Answer:
Most methods of transportation rely on some sort of infrastructure to drive, steer, navigate, or direct at some point or another in a journey. Which category of transportation system is least reliant on infrastructure?(1 point). road
Explanation:
How does the length and direction of a shadow change from sunrise to noon in the northern hemisphere? A: The shadow gets shorter and points more east B: The shadow gets shorter and points more north C: The shadow gets longer and points more north D: The shadow gets shorter and points more west E: The shadow gets longer and points more south F: The shadow gets longer and points more east
Answer: the answer is A
Explanation: because the sun rises in the west and goes down in the east
A frequenter of a pub had observed that the new barman poured in average 0.47 liters of beer into the glass with a standard deviation equal to 0.09 liters instead of a half a liter with the same standard deviation. The frequenter had used a random sample of 47 glasses of beer in his experiment. Consider the one-sided hypothesis test for volume of beer in a glass: H0: u=0.5 against H1: u<0.5. Determine the P-value of this test.
Round your answer to four decimal places (e.g. 98.7654).
Answer:
P-value = 0.0011
Explanation:
Formula for the test statistic is;
z = (x¯ - μ)/(σ/√n)
We have;
Sample mean;x¯ = 0.47
Population mean; μ = 0.5
Standard deviation; σ = 0.09
Sample size; n = 47
Thus;
z = (0.46 - 0.5)/(0.09/√47)
z = -3.05
From z-distribution table attached, the p-value corresponding to z = -3.05 is;
P = 0.00114
To four decimal places gives;
P-value = 0.0011
Someone please help me I’m doing this for science and I only have a few minutes left please
Answer:
A
Explanation:
microorganisms must die, build up the sediment, heat and pressure needed to compress and form fossil fuels
An interior beam supports the floor of a classroom in a school building. The beam spans 26 ft. and the tributary width is 16 ft. Dead load is 20 psf. Find:
a. Basic floor live load Lo in psf
b. Reduced floor live load L in psf
c. Uniformly distributed total load to the beam in lb/ft.
d. Compare the loading in part c with the alternate concentrated load requried by the Code. Which loading is more critical for bending, shear, and deflection.?
Answer:
a. [tex]L_o[/tex] = 40 psf
b. L ≈ 30.80 psf
c. The uniformly distributed total load for the beam = 812.8 ft./lb
d. The alternate concentrated load is more critical to bending , shear and deflection
Explanation:
The given parameters of the beam the beam are;
The span of the beam = 26 ft.
The width of the tributary, b = 16 ft.
The dead load, D = 20 psf.
a. The basic floor live load is given as follows;
The uniform floor live load, = 40 psf
The floor area, A = The span × The width = 26 ft. × 16 ft. = 416 ft.²
Therefore, the uniform live load, [tex]L_o[/tex] = 40 psf
b. The reduced floor live load, L in psf. is given as follows;
[tex]L = L_o \times \left ( 0.25 + \dfrac{15}{\sqrt{k_{LL} \cdot A_T} } \right)[/tex]
For the school, [tex]K_{LL}[/tex] = 2
Therefore, we have;
[tex]L = 40 \times \left ( 0.25 + \dfrac{15}{\sqrt{2 \times 416} } \right) = 30.80126 \ psf[/tex]
The reduced floor live load, L ≈ 30.80 psf
c. The uniformly distributed total load for the beam, [tex]W_d[/tex] = b × [tex]W_{D + L}[/tex] =
∴ [tex]W_d[/tex] = = 16 × (20 + 30.80) ≈ 812.8 ft./lb
The uniformly distributed total load for the beam, [tex]W_d[/tex] = 812.8 ft./lb
d. For the uniformly distributed load, we have;
[tex]V_{max}[/tex] = 812.8 × 26/2 = 10566.4 lbs
[tex]M_{max}[/tex] = 812.8 × 26²/8 = 68,681.6 ft-lbs
[tex]v_{max}[/tex] = 5×812.8×26⁴/348/EI = 4,836,329.333/EI
For the alternate concentrated load, we have;
[tex]P_L[/tex] = 1000 lb
[tex]W_{D}[/tex] = 20 × 16 = 320 lb/ft.
