Answer:
d
Explanation:
OR criteria, AND criteria
In an OR criteria, it doesn't need all the records to be true. Just one record is enough and all other criterion becomes true.
In an AND criteria, it's unlike the OR criteria and works in opposite. It needs every member of the record to be true to be able to adjudge the whole record as true.
And as such, we have
With OR criteria, only one criterion must evaluate true in order for a record to be selected and with AND criteria, all criteria must be evaluate true in order for a record to be selected.
A sign structure on the NJ Turnpike is to be designed to resist a wind force that produces a moment of 25 k-ft in one direction. The axial load is 30 kips. Soil conditions consist of a normally consolidated clay layer with following properties; su=800 psf, andγsat= 110 pcf. Design for a FOS of 3. Assume frost depth to be 3ft below grade
Solution :
Finding the cohesion of the soil(c) using the relation:
[tex]$c = \frac{q_u}{2}$[/tex]
Here, [tex]$q_u$[/tex] is the unconfined compression strength of the soil;
[tex]$c = \frac{800}{2}$[/tex]
= 400 psf
∴ The cohesion value is greater than 0
So the use of the angle of internal friction is 0
Referring to the table relation between bearing capacity factors and angle of internal friction.
For the angle of inter friction [tex]$0^\circ$[/tex]
[tex]$N_c = 5.14$[/tex]
[tex]$N_q = 1.0$[/tex]
[tex]$N_r = 0$[/tex]
Therefore,
[tex]$q_{ult} = (400 \times 5.14 )+(110 \times 3 \times 1.0)+(0.5 \times 100 \times 13 \times 0)$[/tex]
= 2386 psf
∴ Allowable bearing capacity [tex]$q_{a} = \frac{Q_{allow}}{A}$[/tex]
[tex]$=\frac{30}{B^2}$[/tex]
∴ [tex]$q_a = \frac{q_{ult}}{F.O.S}$[/tex]
[tex]$\frac{30}{B^2} = \frac{2386}{3}$[/tex]
∴ B = 0.2 ft
Therefore, the dimension of the square footing is 0.2 ft x 0.2 ft
[tex]$=0.04 \ ft^2$[/tex]
how skateboards works?
Answer
The skateboarder applies pressure to the trucks and gives/releases pressure on the levers. Second, the wheels and the axles are also examples of simple machines. They help the skater ride, spin, grind, and do a bunch of other radical movements on a skateboard.:
Explanation:
The driver of the truck has an acceleration of 0.4g as the truck passes over the top A of the hump in the road at constant speed. The radius of curvature of the road at the top of the hump is 98 m, and the center of mass G of the driver (considered a particle) is 2 m above the road. Calculate the speed v of the truck.
Answer:
19.81 m/s
Explanation:
The total acceleration of the truck (a) is due to the centripetal acceleration and as a result of the linear acceleration. Therefore the total acceleration (a) is given by:
[tex]a^2=a_n^2+a_t^2\\\\where\ a_n=centripetal\ acceleration=\frac{v^2}{r},a_t=linear \ acceleration\\\\But\ since\ the \ speed\ is \ constant, the \ linear \ acceleration(a_t)\ would\ be\ 0.\\\\a^2=a_n^2+a_t^2\\\\a^2=a_n^2\\\\a=a_n=\frac{v^2}{r} \\\\v^2=ar\\\\v=\sqrt{ar} \\\\a=0.4g=0.4*9.81,r=98\ m+2\ m=100\ m\\\\v=\sqrt{0.4*9.81*100} \\\\v=19.81\ m/s[/tex]
I don’t know the answer to this question
Answer:
I dont know the answer either
Explanation:
Answer:
flux
Explanation:
Where does Burj Khalifa located? and how many meters?
Answer:
Burj Khalifa is located in dubai UAE at over 828m
Explanation:
828 metres
Answer:
The Burji Khalifa, known as the Burj Dubai prior to its inauguration in 2010, is a skyscraper in Dubai, United Arab Emirates. With a total hight of 829.8 m and a roof hight of 828 m, the Burji Khalifa has been the tallest structure and building in the world since its topping out in 2009.
a load of 12tonnes is put along a horizontal plane by a force at 30°to and above the flat. if the coefficient of sliding friction is 0.2 find the frictional force
Answer:
20368.917N
Explanation:
Frictional force (F) is the product of the Coefficient of friction and the normal reaction.
F = μN
Coefficient of friction, μ = 0.2
Normal reaction = MgCosθ
Mass, m = 12 tonnes = 12 * 1000 = 12000 kg
N = 12000 * 9.8 * cos30
N = 101844.58
F = 0.2 * 101844.58
F = 20368.917N
Describe how to contribute to
zero/low carbon work outcomes
within the built environment.
Answer:
day if you workout without Zero billing that means you're not sweating. Sweating you're not losing anything that means you have zero outcomes
Explanation:
What current works best when the operator
encounters magnetic arc blow?
•DCEP
•ACEN
•CC
•AC
Answer:
AC
Explanation:
One situation when alternating current would work better than direct current is if the operator is encountering magnetic arc blow.
Current works best when the operator encounters magnetic arc blow is AC
Magnetic arc blow is simply defined as the arc deflection due to the warping of the magnetic field that is produced by electric arc current.
This is caused as a result of the following;
- if the material being welded has residual magnetism at an intolerable level
- When the weld root is being made, and the welding current is direct current which indicates constant direction and maintains constant polarity (either positive or negative).
Since it is caused by DC(Direct Current) which means constant polarity , it means the opposite will be better which is AC(alternating current) because it means that electricity direction will be switching to and fro and as such the polarity will also be revered in response to this back and forth switch manner.Thus, Current works best when the operator encounters magnetic arc blow is AC
Read more at; brainly.in/question/38789815?tbs_match=1
A group of students launches a model rocket in the vertical direction. Based on tracking data, they determine that the altitude of the rocket was 89.6 ft at the end of the powered portion of the flight and that the rocket landed 16.5 s later. The descent parachute failed to deploy so that the rocket fell freely to the ground after reaching its maximum altitude. Assume that g = 32.2 ft/s2.
Determine
(a) the speed v1 of the rocket at the end of powered flight,
(b) the maximum altitude reached by the rocket.
Answer:
[tex]u = 260.22m/s[/tex]
[tex]S_{max} = 1141.07ft[/tex]
Explanation:
Given
[tex]S_0 = 89.6ft[/tex] --- Initial altitude
[tex]S_{16.5} = 0ft[/tex] -- Altitude after 16.5 seconds
[tex]a = -g = -32.2ft/s^2[/tex] --- Acceleration (It is negative because it is an upward movement i.e. against gravity)
Solving (a): Final Speed of the rocket
To do this, we make use of:
[tex]S = ut + \frac{1}{2}at^2[/tex]
The final altitude after 16.5 seconds is represented as:
[tex]S_{16.5} = S_0 + ut + \frac{1}{2}at^2[/tex]
Substitute the following values:
[tex]S_0 = 89.6ft[/tex] [tex]S_{16.5} = 0ft[/tex] [tex]a = -g = -32.2ft/s^2[/tex] and [tex]t = 16.5[/tex]
So, we have:
[tex]0 = 89.6 + u * 16.5 - \frac{1}{2} * 32.2 * 16.5^2[/tex]
[tex]0 = 89.6 + u * 16.5 - \frac{1}{2} * 8766.45[/tex]
[tex]0 = 89.6 + 16.5u- 4383.225[/tex]
Collect Like Terms
[tex]16.5u = -89.6 +4383.225[/tex]
[tex]16.5u = 4293.625[/tex]
Make u the subject
[tex]u = \frac{4293.625}{16.5}[/tex]
[tex]u = 260.21969697[/tex]
[tex]u = 260.22m/s[/tex]
Solving (b): The maximum height attained
First, we calculate the time taken to attain the maximum height.
Using:
[tex]v=u + at[/tex]
At the maximum height:
[tex]v =0[/tex] --- The final velocity
[tex]u = 260.22m/s[/tex]
[tex]a = -g = -32.2ft/s^2[/tex]
So, we have:
[tex]0 = 260.22 - 32.2t[/tex]
Collect Like Terms
[tex]32.2t = 260.22[/tex]
Make t the subject
[tex]t = \frac{260.22}{ 32.2}[/tex]
[tex]t = 8.08s[/tex]
The maximum height is then calculated as:
[tex]S_{max} = S_0 + ut + \frac{1}{2}at^2[/tex]
This gives:
[tex]S_{max} = 89.6 + 260.22 * 8.08 - \frac{1}{2} * 32.2 * 8.08^2[/tex]
[tex]S_{max} = 89.6 + 260.22 * 8.08 - \frac{1}{2} * 2102.22[/tex]
[tex]S_{max} = 89.6 + 260.22 * 8.08 - 1051.11[/tex]
[tex]S_{max} = 1141.0676[/tex]
[tex]S_{max} = 1141.07ft[/tex]
Hence, the maximum height is 1141.07ft
What are the top 4 solar inventions, how they are used, and how they are better than the original way of powering them
Now, you get a turn to practice writing a short program in Scratch. Try to re-create the program that was shown that turns the sprite in a circle. After you have completed that activity, see if you can make one of the improvements suggested. For example, you can try adding a sound. If you run into problems, think about some of the creative problem-solving techniques that were discussed.
When complete, briefly comment on challenges or breakthroughs you encountered while completing the guided practice activity.
Pls help im giving 100 points for this i have this due in minutes
Answer:
u need to plan it out
Explanation:
u need to plan it out
Answer:
use the turn 1 degrees option and put a repeat loop on it
Explanation:
u can add sound in ur loop
A cylindrical bar of metal having a diameter of 15.7 mm and a length of 178 mm is deformed elastically in tension with a force of 49100 N. Given that the elastic modulus and Poisson's ratio of the metal are 67.1 GPa and 0.34 respectively, Determine the following:(a) The amount by which this specimen will elongate (in mm) in the direction of the applied stress. The entry field with incorrect answer now contains modified data.(b) The change in diameter of the specimen (in mm). Indicate an increase in diameter with a positive number and a decrease with a negative number.
Answer:
0.6727
-0.02017
Explanation:
diameter = 15.7
l = 178
E =elastic modulus = 67.1 Gpa
poisson ratio = 0.34
p = force = 49100N
first we calculate the area of the cross section
[tex]A=\frac{\pi }{4} d^{2}[/tex]
[tex]A=\frac{\pi }{4} (15.7)^{2} \\A = \frac{774.683}{4} \\[/tex]
A = 193.6mm²
1. Change in directon of the applied stress
[tex]= \frac{pl}{AE}[/tex]
= 49100*178/193.6*67.1*10³
= [tex]=\frac{8739800}{12990560}[/tex]
δl = 0.6727 mm
2. change in diameter of the specimen
equation for poisson distribution =
m = -(δd/d) / (δl/l)
0.34 = (δd/15.7) / (0.6727/178)
0.34 = (-δd * 178) / 15.7 * 0.6727
0.34 = -178δd / 10.56139
we cross multiply
10.56139*0.34 =-178δd
3.5908726 = -178δd
δd = 3.5908726/-178
δd = -0.02017 mm
the change in dimeter has a negative sign. it decreases
Convert an acceleration of 12m/s² to km/h²
Lab scale tests performed on a cell broth with a viscosity of 5cP gave a specific cake resistance of 1 x1011 cm/g and a negligible medium resistance. The cake solids (dry basis) per volume of filtrate was 20 g/liter. It is desired to operate a larger rotary vacuum filter (diameter 8 m and length 12 m) at a vacuum pressure of 80 kPA with a cake formation time of 20 s and a cycle time of 60 s. Determine the filtration rate in volumes/hr expected for the rotary vacuum filter.
Answer:
5.118 m^3/hr
Explanation:
Given data:
viscosity of cell broth = 5cP
cake resistance = 1*1011 cm/g
dry basis per volume of filtrate = 20 g/liter
Diameter = 8m , Length = 12m
vacuum pressure = 80 kpa
cake formation time = 20 s
cycle time = 60 s
Determine the filtration rate in volumes/hr expected fir the rotary vacuum filter
attached below is a detailed solution of the question
Hence The filtration rate in volumes/hr expected for the rotary vacuum filter
V' = ( [tex]\frac{60}{20}[/tex] ) * 1706.0670
= 5118.201 liters ≈ 5.118 m^3/hr
A storm with a duration of about 24 hours resultsin the following hydrograph at a gaging station on a river. The flow was 52 cubic meters per second (cms) before the rain began. The drainage area above the gaging station is 1,450 square kilometers. Use the observed hydrograph to develop a 24-hour rainfall duration unit hydrograph for this watershed.
Time 0 6 12 18 24 30 36 42 48 54 60 66 72 78 84
(Hours)
Flow 52 52 55 56 97 176 349 450 442 370 328 307 280 259 238
(m3/s)
Time 90 66 102 108 114 120 126 132 138 144 155 156 162 168 176
(Hours)
(m3/s) 214 193 173 145 135 124 114 107 97 86 79 66 62 58 52
Answer:
attached below
Explanation:
Given data:
Gaged flow = 52 m^3 / sec
Depth covering drainage area = 1450 km^2
Develop a 24-hour rainfall duration unit hydrograph for the watershed using observed hydorgraph
Runoff flow = gaged flow - base flows
= 52 - 52 = 0 m^3/sec
For 18 hours time duration
Direct runoff volume ( Vr ) = ∑ ( QΔt1 )
where Δt = 6
∑Q = 3666 m^3/sec
hence Vr = Δt * ∑Q = 79185600 m^3
Next we will convert the Direct runoff volume to its equivalent depth covering the drainage area
= Vr / drainage area depth
= 79185600 / 1450000000 = 5.46 cm
Next we will find the unit hydrograph flows by applying the relation below
[tex]Q_{1.0cm} = Q_{5.46cm} (\frac{1.0cm}{5.46cm} )[/tex]
where 14m^3/sec = [tex]Q_{5.6cm}[/tex]
input value back to the above relation
[tex]Q_{1.0cm} = 2.57 m^3/sec[/tex]
Attached below is The remaining part of the solution tabulated below
and A graph of the unit hydorgraph for the given watershed
Note :
Base flow total = 1560
UH total = 671.30
A single crystal of a metal that has the FCC crystal structure is oriented such that a tensile stress is applied parallel to the [100] direction. If the critical resolved shear stress for this material is 2.00 MPa, calculate the magnitude of applied stress necessary to cause slip to occur on the (111) plane in the direction.
Answer:
Explanation:
From the given information:
The equation for applied stress can be expressed as:
[tex]\sigma_{app} = \dfrac{\tau_{CRSS}}{cos \phi \ cos \lambda}[/tex]
where;
[tex]\phi[/tex] = angle between the applied stress [100] and [111]
To determine the [tex]\phi[/tex] and [tex]\lambda[/tex] for the system
Using the equation:
[tex]\phi= cos^{-1}\Big [\dfrac{l_1l_2+m_1m_2+n_1n_2}{\sqrt{(l_1^2+m_1^2+n_1^2)(l_2^2+m_2^2+n_2^2)}}\Big][/tex]
for [100]
[tex]l_1 = 1, m_1 = 0, n_1 = 0[/tex]
for [111]
[tex]l_1 = 1 , m_1 = 1, n_1 = 1[/tex]
Thus;
[tex]\phi= cos^{-1}\Big [\dfrac{1*1+0*1+0*1}{\sqrt{(1^2+0^2+0^2)(1^2+1^2+1^2)}}\Big][/tex]
[tex]\phi= cos^{-1}\Big [\dfrac{1}{\sqrt{(3)}}\Big][/tex]
[tex]\phi= 54.74^0[/tex]
To determine [tex]\lambda[/tex] for [tex][1 \overline 1 0][/tex]
where;
for [100]
[tex]l_1 = 1, m_1 = 0, n_1 = 0[/tex]
for [tex][1 \overline 1 0][/tex]
[tex]l_1 = 1 , m_1 = -1, n_1 = 0[/tex]
Thus;
[tex]\lambda= cos^{-1}\Big [\dfrac{1*1+0*1+0*0}{\sqrt{(1^2+0^2+0^2)(1^2+(-1)^2+0^2)}}\Big][/tex]
[tex]\phi= cos^{-1}\Big [\dfrac{1}{\sqrt{(2)}}\Big][/tex]
[tex]\phi= 45^0[/tex]
Thus, the magnitude of the applied stress can be computed as:
[tex]\sigma_{app} = \dfrac{\tau_{CRSS}}{cos \phi \ cos \lambda }[/tex]
[tex]\sigma_{app} = \dfrac{2.00}{cos (54.74) \ cos (45) }[/tex]
[tex]\mathbf{\sigma_{app} =4.89 \ MPa}[/tex]
who was part of dempwolf his firm when he first started
Explanation:
Dempwolf created by John Augustus, Among the most prominent innovative solutions in Southern California Pennsylvania was established by Dempwolf with brother Reinhardt or uncle's son Frederick entered the company of J.A. Dozens of structures in 10 states were engineered by Dempwolf.