac{\pi}{4}c^4}[/tex]
[tex]\sigma_A = \dfrac{2500}{\pi c^2}+\dfrac{72000c}{\pi c^3}}[/tex]
[tex]\tau_A = \dfrac{T_c}{\tau}[/tex]
where;
[tex]\tau = \dfrac{\pi c^4}{2}[/tex]
[tex]\tau_A = \dfrac{T_c}{\dfrac{\pi c^4}{2}}[/tex]
[tex]\tau_A = \dfrac{2300*12 c}{\dfrac{\pi c^4}{2}}[/tex]
[tex]\tau_A = \dfrac{55200 }{\pi c^3}}[/tex]
[tex]\sigma_{1,2} = \dfrac{\sigma_x+\sigma_y}{2} \pm \sqrt{\dfrac{(\sigma_x - \sigma_y)^2}{2}+ \tau_y^2}[/tex]
[tex]\sigma_{1,2} = \dfrac{2500+72000}{2 \pi c ^3} \pm \sqrt{\dfrac{(2500 +72000)^2}{2 \pi c^3}+ \dfrac{55200}{\pi c^3}} \ \ \ \ \ ------(1)[/tex]
Let say :
[tex]|\sigma_1 - \sigma_2| = \sigma_y[/tex]
Then :
[tex]2\sqrt{( \dfrac{2500c + 72000}{2 \pi c^3})^2+ ( \dfrac{55200}{\pi c^3})^2 } = 100(10^3)[/tex]
[tex](2500 c + 72000)^2 +(110400)^2 = 10000*10^6 \pi^2 c^6[/tex]
[tex]6.25c^2 + 360c+ 17372.16-10,000\ \pi^2 c^6 =0[/tex]
According to trial and error;
c = 0.75057 in
Replacing c into equation (1)
[tex]\sigma_{1,2} = \dfrac{2500+72000}{2 \pi (0.75057) ^3} \pm \sqrt{\dfrac{(2500 +72000)^2}{2 \pi (0.75057)^3}+ \dfrac{55200}{\pi (0.75057)^3}}[/tex]
[tex]\sigma_{1,2} = \dfrac{2500+72000}{2 \pi (0.75057) ^3} + \sqrt{\dfrac{(2500 +72000)^2}{2 \pi (0.75057)^3}+ \dfrac{55200}{\pi (0.75057)^3}} \ \ \ OR \\ \\ \\ \sigma_{1,2} = \dfrac{2500+72000}{2 \pi (0.75057) ^3} - \sqrt{\dfrac{(2500 +72000)^2}{2 \pi (0.75057)^3}+ \dfrac{55200}{\pi (0.75057)^3}}[/tex]
[tex]\sigma _1 = 22193 \ Psi[/tex]
[tex]\sigma_2 = -77807 \ Psi[/tex]
The required diameter d = 2c
d = 1.50 in or 0.125 ft
two opposite poles repel each other
Answer:
South Pole and South Pole or North Pole and North Pole.
: Explain why testing can only detect the presence of errors, not their absence?
Answer:
The goal of the software is to observe the software behavior to meet its requirement expectation. In software engineering, validating software might be harder since client's expectation may be vague or unclear.
Explanation:
xpress the negative value -22 as a 2's complement integer, using eight bits. Repeat it for 16 bits and 32 bits. What does this illustrate with respect to the properties of sign extension as they pertain to 2's complement representation? 8 bit The 8-bit binary representation of 22 is 00010110. So, -22 in 2’s complement form is (NOT (00010110) + 1) = (11101001 + 1) = 11101010
Answer:
Explanation:
A negative binary number is represeneted by its 2's complement value. To get 2's complement, you just need to invert the bits and add 1 to it. So the formula is:
twos_complement = ~val + 1
So you start out with 22 and you want to make it negative.
22₁₀ = 0001 0110₂
~22₁₀ = 1110 1001₂ inverting the bits
~22₁₀ + 1 = 1110 1010₂ adding 1 to it.
so -22₁₀ == ~22₁₀ + 1 == 1110 1010₂
Do the same process for 16-bits and 32-bits and you'll find that the most significant bits will be padded with 1's.
-22₁₀ = 1110 1010₂ 8-bits
-22₁₀ = 1111 1111 1110 1010₂ 16-bits
-22₁₀ = 1111 1111 1111 1111 1111 1111 1110 1010₂ 32-bits
Suppose you used the pipette to make 10 additions to a flask, and suppose the pipette had a 10% random error in the amount delivered with each delivery. Use equation 1 on page 25 to calculate the percent error in the total volume delivered to the flask using the number of clicks you were permitted to make. Report that total percentage below.