[tex]V_{max}[/tex] = 1,000 + 320 × 26/2 = 5,160 lbs
[tex]M_{max}[/tex] = 1,000 × 26/4 + 320 × 26²/8 = 33,540 ft-lbs
[tex]v_{max}[/tex] = 1,000 × 26³/(48·EI) + 5×320×26⁴/348/EI = 2,467,205.74713/EI
Therefore, the loading more critical to bending , shear and deflection, is the alternate concentrated load
You have available three blocks of different material, at various temperatures. They are, respectively, a 2 kg block of iron at 600 K, a 3 kg block of copper at 800 K and a 10 kg block of granite at 300 K. The heat capacities for the three materials are 0.460 (iron), 0.385 (copper), and 0.790 (granite), in kj/(kg*K), all independent of temperature. For solids, the heat capacities at constant pressure and constant volume can be assumed to be equal, Cp=Cv. what is the minimum temperature that could be obtained in any one of the block? what is the maximum temperature that could be obtained? no heat or work interactions with the enviroment are allowed.
Answer:
max temp = 711.32 k
mini temp = 331.29 k
Explanation:
Given data:
2kg block of Iron : temperature = 600k , C = 0.460 kJ/kgk
3 kg block of copper : temp = 800k , C = 0.385 KJ /kgk
10 kg block of granite : temp = 300k , C = 0.790 KJ/kgk
Cp = Cv at constant pressure and constant volume
Determine the minimum temperature that is obtained in any one of the block
considering the heat transfer equation
Q = mC ( T2 - T1 )
attached below is a detailed solution of the problem
The steam requirements of a manufacturing facility are being met by a boiler whose rated heat input is 5.5 x 3^106 Btu/h. The combustion efficiency of the boiler is measured to be 0.7 by a hand-held flue gas analyzer. After tuning up the boiler, the combustion efficiency rises to 0.8. The boiler operates 4200 hours a year intermittently. Taking the unit cost of energy to be $4.35/10^6 Btu, determine the annual energy and cost savings as a result of tuning up the boiler.
Answer:
Energy Saved = 6.93 x 10⁹ Btu
Cost Saved = $ 30145.5
Explanation:
The energy generated by each boiler can be given by the following formula:
[tex]Annual\ Energy = (Heat\ In)(Combustion\ Efficiency)(Operating\ Hours)[/tex]
Now, the energy saved by the increase of efficiency through tuning will be the difference between the energy produced before and after tuning:
[tex]Energy\ Saved = (Heat\ In)(Efficiency\ After\ Tune - Efficiency\ Before\ Tune)(Hours)[/tex][tex]Energy\ Saved = (5.5\ x\ 3\ x\ 10^{6}\ Btu/h)(0.8-0.7)(4200\ h)[/tex]
Energy Saved = 6.93 x 10⁹ Btu
Now, for the saved cost:
[tex]Cost\ Saved = (Energy\ Saved)(Unit\ Cost)\\Cost\ Saved = (6.93\ x\ 10^{9}\ Btu)(\$4.35/10^{6}Btu)\\[/tex]
Cost Saved = $ 30145.5
You have been hired to design a control system for a nuclear reactor. The system monitors the temperature of the reactor using sensors at two points, A & C. If the temperature at either of these points rises past a certain level the sensor will toggle high. A sudden blackout would be disruptive so an override lever, B, is installed to keep the plant running when one of the sensors are tripped until a maintenance team can come in. If both sensors trip, the plant will shut down regardless of the override lever. Assume that the output - a plant shutdown - is represented as a logic
Assume that it takes only one sensor to toggle high for the plant shutdown if the override lever is not activated. Assume B=1 indicates the override lever has been activated.
Construct the truth table for the control system.
Solution :
B A C Shutdown (V)
0 0 0 0
0 0 1 0
0 1 0 0
0 1 1 1
1 0 0 0
1 0 1 1
1 1 0 1
1 1 1 1
[tex]$Y = A + A\bar B \bar C+\bar A \bar B C$[/tex]
[tex]$Y = \bar B A C + BAC+ \bar B A \bar C + \bar B \bar A C$[/tex]
Build a 32-bit accumulator circuit. The circuit features a control signal inc and enable input en. If en is 1 and inc is 1, the circuit increments the stored value by an amount specified by an input A[31:0] on the next clock cycle. If en is 1 and inc is 0 the circuit decrements the stored value by the amount specified in the input A on the next clock cycle. If en is 0, the circuit simply stores its current value without modification. The circuit has the following interface:______.
Input clock governs the state transitions in the circuit upon each rising edge.
Input clear is used as a synchronous reset for the stored value.
Input inc controls whether the value stored is to be incremented or decremented.
Input en is a control signal that activates the values increment/decrement
Input A determines how much to increment or decrement by
Output value is a 32-bit signal that can be used to read the stored value at any time.
* Note: Use any combination of combinational or sequential logic. It may be helpful to look into D Flip Flops and Registers.
Sorry need.points I'm new
The natural material in a borrow pit has a mass unit weight of 110.0 pcf and a water content of 6%, and the specific gravity of the soil solids is 2.63. The specifications require that the soil be compacted to a dry unit weight of 122 lb. per cf and the water content be held to 5.5%.
A. How many cubic yards of borrow are required to construct an embankment having a net section volume of 245,000 cy?
B. How many gallons of water must be added per cubic yard of borrow material assuming no loss by evaporation?
Answer:
A. 288,030.91 cy
B. The amount of water that must be removed from the natural material is 483541.04254 gallons of water
Explanation:
The natural material in the barrow properties are;
The mass unit weight, γ = 110.0 pcf
The water content, w = 6%
The specific gravity of the soil solids, [tex]G_s[/tex] = 2.63
The desired dry unit weight, [tex]\gamma _d[/tex] = 122 pcf
The water content, w₁ = 5.5 %
The net section volume, [tex]V_T[/tex] = 245,000 cy = 6,615,000 ft³
A. [tex]\gamma _d[/tex] = [tex]W_s[/tex]/[tex]V_T[/tex]
∴ [tex]W_s[/tex] = [tex]V_T[/tex] × [tex]\gamma _d[/tex] = 6,615,000 ft³ × 122 lb/ft³ = 807030000 lbs
w = ([tex]W_w[/tex]/[tex]W_s[/tex]) × 100
∴ [tex]W_w[/tex] = (w/100) × [tex]W_s[/tex] = (6/100) × 807030000 lbs = 48421800 lbs
The weight of solids
W = [tex]W_s[/tex] + [tex]W_w[/tex] = 807030000 lbs + 48421800 lbs = 855451800 lbs
V = W/γ = 855451800 lbs/(110.0 lb/ft.³) = 7776834.54545 ft.³ = 288,030.91 cy
V = 288,030.91 cy
The amount of cubic yards of borrow required = 288,030.91 cy
B. The volume of water in the required soil is found as follows;
[tex]W_{w1}[/tex] = (w₁/100) × [tex]W_s[/tex] = (5.5/100) × 807030000 lbs = 44386650 lbs
The amount of water that must be added = [tex]W_{w1}[/tex] - [tex]W_w[/tex] = 44386650 lbs - 48421800 lbs = -4,035,150 lbs
Therefore, 4,035,150 lbs of water must be removed
The density of water, ρ = 8.345 lbs/gal
Therefore, V = 4,035,150 lbs/(8.345 lbs/gal) = 483541.04254 gal of water must be removed from the natural material
What is the tallest building ever made
Answer:
Burj Khalifa
Explanation:
The world's tallest artificial structure is the 829.8-metre-tall (2,722 ft) Burj Khalifa in Dubai (of the United Arab Emirates). The building gained the official title of "tallest building in the world" and the tallest self-supported structure at its opening on January 9, 2010.
a brainliest would be appriciated
Answer:
Burj khalifa
Explanation:
829.8 meters tall
A runner ran a 600 m race in 2 min 17 seconds. Calculate his average speed in m/sec.
I just need help on problem B
F12-33. The ca . R has a speed of 55 ft/s. Determine the angular velocity 8 of the radial line OA at this instant.
Answer:
0.1375 rad/s
Explanation:
Speed of car = 55 ft/s
We are to find angular velocity, which is θ.
There are mistakes in this question. It should be θ not 8 and r = 400 ft
Radius r = 400ft
Speed = velocity = 55 ft/s
We Express transverse of velocity
Vθ = rθ
Vθ = 400θ
Then magnitude
V = √(Vr)²+(Vθ)²
55 = √0² + 400θ²
55 = √160000θ
55 = 400θ
We find the value of θ
θ = 55/400
= 0.1375rad/s
The angular velocityθ = 0.1375 rad/s
To convert a measurement in centimeters to
meters, you simply move the decimal point
a. two places to the left
b. three places to the right
c. three places to the left
do four places to the right
Answer:
a
Explanation:
To convert a measurement in centimeters to meters, simply move the decimal point two places to the left. Thus, the correct option is A). two places to the left.
What is the way to convert centimeters to meters and decimals?There are 100 centimeters in every meter, which means that dividing the measurement in centimeters by 100 will convert it to meters. This conversion is very quick and easy by simply moving the decimal point in the measurement 2 spaces to the left.
The centimeter to meter conversion (cm to m) is basically, the conversion from centimeters to meters. One centimeter is approximately equal to 0.01 meter or we can say that one meter equals to 100 centimeters.
Simply, in order to convert cm to m, multiply the given centimeter value by 0.01 m. For example:- 5 cm = 5 x 0.01 m
5 cm = 0.05 m
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You would like to use a 16-bit ADC to sample and digitally store a stack of vinyl music records you found in your parents’ basement. If the ADC is capable of sampling up to 25 kSamples/sec:________.
a) What is the minimum time required for each sample conversion?
b) How many bytes of storage would be required to record "Uptown Funk" by Bruno Mars and Mark Ronson (4 min, 22 sec) if the ADC is operating at the maximum sample rate?
c) If the highest frequency in "Uptown Funk" is 10 kHz, what is the minimum sampling rate that would need to be used to accurately reconstruct the song?
d) How many bytes of storage are required at this lower sampling rate?
Answer:
a) 40 micro sec
b) 13,100,000 bytes
c) 20,000 samples / sec
d) 10.48 Mbytes
Explanation:
a) The minimum time required for each sample conversion
minimum sample conversion time = max performance of ADC ( 25 k samples /sec )
hence : 1 sample conversion time = 1 / 25k = 1/ 25000
hence minimum time required for each sample conversion time = 40 microsec
b) Determine the number of bytes of storage required to record if the ADC is operating at maximum sample rate
In 1 second the ADC will produce 25000 * 2 bytes of data = 50000 bytes
audio length = 4 min 22 secs = 262 seconds
Hence the number of bytes of storage required =
= 262 * 50000 bytes
= 13,100,000 bytes
c) Determine the minimum sampling rate that would need to be used to accurately reconstruct the song
highest frequency = 10 kHz
hence the minimum sampling rate to reconstruct the song
= 2 * 10 kHz = 20,000 samples / sec
d) Determine The number of bytes of storage required at the lower sampling rate
given that the audio is sampled at 20kHz per sec then each sample will be encoded into a 2 bytes
therefore each each second of the audio will be generated by : 20,000 * 2 bytes
∴ the number of bytes of storage required at this lower sampling rate
= 262 secs * 40,000
= 10.48 Mbytes
Calculate the resistance of a lamp if the current through it is 0.4 A and the voltage across it is 8 V.
Answer:
Answer is 3.2 Ω (Ohms)
Explanation:
From Ohms Law I = V/R
R = V(I)
R = 8(0.4)
R = 3.2
The resistance of a lamp if the current through it is 0.4 A and the voltage across it is 8 V is 3.2 ohm.
What is Ohm's Law?According to Ohm's law, when all other physical parameters, including temperature, are held constant, the voltage across a conductor is directly proportional to the current flowing through it.
According to Ohm's Law, the electrical current I flowing through a particular conductor is precisely proportional to the potential difference (voltage) V across its ends (assuming that the conductor's physical properties, such as its temperature and pressure, stay constant). where R is a proportionality constant.
Given:
Current, I= 0.4 A
Voltage, V= 8 V
Using Ohm's Law
V= IR
I = V/R
R = V(I)
R = 8(0.4)
R = 3.2 ohm
Hence, the resistance of a lamp if the current through it is 0.4 A and the voltage across it is 8 V is 3.2 ohm.
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[3] (25%) Superheated steam at 9 MPa and 480oC leaves the steam generator of a vapor power plant. Heat transfer and frictional effects in the line connecting the steam generator and the turbine reduce the pressure and temperature at the turbine inlet to 8.6 MPa and 440oC, respectively. The pressure at the exit of the turbine is 10 kPa, and the turbine operates adiabatically. Liquid leaves the condenser at 8 kPa, 35 oC. The pressure is increased to 9.6 MPa across the pump. The turbine and pump isentropic efficiencies are 85%. The mass flow rate of steam is 80 kg/s. Determine (a) the net power output, in kW. (b) the thermal efficiency (c) the rate of heat transfer from the line connecting the steam generator and the turbine, in kW. (d) the mass flow rate of the condenser cooling water, in kg/s, if the cooling water enters at 15oC and exits at 35oC with negligible pressure change.
