Answer:3000 yd3
Explanation:
so i got the 50000 divided it by 100 times that by 13 and got 4500 yd3
so now the soil is at 9 so i need to get the 4 percent back up so i yet again got 50000 yd3 and divided by 100 and times it by 9 to get 1500 yd3 to then minus my 1500 and 4500 yd3 and got 3000 yd3 i hope it is right.
The point of contact of two pitch circles of mating gears is called?
Answer:
the pitch point im pritty sure what is is
The point of contact of two pitch circles of mating gears is called the pitch point.
What is the pitch point?
A pitch point is where two pitch circles frequently come into touch. A common point of contact between two pitch circles of mating gears is called a pitch point. The pitch circle and is centered on the top of the teeth. Dedendum circle, also known as the root circle, is the circle traced through the base of the tooth.
The pressure angle, often known as the “tooth shape,” is the angle at which pressure from one gear's tooth is transferred to another gear's tooth. Pure rolling motion will occur at pitch point.
Therefore, it is pressure point.
Learn more about the pressure point, refer to:
https://brainly.com/question/2510654
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In your own words how does an airplane take off?
Answer:
Explanation:
How does an aircraft take off?
They are lift, weight, thrust and drag. Lift pushes the airplane up. The way air moves around the wings gives the airplane lift. The shape of the wings helps with lift, too.
HOPE THIS HELPS!
PLEASE HELP THIS IS A TEST I WAS SUPPOSE TO DO 8 HOURS AGO.
An object has a width of 1.0 in. if you were to create a 1:4 scaled drawing of this object, what would the dimensions of the drawing be?
Convenience items like liquid egg are a lot more expensive?
True
Or False
Answer:
True
Explanation:
It's just because of the processing
A cone penetration test was carried out in normally consolidated sand, for which the results are summarized below: Depth (m) Cone resistance, qc (MN/m2 ) 2.0 3.12 3.5 4.25 5.0 5.14 6.5 9.23 8.0 12.2 The average unit weight of the sand is 16.5 kN/m3 . Determine the friction angle at each depth using Eq. (3.52)
Answer:
hello your question is incomplete attached below is the missing equation related to the question
answer : 40.389° , 38.987° , 38° , 39.869° , 40.265°
Explanation:
Determine the friction angle at each depth
attached below is the detailed solution
To calculate the vertical stress = depth * unit weight of sand
also inverse of Tan = Tan^-1
also qc is in Mpa while σ0 is in kPa
Friction angle at each depth
2 meters = 40.389°
3.5 meters = 38.987°
5 meters = 38.022°
6.5 meters = 39.869°
8 meters = 40.265°
(25%) A well-insulated compressor operating at steady state takes in air at 70 oF and 15 psi, with a volumetric flow rate of 500 ft3 /min, and compresses it to 60 psi. The power input to the compressor is 80 horsepower. Kinetic and potential energy changes from the inlet to exit can be neglected. Determine the exit temperature, in oR.
Answer:
You can look it up
Explanation: if you don't know what it is look it up on .
Please answer ASAP!!
Which statement correctly describes why the electric pole has bent?
A.) Tensile stress acting on the pole caused it to bend.
В.) Compressive stress acting on the pole caused it to bend
С.) The force acting on the pole exceeded its yield strength.
D.) The force acting on the pole exceeded its compressive strength.
Answer:
С.) The force acting on the pole exceeded its yield strength.
Explanation:
Answer:
The correct answer is C. The force acting on the pole exceeded its yield strength.
Explanation:
I got it right on the Plato test.
on the same scale for stress, the tensile true stress-true strain curve is higher than the engineeringstress-engineering strain curve. Explain whether this condition also holds for a compression test.
Answer:
The condition does not hold for a compression test
Explanation:
For a compression test the engineering stress - strain curve is higher than the actual stress-strain curve and this is because the force needed in compression is higher than the force needed during Tension. The higher the force in compression leads to increase in the area therefore for the same scale of stress the there is more stress on the Engineering curve making it higher than the actual curve.