Here is the equation: random error of average= error in one measurement/n^1/2
Answer:
The total percentage is 3.16237%
Explanation:
Solution
Now,
We have to know what a random error is.
A random error is an error in measured caused by factors or elements which varies from one measurement to another.
The random error is shown as follows:
The average random error is = the error found in one measurement/n^1/2
Where
n =Number ( how many times the experiment was done)
Now that we added 10 times we have that,
n → 10
Thus,
The error in one measurement = 10%
So,
The average random error = 10 %/(10)^1/2
= (10)^1/2 %
√10%
The total percentage is = 3.16237%
Under the normal sign convention, the distributed load on a beam is equal to the:_______A. The rate of change of the bending moment with respect to the shear force. B. The second derivative of the bending moment with respect to the length of the beam. C. The rate of change of the bending moment with respect to the length of the beam. D. Negative of the rate of change of the shear force with respect to the length of the beam.
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
Water vapor initially at 3.0 MPa and 300°C (state 1) is contained within a piston- cylinder. The water is cooled at constant volume until its temperature is 200°C (state 2). The water is then compressed isothermally to a state where the pressure is 2.5 MPa (state 3).a. Locate states 1, 2, and 3 on a T-v and P-v diagram.b. Determine the specific volume at all three states.c. Calculate the compressibility factor Z at state 1 and comment.d. Find the quality (if applicable) at all three states.
Answer:
a. T-V and P-V diagram are included
b. State 1: Specific volume = 0.0811753 m³/kg
State 2: Specific volume = 0.0811753 m³/kg
State 3: Specific volume = 0.0804155 m³/kg
c. Z = 51.1
d. Quality for state 1 = 100%
Quality for state 2 = 63.47%
Quality for state 3 = 100%
Explanation:
a. T-V and P-V diagram are included
b. State 1: Water vapor
P₁ = 3.0 MPa = 30 bar
T₁ = 300°C = 573.15
Saturation temperature = 233.86°C Hence the steam is super heated
Specific volume = 0.0811753 m³/kg
State 2:
Constant volume formula is P₁/T₁ = P₂/T₂
Specific volume = 0.0811753 m³/kg
T₂ = 200°C = 473.15
Therefore, P₂ = P₁/T₁ × T₂ = 3×473.15/573.15 = 2.4766 MPa
At T₂ water is mixed water and steam and the [tex]v_f[/tex] = 0.00115651 m³/kg
[tex]v_g[/tex] = 0.127222 m³/kg
State 3:
P₃ = 2.5 MPa
T₃ = 200°C
Isothermal compression P₂V₂ = P₃V₃
V₃ = P₂V₂ ÷ P₃ = 2.4766 × 0.0811753/2.5 = 0.0804155 m³/kg
Specific volume = 0.0804155 m³/kg
2) Compressibility factor is given by the relation;
[tex]Z = \dfrac{PV}{RT} = \dfrac{3\times 10^6 \times 0.0811753 }{8.3145 \times 573.15} = 51.1[/tex]
Z = 51.1
3) Gas quality, x, is given by the relation
[tex]x = \dfrac{Mass_{saturated \, vapor}}{Total \, mass} = \dfrac{v - v_f}{v_g - v_f}[/tex]
Quality at state 1 = Saturated quality = 100%
State 2 Vapor + liquid Quality
Gas quality = (0.0811753 - 0.00115651)/ (0.127222-0.00115651) = 63.47%
State 3: Saturated vapor, quality = 100%.
A float valve, regulating the flow of water into a reservoir, is shown in the figure. The spherical float (half of the sphere is submerged) is 0.1553 m in diameter. AOB is the weightless link carrying the float at one end, and a valve at the other end which closes the pipe through which flows into the reservoir. The link is mounted on a frictionless hinge at O, and the angle AOB is 135o. The length of OA is 253 mm and the distance between the center of the float and the hinge is 553 mm. When the flow is stopped AO will be vertical. The valve is to be pressed on to the seat with a force of 10,53 N to be completely stop the flow in the reservoir. It was observed that the flow of water is stopped, when the free surface of water in the reservoir is 353 mm below the hinge at O. Determine the weight of the float sphere.
Answer:
9.29 N . . . . weight of 0.948 kg sphere
Explanation:
The sum of torques on the link BOA is zero, so we have ...
(right force at A)(OA) = (up force at B)(OB·sin(135°))
Solving for the force at B, we have ...
up force at B = (10.53 N)(253 mm)/((553 mm)/√2) ≈ 6.81301 N
This force is due to the difference between the buoyant force on the float sphere and the weight of the float sphere. Dividing by the acceleration due to gravity, it translates to the difference in mass between the water displaced and the mass of the sphere.