What is the primary tool material used for CNC turning?
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A continuous and aligned fiber-reinforced composite is to be produced consisting of 30 vol% aramid fibers and 70 vol% of a polycarbonate matrix; mechanical characteristics of these two materials are as follows:
Modulus of Elasticity [GPa] Tensile Strength [MPa] Aramid fiber 131 3600 Polycarbonate 2.4 65
Also, the stress on the polycarbonate matrix when the aramid fibers fail is 45 MPa. For this composite, compute the following:
(a) the longitudinal tensile strength, and
(b) the longitudinal modulus of elasticity
Answer:
1. 1111.5MPa
2. 56.1GPa
Explanation:
1. Longitudinal tensile stress can be obtained by obtaining the strength and volume of the fiber reinforcement. The derived formula is given by;
σcl = σm (1 - Vf) + σfVf
Substituting the figures, we will have;
45(1 - 0.30) + 3600(0.30)
45(0.70) + 1080
31.5 + 1080
= 1111.5MPa
2. Longitudinal modulus of elasticity or Young's modulus is the ability of an object to resist deformation. The derived formula is given by;
Ecl = EmVm + EfVf
Substituting the formula gives;
= 2.4 (1 - 0.30) + 131 (0.30)
= 2.4(0.70) + 39.3
= 16.8 + 39.3
= 56.1GPa
Using the appropriate relation, the longitudinal tensile stress and the longitudinal modulus are 1111.50 and 56.10 respectively.
Longitudinal tensile stress can be obtained using the relation :
σcl = σm (1 - Vf) + σfVfSubstituting the values into the relation:
45(1 - 0.30) + 3600(0.30)
45 × 0.70 + 1080
31.5 + 1080
= 1111.50 MPa
2.)
Longitudinal modulus of elasticity is obtained using the relation :
Ecl = EmVm + EfVfSubstituting the values thus :
2.4 (1 - 0.30) + 131 (0.30)
= 2.4 × 0.70 + 39.3
= 16.8 + 39.3
= 56.10 GPa
Hence, the longitudinal tensile stress and the longitudinal modulus are 1111.50 and 56.10 respectively.
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what are difference between conic sectional and solids?
Answer:
Conic Sections
a conic section is a curve which is obtained when a surface performs an intersection with a plane. The types of conic sections include hyperbola, parabola and ellipse. A circle can also be considered as a conic section.
Conic Solids on the other hand are the set of points on any segment between a region and a point which is not present in the plane of the base. They are solids with one base.
Another name for a load-center distribution system is a A. primary radial system. B. complex radial system. C. split-radial system. D. dual-radial system.
Answer:
A
Explanation:
Primary radial system
Hard steering can be caused by
Answer:
Lack of fluid oil – lack of fluid oil in your vehicle, or a fluid leakage, can lead to heavy steering. If there is a lack of fluid oil, or a leak, this can reduce the pressure in the system, meaning the steering wheel does not receive enough supply of fluid to perform freely.
Please help me it’s for science I only have a few minutes
Answer:
Rocks
Explanation:
I am not sure tho bc they are made out of coal and I think coal is a kind of rock
Answer:
I'm taking a guess for ya I shall say Metal or and Minerals
Explanation:
Good luck
pls help me it’s due today
Answer:
C. 14.55
Explanation:
12 x 10 = 120
120 divded by 10 is 12
so now we do the left side
7 x 3 = 21 divded by 10 is 2
so now we have 14
and the remaning area is 0.55
so 14.55
All of these are true about using adhesive EXCEPT:
Answer:
Except what? I'm confused
All of these are true about using adhesive except Bilateral. A bilateral contract is defined as an agreements between two parties in which each side agrees to fulfill his or her side of the bargain.
What is bilateral contract?A bilateral contract is defined as an agreements between two parties in which each side agrees to fulfill his or her side of the bargain. According to my research on the different terms used when referencing an insurance contract, I can say that all of the answers provided except for Bilateral are considered typical characteristics describing the nature of an insurance contract.
Since an insurance contract is a fund that the insurance company pays in the case of an accident in which the person is injured, there is only one party that agrees to fulfill their side of the bargain and that is the insurance company.
Therefore, All of these are true about using adhesive except Bilateral. A bilateral contract is defined as an agreements between two parties in which each side agrees to fulfill his or her side of the bargain.
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QUESTION 4:
4.1
Name FOUR principles of kinetic friction
Answer:
The force of friction always acts in a direction, opposite to that in which the body is moving.
The magnitude of kinetic friction bears a constant ratio to the normal reaction between the two surfaces. ...
For moderate speeds, the force of friction remains constant.
Answer:
Explanation:Kinetic friction is a force that acts between moving surfaces. An object that is being moved over a surface will experience a force in the opposite direction as its movement. The magnitude of the force depends on the coefficient of kinetic friction between the two kinds of material.
A contractor excavates 10,000 m3 soil at moist unit weight of 17.5 kN/m3 and moisture content of 10% from a borrow pit and transports it to a project site. The project has an area of 20,000 m2 to be filled with this compacted soil. If the required dry unit weight and moisture content of the compacted soil are 18.3 kN/m3 and 12.5% (assume there is no soil loss during transportation and compaction), what is the thickness of the compacted soil and how much water needs to be added?
Answer:
Part A
The thickness of the compacted soil is approximately 4.3467 × 10⁻¹ m
Part B
The weight of water to be added is approximately 19886.[tex]\overline{36}[/tex] kN, the volume of the water added is approximately 2,027.77 m³
Explanation:
The parameters of the soil are;
The volume of sol the excavator excavates, [tex]V_T[/tex] = 10,000 m³
The moist unit weight, W = 17.5 kN/m³
The moisture content = 10%
The area of the project, A = 20,000 m²
The required dry unit weight = 18.3 kN/m³
The required moisture content = 12.5%
Part A
Therefore, we have;
The moist unit weight = Unit weight = ([tex]W_s[/tex] + [tex]W_w[/tex])/[tex]V_T[/tex]
The moisture content, MC = 10% = ([tex]W_w[/tex]/[tex]W_s[/tex]) × 100
∴ [tex]W_w[/tex] = 0.1·[tex]W_s[/tex]
∴ The moist unit weight = 17.5 kN/m³ = ([tex]W_s[/tex] + 0.1·[tex]W_s[/tex])/(10,000 m³)
1.1·[tex]W_s[/tex] = 10,000 m³ × 17.5 kN/m³ = 175,000 kN
[tex]W_s[/tex] = 175,000 kN/1.1 = 159,090.[tex]\overline{09}[/tex] kN
For the required soil, we have;
The required dry unit weight = 18.3 kN/m³ = [tex]W_s[/tex]/[tex]V_T[/tex] = 159,090.[tex]\overline{09}[/tex] kN/[tex]V_T[/tex]
[tex]V_T[/tex] = 159,090.[tex]\overline{09}[/tex] kN/(18.3 kN/m³) ≈ 8,693.4923 m³
The total volume of the required soil ≈ 8,693.4923 m³
Volume [tex]V_T[/tex] = Area, A × Thickness, d
∴ d = [tex]V_T[/tex]/A
d = 8,693.4923 m³/(20,000 m²) ≈ 4.3467 × 10⁻¹ m
The thickness of the compacted soil ≈ 4.3467 × 10⁻¹ m
Part A
The moisture content, MC = 12.5% = ([tex]W_w[/tex]/[tex]W_s[/tex]) × 100
[tex]W_w[/tex] = [tex]W_s[/tex] × MC/100 = 159,090.[tex]\overline{09}[/tex] kN × 12.5/100 = 19886.[tex]\overline{36}[/tex] kN
The weight of water to be added, [tex]W_w[/tex] = 19886.[tex]\overline{36}[/tex] kN
Where the density of water, ρ = 9.807 kN/m³
Therefore, we have;
The volume of water, V = [tex]W_w[/tex]/ρ
∴ V = 19886.[tex]\overline{36}[/tex] kN/(9.807 kN/m³) ≈ 2027.77 m³
The volume of water, V ≈ 2027.77 m³
if you are running and you fall and everyone↓↓↓↓↓↓↓ passes you how can you still be in first place??
ik the answer but lets see if you know it twooo
The alternator must be operated with the battery disconnected or with the terminals at the back of the alternator
disconnected.
True or false
Answer:
true
Explanation:
What causes the charging system warning lamp to go out when the engine starts up?
Select one:
a. It turns off when ground is supplied to the lamp.
b. It turns off because voltage is applied to both sides of the lamp.
c. It turns off automatically after about 5 seconds.
d. It turns off because voltage is applied to one side of the bulb and ground to the other side.