Hence the condition of : on the same scale for stress, the tensile true stress-true strain curve is higher than the engineering stress-engineering strain curve. does not hold for compression test
What is "Engineering"?
Were you surprised by the “pie data”? Is it true for you, your family, and your friends? Why or why not?
this is a coding question, and pie data is a misprint. It is actulaly big data.
Answer:
No because it is something to gain knowledge of peoples lives.
Answer: i was not suprised because, it is something all people need to know to gai knowledege about it.
The answer is all income (&) expenses
Answer:
NICE
Explanation:
Bob would like to run his house off the grid, therefore he needs to find out how many solar panels and batteries he needs to buy.
The power output of a 250 W solar panel is p=250−10(t−5)2 watts for the average 10 hour day. He plans on using 12 V, 120 A⋅h batteries and would like to able to store 3 days worth of energy. Bob's house uses 1100 kW⋅h per month (30 days).
Answer:
stink
Explanation:
what is an unit resistance
Answer: Ohm Ω
Explanation:
The resistance can be defined as the hindrance offered to the flow of current. Lower the current the higher is the resistance. The SI unit of resistance is ohm Ω. When one ampere current flows from an object or wire the application of one volt of potential difference on it the resistance so produce is one ohm. Resistance can be achieved by the application of the certain gadgets in the circuit which can prevent the flow of excess current from devices and can prevent shock circuit.
Consider a single crystal of some hypothetical metal that has an FCC crystal structure and is oriented such that a tensile stress is applied along a[102] direction. If slip occurs on a( 111) plane and in a [ 101] direction, compute the stress at which the crystal yields if its critical resolved shear stress is 12.76 MPa.
a. 13.96 MPa
b. 52.1 MPa
c. 0 MPa
d. 30.9 MPa
e. 3.1 MPa
Draw a FBD of the beam with reactions at A & B. A is a pin, B is a roller. Try to guess intuitively which way the vertical components of A & B are pointing. Do not show the 6 kN forces in your FBD. Only show the couple moment, or pure moment
Answer:
kindly check the drawing of the FBD of the beam with reactions at A & B. A is a pin, B is a roller in the attached picture.
Explanation:
Without further ado, let's dive straight into the solution to the question above. From the diagram of the FBD of the beam with reactions at A & B it can be shown that the reaction moment is anticlockwise while the moment is clockwise.
The system is at equilibrium and the it does not matter where you place the couple (pure) moment.
The distance from A to C can either be equal or not. If AY = 2.15 kN and M = 25.8. Then, the distance between A and B = 25.8/2.15 = 12m.
Convert the following measurement to their unit equivalents.
14.5 ft =___inch
15. 60 inch=___feet
16. 10 inch = __centimeter
17. 5 meter =__decimeter
18. 8 meter = ___feet
Explanation:
(1) 174 inch
(2)1 foot 3.6 inches
(3) 39.649 centimetre
(4)175 decimeter
(5)26.25 feet
what car has autopilot?
tesla lol how did you not know that?
Answer:
Tesla is a car that is able to drive automatically.
What is a kind of useful car?
Rigid bar ABC is supported by three symmetrically-positioned vertical rods, which are initially unstrained. After load P is applied, the normal strain in rods (2) is 0.0010 mm/mm. Determine the normal strain in rod (1) if there is a gap of 1.0 mm in the connection between rod (1) and the rigid bar at B. Report the strain in mm/mm
Answer:
The answer is below
Explanation:
The lengths of the rods are not given.
Let us assume the length of rod 1 = 1500 mm and the length of rod 2 = 800 mm
Solution:
The normal strain is defined as the change in member length δ divided by the initial member length L. The normal strain (ε) is:
ε = δ / L
δ = εL
For rod 1:
[tex]\delta_2=\epsilon_2 L_2\\\\\delta_2=0.0010\ mm/mm*1500\ mm=1.5\ mm[/tex]
The axial elongation of rod 2 is 1.5 mm. Since rigid bar ABC is attached to rod 2, the rigid bar move down by same amount.
The rigid bar moves down 1.8 mm but rods 1 will not be stretched by this amount. Because there is a gap between rod (1) and the rigid bar at B, the first deflection of 1 mm would not cause an elongation in rod 1. Therefore, the elongation in rods (1) is:
[tex]\delta_1=1.5\ mm-1\ mm=0.5\ mm[/tex]
The normal strain in rod 1 is:
[tex]\epsilon_1=\frac{\delta_1}{L_1} =\frac{0.5\ mm}{800\ mm} \\\\\epsilon_1=0.000625\ mm/mm[/tex]
Jumbo Eggs are the most commonly used size egg when making a recipe because they are the highest produced size egg?
True or
False
Answer:
True
Explanation:
Many chickens lay eggs of that size and eggs from chickens are the most commonly used
Consider a continuous-flow, indraft supersonic wind tunnel, which uses a vacuum pump to draw atmospheric air from outside of the wind tunnel building through a converging-diverging nozzle, to establish the supersonic flow in the test section. Assuming a standard day in Morgantown, WV (elevation of 961 feet) and a Mach number of 2.0 in the test section, compute the following:_____
a) The test section temperature
b) The test section pressure, assuming isentropic flow
c) The test section velocity
d) The change in the specific enthalpy of the air
Question Completion Status:
The point of application of the total pressure on surface is
O a. None of the above
O b. Centroid of the surface
O c. Centre of pressure
Answer: c. Centre of pressure
Explanation:
Pressure is applied on a surface when a force is exerted on a particular point on that surface by another object when the two come into contact with each other.
The point where the pressure is applied is known as the centre of the pressure with the force then spreading out from this point much like an epicentre in an earthquake.
Rigid bar ACB is supported by an elastic cir-cular strut DC having an outer diameter of 15 in. and inner diameter of 14.4 in. The strut is made of steel with a modulus elasticity of load P 5kips5 E 5 29,000 ksi. Point is applied at B. Calculate the change in length of the circular strut DC. What is the vertical displacement of the rigid bar at point B
Answer:
The change in length of the circular strut DC = 0.0028 in.
The vertical displacement of the rigid bar at point B = 0.00378 in.
Explanation:
We have the following parameters or information in the question given above:
=> The outer diameter = 15 in., the inner diameter = 14.4 in., the modulus elasticity of E = 29,000 ksi, and the Point load P = 5kips.
The diagram showing the rigid bar ACB is supported by an elastic cir-cular strut DC is given in the attached picture below.
According to Newton's law of motion, it can be seen that the force on CD, that is FCD is equal and opposite to ACD. Hence, FCD = ACD.
Where FCD = p × [4 + 5] ÷ [sin Ф × 4].
kindly note that from the diagram sin Ф = 3/5, cos Ф = 4/5 and tan Ф = 3/4. Also p =5.
Hence, FCD =[ 5 × 9] ÷ [3/5 × 4] = 18.75 kip. So FCD = ACD.
The next thing here is to determine the area and length of CD, say the area of CD is G, thus, G = π/4 × [ 15² - 14.4²] = 13.854 in².
The lenght of CD is = √[4² + 3²] = √[16 + 9] = 5ft. Thus, 5 × 12 = 60in.
Hence, the change in length of the circular strut DC = [18.75 × 60] ÷ 13.854 × 29000 = 0.0028 in.
The vertical deflection of CD = 0.0028 × 3/5 = 0.00168 in.
We have that; 4 /CV = 9BV. Hence, BV = 9/4× CV.
(CV = vertical deflection of CD).
The vertical displacement of the rigid bar at point B = 9/ 4 × 0.00168 in = 0.00378 in.