∆mass = (6.81301 N)/(9.8 m/s^2) = 0.695205 kg
__
The center of the sphere of diameter 0.1553 m is below the waterline by ...
(553 mm)cos(45°) -(353 mm) = 38.0300 mm
The volume of the spherical cap of radius 155.3/2 = 77.65 mm and height 77.65+38.0300 = 115.680 mm can be found from the formula ...
V = (π/3)h^3(3r -h) = (π/3)(115.680^2)(3·77.65 -115.68) mm^3 ≈ 1.64336 L
So the mass of water contributing to the buoyant force is 1.64336 kg. For the net upward force to correspond to a mass of 0.695305 kg, the mass of the float sphere must be ...
1.64336 kg -0.695205 kg ≈ 0.948 kg
The weight of the float sphere is then (9.8 m/s^2)·(0.948 kg) = 9.29 N
The weight of the 0.948 kg float sphere is about 9.29 N.
Liquid benzene and liquid n-hexane are blended to form a stream flowing at a rate of 1700 lbm/h. An on-line densitometer (an instrument used to determine density) indicates that the stream has a density of 0.810 g/mL. Using specific tractors from Table B.1, estimate the mass and volumetric feed rates of the two hydrocarbons to the mixing vessel (in U.S. customary units). State at least two assumptions required to obtain the estimate from the recommended date.
When an electrical signal travels through a conductive wire, it produces an electromagnetic (EM) field. Likewise, when an EM field encounters a conductive wire, it produces a proportional electrical current.
A. True
B. False
Answer:
A. True
Explanation:
When an electromagnetic field wave strikes a conductor, say a wire, it induces an alternating current that is proportional to the wave in the conductor. This is a reversal of generating electromagnetic wave from accelerating a charged particle. This phenomenon is used in radio antena for receiving radio wave signals and also use in medicine for body scanning.
there is usually a positive side and a negative side to each new technological improvement?
Answer:
positive sides:
low cost improves production speedless timeeducational improvementsnegative sides:
unemployment lot of space required increased pollution creates lots of ethical issuesAn 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
A particle oscillates between the points x=40 mm and x=160 mm with
an acceleration a =
k(100 - x), where a and x are expressed in mm/s2 and
respectively, and k is a constant. The velocity of the particle is 18 mm/s when x = 100 mm
and is zero at both x = 40 mm and x = 160 mm. Determine (a) the value of k,
(b) the velocity when x = 120 mm.
Answer:
(a) k = 0.09 s⁻¹
(b) The velocity= ± 16.97 mm/s
Explanation:
(a) Given that the acceleration = a = k(100 - x)
Therefore;
[tex]a = \dfrac{dv}{dt} = \dfrac{dv}{dx} \times \dfrac{dx}{dt} = \dfrac{dv}{dx} \times v = k(100 - x)[/tex]
When x = 40 mm, v = 0 mm/s hence;
[tex]\int\limits^v_0 {v } \, dv = \int\limits^x_{40} {k(100 - x)} \, dx[/tex]
[tex]\dfrac{1}{2} v^2 = k \cdot \left [100\cdot x-\frac{1}{2}\cdot x^{2} \right ]_{x}^{40}[/tex]
[tex]\dfrac{1}{2} v^2 = -\dfrac{ k\cdot \left (x^{2}-200\cdot x+6400 \right ) }{2}[/tex]
At x = 100 mm, v = 18 mm/s hence we have;
[tex]\dfrac{1}{2} 18^2 = -\dfrac{ k\cdot \left (100^{2}-200\times 100+6400 \right ) }{2} = 1800\cdot k[/tex]
[tex]\dfrac{1}{2} 18^2 =162 = 1800\cdot k[/tex]
k = 162/1800 = 9/100 = 0.09 s⁻¹
(b) When x = 120 mm, we have
[tex]\dfrac{1}{2} v^2 = -\dfrac{ 0.09\times \left (120^{2}-200\times 120+6400 \right ) }{2} = 144[/tex]
Therefore;
v² = 2 × 144 = 288
The velocity, v = √288 = ±12·√2 = ± 16.97 mm/s.
Consider a classroom for 56 students and one instructor, each generating heat at a rate of 100 W. Lighting is provided by 18 fluorescent lightbulbs, 40 W each, and the ballasts consume an additional 10 percent. Determine the rate of internal heat generation in this classroom when it is fully occupied. The rate of internal heat generation in this classroom when it is fully occupied is W.
Answer:
What is the probability of selecting the 4 of spade or black diamond from a deck of 52 playing cards?
a) 2/52
b) 4/52
c) 3/52
d) 1/5
Explanation:
